#include <stdio.h>
int main(void)
{
int i = 0;
i = i++ + ++i;
printf("%d\n", i); // 3
i = 1;
i = (i++);
printf("%d\n", i); // 2 Should be 1, no ?
volatile int u = 0;
u = u++ + ++u;
printf("%d\n", u); // 1
u = 1;
u = (u++);
printf("%d\n", u); // 2 Should also be one, no ?
register int v = 0;
v = v++ + ++v;
printf("%d\n", v); // 3 (Should be the same as u ?)
int w = 0;
printf("%d %d\n", ++w, w); // shouldn't this print 1 1
int x[2] = { 5, 8 }, y = 0;
x[y] = y ++;
printf("%d %d\n", x[0], x[1]); // shouldn't this print 0 8? or 5 0?
}
Why are these constructs using pre and post-increment undefined behavior?
C has the concept of undefined behavior, i.e. some language constructs are syntactically valid but you can't predict the behavior when the code is run.
As far as I know, the standard doesn't explicitly say why the concept of undefined behavior exists. In my mind, it's simply because the language designers wanted there to be some leeway in the semantics, instead of i.e. requiring that all implementations handle integer overflow in the exact same way, which would very likely impose serious performance costs, they just left the behavior undefined so that if you write code that causes integer overflow, anything can happen.
So, with that in mind, why are these "issues"? The language clearly says that certain things lead to undefined behavior. There is no problem, there is no "should" involved. If the undefined behavior changes when one of the involved variables is declared volatile, that doesn't prove or change anything. It is undefined; you cannot reason about the behavior.
Your most interesting-looking example, the one with
u = (u++);
is a text-book example of undefined behavior (see Wikipedia's entry on sequence points).
@PiX: Things are undefined for a number of possible reasons. These include: there is no clear "right result", different machine architectures would strongly favour different results, existing practice is not consistent, or beyond the scope of the standard (e.g. what filenames are valid). –
Gregor
Just to confuse everyone, some such examples are now well-defined in C11, e.g.
i = ++i + 1; . –
Elspeth I've edited the question to add the UB in evaluation of function arguments, as this question is often used as a duplicate for that. (The last example) –
Hesperidium
Reading the Standard and the published rationale, It's clear why the concept of UB exists. The Standard was never intended to fully describe everything a C implementation must do to be suitable for any particular purpose (see the discussion of the "One Program" rule), but instead relies upon implementors' judgment and desire to produce useful quality implementations. A quality implementation suitable for low-level systems programming will need to define the behavior of actions that wouldn't be needed in high-end number crunching.applications. Rather than try to complicate the Standard... –
Fashion
...by getting into extreme detail about which corner cases are or are not defined, the authors of the Standard recognized that implementors should be better paced to judge which kinds of behaviors will be needed by the kinds of programs they're expected to support. Hyper-modernist compilers pretend that making certain actions UB was intended to imply that no quality program should need them, but the Standard and rationale are inconsistent with such a supposed intent. –
Fashion
@jrh: I wrote that answer before I'd realized how out of hand the hyper-modernist philosophy had gotten. What irks me is the progression from "We don't need to officially recognize this behavior because the platforms where it's needed can support it anyway" to "We can remove this behavior without providing a usable replacement because it was never recognized and thus any code needing it was broken". Many behaviors should have been deprecated long ago in favor of replacements that were in every way better, but that would have required acknowledging their legitimacy. –
Fashion
Undefined behavior basically allows the compiler to make more assumption about conditions which can only be verified at runtime, e.g. assume that in the expression
*ptr the pointer is valid, because if it is null the program is allowed to do anything and so it is not necessary to add code to the program to check for that condition and ensure a defined behavior. –
Pooka A the time that C was standardized (1989), many C compilers existed, and each one played by slightly different rules. The primary goal of the ANSI (and later ISO) committee was to codify existing practice. Thus in many cases where multiple compilers disagreed on the "correct" semantic behavior for obviously ambiguous cases (mostly having to do with the evaluation order of expression operators), the committee (wisely) chose to deem such cases as "undefined behavior" or "implementation defined behavior". –
Hufnagel
@Scribner For
u=1; u=u++;, is it true that what's undefined is the value of u after the second statement is executed? I mean, by the rules of sequencing of value evaluation (as opposed to the side-effect evaluation), the expression u=u++ must be guaranteed to evaluate to 1, right? –
Constitutional @stillanoob: No, because the behavior of any code containing that expression is undefined, meaning that it can do literally anything. It might always evaluate to 42, except on Sundays when the moon is waxing gibbous. It might get stuck in an infinite loop instead of evaluating to anything at all. It might jump to a random location in your code. It might crash the process. It might even make your computer catch fire and make demons fly out of your nose, and the C standard still wouldn't care. –
Luba
@Rajesh Because the operator
+ is not a sequence point. Please read the Wikipedia page. –
Scribner Not to be offensive, but IMHO this answer addresses more an explanation of undefined behavior itself instead of why the code in the question actually invokes undefined behavior. Thus, I think it would be more appropriate as answer to this question. - For the reason why (due to the question title), you actually only link to the "Sequence point"-Wikipedia page, which is IMHO a little too less for being the accepted answer to that question. - Just my personal opinion. –
Balls
@Elspeth
i = ++i is significantly different (regarding the definedness) from i = i++. IMHO the first is just ++i, while the second is "++i maybe". –
Acaleph @Constitutional I think the important concept of undefinedness is that it propagates: Using an undefined statement or expression causes the surrounding statement or expression to be undefined as well. In the end it makes the whole program effect being undefined. –
Acaleph
@U.Windl in C they are both undefined behaviour. There are no degrees of "definedness" of undefined behaviour. If execution reaches either statement then the there is no defined behaviour for the whole program. –
Elspeth
@Elspeth execution does not need to reach the malformed code to cause undefined behavior. For concrete examples see: Undefined behavior can result in time travel (among other things, but time travel is the funkiest). –
Histamine
@MarkRansom The time travel can't "start" unless the code would be reached, e.g.
if (0) 1/0; can't time-travel to blow up the whole program. –
Elspeth @Elspeth did you read the whole article? The reference to "time travel" is about bad effects that occur before you reach the bad code. –
Histamine
@MarkRansom yes, I understand the subject well. If the behaviour of the program is undefined, effects can be seen at any point in the execution. If the behaviour is not undefined then there cannot be such effects. A program that would never reach an erroneous construct when executed according to the rules of execution in the standard, does not have undefined behaviour.
if (0) 1/0; being a simple demonstration of this. –
Elspeth @Elspeth read the article again. There's an example very similar to the one you keep repeating, and it fails. –
Histamine
@MarkRansom No there isn't. The article doesn't even have any complete programs; e.g. if
unwitting() is never called then the program's behaviour is not undefined, and the effects described cannot happen, time-travelling or not. In any case, comments are not the place for this sort of discussion -- if you think if (0) 1/0; can time-travel then start a new question (referencing my comment) and I will respond. –
Elspeth To put it another way - there's no time travel in the abstract machine. C is defined as an abstract machine with statements executed sequentially according to the rules of the standard. The behaviour of a program only becomes undefined if the abstract machine's execution reaches the erroneous expression. This occurrence then cancels the requirement for the real machine's output to match the abstract machine's output , meaning that the real machine is permitted to do anything at all, including display apparent time-travel effects . –
Elspeth
The abstract machine must execute the code exactly as written -- the "transformations" described by the article are operations the real machine undertakes in order to produce the same observable behaviour as the abstract machine specifies, but faster. And would be non-conforming if they caused a program without UB to display effects of UB. The examples in the article are all supposed to be taken in the context of execution of a program where the abstract machine actually reaches the erroneous expression. –
Elspeth
Most of the answers here quoted from C standard emphasizing that the behavior of these constructs are undefined. To understand why the behavior of these constructs are undefined, let's understand these terms first in the light of C11 standard:
Sequenced: (5.1.2.3)
Given any two evaluations
AandB, ifAis sequenced beforeB, then the execution ofAshall precede the execution ofB.
Unsequenced:
If
Ais not sequenced before or afterB, thenAandBare unsequenced.
Evaluations can be one of two things:
- value computations, which work out the result of an expression; and
- side effects, which are modifications of objects.
Sequence Point:
The presence of a sequence point between the evaluation of expressions
AandBimplies that every value computation and side effect associated withAis sequenced before every value computation and side effect associated withB.
Now coming to the question, for the expressions like
int i = 1;
i = i++;
standard says that:
6.5 Expressions:
If a side effect on a scalar object is unsequenced relative to either a different side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined. [...]
Therefore, the above expression invokes UB because two side effects on the same object i is unsequenced relative to each other. That means it is not sequenced whether the side effect by assignment to i will be done before or after the side effect by ++.
Depending on whether assignment occurs before or after the increment, different results will be produced and that's the one of the case of undefined behavior.
Lets rename the i at left of assignment be il and at the right of assignment (in the expression i++) be ir, then the expression be like
il = ir++ // Note that suffix l and r are used for the sake of clarity.
// Both il and ir represents the same object.
An important point regarding Postfix ++ operator is that:
just because the
++comes after the variable does not mean that the increment happens late. The increment can happen as early as the compiler likes as long as the compiler ensures that the original value is used.
It means the expression il = ir++ could be evaluated either as
temp = ir; // i = 1
ir = ir + 1; // i = 2 side effect by ++ before assignment
il = temp; // i = 1 result is 1
or
temp = ir; // i = 1
il = temp; // i = 1 side effect by assignment before ++
ir = ir + 1; // i = 2 result is 2
resulting in two different results 1 and 2 which depends on the sequence of side effects by assignment and ++ and hence invokes UB.
I think the relevant parts of the C99 standard are 6.5 Expressions, §2
Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.
and 6.5.16 Assignment operators, §4:
The order of evaluation of the operands is unspecified. If an attempt is made to modify the result of an assignment operator or to access it after the next sequence point, the behavior is undefined.
Would the above imply that 'i=i=5;" would be Undefined Behavior? –
Fashion
@Fashion as far as I know
i=i=5 is also undefined behavior –
Pylorectomy @Zaibis: The rationale I like to use for most places rule applies that in theory a mutli-processor platform could implement something like
A=B=5; as "Write-lock A; Write-Lock B; Store 5 to A; store 5 to B; Unlock B; Unock A;", and a statement like C=A+B; as "Read-lock A; Read-lock B; Compute A+B; Unlock A and B; Write-lock C; Store result; Unlock C;". That would ensure that if one thread did A=B=5; while another did C=A+B; the latter thread would either see both writes as having taken place or neither. Potentially a useful guarantee. If one thread did I=I=5;, however, ... –
Fashion ... and the compiler didn't notice that both writes were to the same location (if one or both lvalues involve pointers, that may be hard to determine), the generated code could deadlock. I don't think any real-world implementations implement such locking as part of their normal behavior, but it would be permissible under the standard, and if hardware could implement such behaviors cheaply it might be useful. On today's hardware such behavior would be way too expensive to implement as a default, but that doesn't mean it would always be thus. –
Fashion
@Fashion but wouldn't the sequence point access rule of c99 alone be enough to declare it as undefined behavior? So it doesn't matter what technically the hardware could implement? –
Pylorectomy
@Zaibis: Rules which characterize actions as Undefined Behavior aren't supposed to exist merely to allow implementations to behave in hostile fashion. They're supposed to exist to allow implementers to either do something more efficiently or more usefully than would be possible in their absence. To understand why the specs characterize something as UB, it's helpful to identify something useful the rule would allow implementations to do which they otherwise could not. –
Fashion
@Fashion I absolutly agree to that what you say about the behavior of undefined behavior(^^). But this doesn't change the point that if something is in the standard listed as UB you can expect, it is well defined just because it would be easy to implement as well defined construct. If the standard says it is UB, then the answer to the question is it UB? is "Yes!", and not "It could... [...]". –
Pylorectomy
@Zaibis: The answer to almost any question of the form "Why is X in language/framework Y Undefined Behavior" is "Because that's what the standard for Y says", but that's hardly enlightening. In most cases, however, what someone asking such a question really wants to know is "Why did the makers of the standard specify that". In most cases, things are specified as UB (rather than partially-specified behaviors) to allow for the possibility of an implementation which might do something unexpected. For example, the spec could have said that
p1=malloc(4); p2=malloc(5); r=p1>p2;... –
Fashion ...may result in
r arbitrarily holding 1 or 0, with no guarantee that the value will relate in any way to future comparisons among the same or different operands. Such a spec (returning an arbitrary 0 or 1) would have allowed an efficient memmove to be written in portable fashion [if dest > src, apply a top-down copy, else bottom-up; if the regions don't overlap, either will work so the comparison result wouldn't matter]. I believe the standard says such comparison is UB, however; if every machine could easily--at worst--arbitrarily yield a 0 or 1, there'd be no reason not to say so. –
Fashion Just compile and disassemble your line of code, if you are so inclined to know how exactly it is you get what you are getting.
This is what I get on my machine, together with what I think is going on:
$ cat evil.c
void evil(){
int i = 0;
i+= i++ + ++i;
}
$ gcc evil.c -c -o evil.bin
$ gdb evil.bin
(gdb) disassemble evil
Dump of assembler code for function evil:
0x00000000 <+0>: push %ebp
0x00000001 <+1>: mov %esp,%ebp
0x00000003 <+3>: sub $0x10,%esp
0x00000006 <+6>: movl $0x0,-0x4(%ebp) // i = 0 i = 0
0x0000000d <+13>: addl $0x1,-0x4(%ebp) // i++ i = 1
0x00000011 <+17>: mov -0x4(%ebp),%eax // j = i i = 1 j = 1
0x00000014 <+20>: add %eax,%eax // j += j i = 1 j = 2
0x00000016 <+22>: add %eax,-0x4(%ebp) // i += j i = 3
0x00000019 <+25>: addl $0x1,-0x4(%ebp) // i++ i = 4
0x0000001d <+29>: leave
0x0000001e <+30>: ret
End of assembler dump.
(I... suppose that the 0x00000014 instruction was some kind of compiler optimization?)
how do i get the machine code? I use Dev C++, and i played around with 'Code Generation' option in compiler settings, but go no extra file output or any console output –
Davie
@ronnieaka
gcc evil.c -c -o evil.bin and gdb evil.bin → disassemble evil, or whatever the Windows equivalents of those are :) –
Favorite is -0x4(%ebp) = 4 at the end? –
Foliole
This answer does not really address the question of
Why are these constructs undefined behavior?. –
Mayan @ShafikYaghmour I'm addressing the questions in the question body ("why am I not getting the results I am getting?"), see the comments in the code. Given that this is undefined behaviour, I can only show how to get the actual assembly he's compiled. –
Favorite
Perhaps the answer is in there but I think most would not be able to figure it out without some elaboration. Just add some explanatory text and it becomes an answer. –
Mayan
@ShafikYaghmour I must admit that the assembly is kinda baffling me; especially the instruction at +20. But why am I trying to make sense of it? –
Favorite
As an aside, it'll be easier to compile to assembly (with
gcc -S evil.c), which is all that's needed here. Assembling then disassembling it is just a roundabout way of doing it. –
Dictate For the record, if for whatever reason you're wondering what a given construct does -- and especially if there's any suspicion that it might be undefined behavior -- the age-old advice of "just try it with your compiler and see" is potentially quite perilous. You will learn, at best, what it does under this version of your compiler, under these circumstances, today. You will not learn much if anything about what it's guaranteed to do. In general, "just try it with your compiler" leads to nonportable programs that work only with your compiler. –
Wedgwood
@Davie I'd suggest using godbolt.org to quickly see generated assembly code for many different compilers/architectures/optimization levels. Links can also be shared easily. –
Pharmacist
The behavior can't really be explained because it invokes both unspecified behavior and undefined behavior, so we can not make any general predictions about this code, although if you read Olve Maudal's work such as Deep C and Unspecified and Undefined sometimes you can make good guesses in very specific cases with a specific compiler and environment but please don't do that anywhere near production.
So moving on to unspecified behavior, in draft c99 standard section6.5 paragraph 3 says(emphasis mine):
The grouping of operators and operands is indicated by the syntax.74) Except as specified later (for the function-call (), &&, ||, ?:, and comma operators), the order of evaluation of subexpressions and the order in which side effects take place are both unspecified.
So when we have a line like this:
i = i++ + ++i;
we do not know whether i++ or ++i will be evaluated first. This is mainly to give the compiler better options for optimization.
We also have undefined behavior here as well since the program is modifying variables(i, u, etc..) more than once between sequence points. From draft standard section 6.5 paragraph 2(emphasis mine):
Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.
it cites the following code examples as being undefined:
i = ++i + 1;
a[i++] = i;
In all these examples the code is attempting to modify an object more than once in the same sequence point, which will end with the ; in each one of these cases:
i = i++ + ++i;
^ ^ ^
i = (i++);
^ ^
u = u++ + ++u;
^ ^ ^
u = (u++);
^ ^
v = v++ + ++v;
^ ^ ^
Unspecified behavior is defined in the draft c99 standard in section 3.4.4 as:
use of an unspecified value, or other behavior where this International Standard provides two or more possibilities and imposes no further requirements on which is chosen in any instance
and undefined behavior is defined in section 3.4.3 as:
behavior, upon use of a nonportable or erroneous program construct or of erroneous data, for which this International Standard imposes no requirements
and notes that:
Possible undefined behavior ranges from ignoring the situation completely with unpredictable results, to behaving during translation or program execution in a documented manner characteristic of the environment (with or without the issuance of a diagnostic message), to terminating a translation or execution (with the issuance of a diagnostic message).
Another way of answering this, rather than getting bogged down in arcane details of sequence points and undefined behavior, is simply to ask, what are they supposed to mean? What was the programmer trying to do?
The first fragment asked about, i = i++ + ++i, is pretty clearly insane in my book. No one would ever write it in a real program, it's not obvious what it does, there's no conceivable algorithm someone could have been trying to code that would have resulted in this particular contrived sequence of operations. And since it's not obvious to you and me what it's supposed to do, it's fine in my book if the compiler can't figure out what it's supposed to do, either.
The second fragment, i = i++, is a little easier to understand. It looks like someone is trying to increment i, and assign the result back to i. But there are a couple ways of doing this in C. The most basic way to take i's value, add 1, and assign the result back to i, is the same in almost any programming language:
i = i + 1
C, of course, has a handy shortcut:
i++
This also means, "take i's value, add 1, and assign the result back to i". So if we construct a hodgepodge of the two, by writing
i = i++
what we're really saying is "take i's value, add 1, assign the result back to i, and assign the result back to i". We're confused, so it doesn't bother me too much if the compiler gets confused, too.
Realistically, the only time these crazy expressions get written is when people are using them as artificial examples of how ++ is supposed to work. And of course it is important to understand how ++ works. But one practical rule for using ++ is, "If it's not obvious what an expression using ++ means, don't write it."
We used to spend countless hours on comp.lang.c discussing expressions like these and why they're undefined. Two of my longer answers, that try to really explain why, are archived on the web:
- Why doesn't the Standard define what these do?
- Doesn't operator precedence determine the order of evaluation?
See also question 3.8 and the rest of the questions in section 3 of the C FAQ list.
A rather nasty gotcha with regard to Undefined Behavior is that while it used to be safe on 99.9% of compilers to use
*p=(*q)++; to mean if (p!=q) *p=(*q)++; else *p= __ARBITRARY_VALUE; that is no longer the case. Hyper-modern C would require writing something like the latter formulation (though there's no standard way of indicating code doesn't care what's in *p) to achieve the level of efficiency compilers used to provide with the former (the else clause is necessary in order to let the compiler optimize out the if which some newer compilers would require). –
Fashion @Fashion I now believe that any compiler that's "smart" enough to perform that sort of optimization must also be smart enough to peek at
assert statements, so that the programmer can precede the line in question with a simple assert(p != q). (Of course, taking that course would also require rewriting <assert.h> to not delete assertions outright in non-debug versions, but rather, turn them into something like __builtin_assert_disabled() that the compiler proper can see, and then not emit code for.) –
Wedgwood what we're really saying is "add 1 to i, and assign the result back to i, and assign the result back to i". --- I think there is one "and assign the result back to i" too much. –
Balls
@RobertSsupportsMonicaCellio It's admittedly a bit confusing the way it's written. Read it as "Add 1 to the value fetched from
i, assign the result back to i, and assign the result back to i". –
Wedgwood "C, of course, has a handy shortcut: i++" Haha, nice. Or worst case, i += 1. –
Eurus
@chux Yes, but you know C, and you're comfortable with the proper definitions of the autoincrement operators. The people who are confused about these operators are not! In particular, I believe it becomes easier to understand the misunderstandings of some beginners if you imagine that they imagine that
++i and i++ are basically both shortcuts for i + 1. Nevertheless, I've softened the sentence you commented on. –
Wedgwood Just like the prisoner’s dilemma thought experiment is not a means of studying the psychology of literal prisoners, merely because nobody sensible would write
i = i++ + ++i; literally in their code doesn’t make it a worthless example to study. Someone inclined to write their code in a functional (i.e. expression-based) style might end up writing code with similar UB, ultimately boiling down to reasons visible in examples as simple as this. For something more realistic, see <https://mcmap.net/q/23912/-order-of-evaluation-in-v-std-exchange-v-predecessor-v/3840170>. (To pedants who say ‘C/C++ iS nOt a lAnGuAgE’: don’t bother.) –
Looksee Often this question is linked as a duplicate of questions related to code like
printf("%d %d\n", i, i++);
or
printf("%d %d\n", ++i, i++);
or similar variants.
While this is also undefined behaviour as stated already, there are subtle differences when printf() is involved when comparing to a statement such as:
x = i++ + i++;
In the following statement:
printf("%d %d\n", ++i, i++);
the order of evaluation of arguments in printf() is unspecified. That means, expressions i++ and ++i could be evaluated in any order. C11 standard has some relevant descriptions on this:
Annex J, unspecified behaviours
The order in which the function designator, arguments, and subexpressions within the arguments are evaluated in a function call (6.5.2.2).
3.4.4, unspecified behavior
Use of an unspecified value, or other behavior where this International Standard provides two or more possibilities and imposes no further requirements on which is chosen in any instance.
EXAMPLE An example of unspecified behavior is the order in which the arguments to a function are evaluated.
The unspecified behaviour itself is NOT an issue. Consider this example:
printf("%d %d\n", ++x, y++);
This too has unspecified behaviour because the order of evaluation of ++x and y++ is unspecified. But it's perfectly legal and valid statement. There's no undefined behaviour in this statement. Because the modifications (++x and y++) are done to distinct objects.
What renders the following statement
printf("%d %d\n", ++i, i++);
as undefined behaviour is the fact that these two expressions modify the same object i without an intervening sequence point.
Another detail is that the comma involved in the printf() call is a separator, not the comma operator.
This is an important distinction because the comma operator does introduce a sequence point between the evaluation of their operands, which makes the following legal:
int i = 5;
int j;
j = (++i, i++); // No undefined behaviour here because the comma operator
// introduces a sequence point between '++i' and 'i++'
printf("i=%d j=%d\n",i, j); // prints: i=7 j=6
The comma operator evaluates its operands left-to-right and yields only the value of the last operand. So in j = (++i, i++);, ++i increments i to 6 and i++ yields old value of i (6) which is assigned to j. Then i becomes 7 due to post-increment.
So if the comma in the function call were to be a comma operator then
printf("%d %d\n", ++i, i++);
will not be a problem. But it invokes undefined behaviour because the comma here is a separator.
For those who are new to undefined behaviour would benefit from reading What Every C Programmer Should Know About Undefined Behavior to understand the concept and many other variants of undefined behaviour in C.
This post: Undefined, unspecified and implementation-defined behavior is also relevant.
This sequence
int a = 10, b = 20, c = 30; printf("a=%d b=%d c=%d\n", (a = a + b + c), (b = b + b), (c = c + c)); appears to give stable behavior (right-to-left argument evaluation in gcc v7.3.0; result "a=110 b=40 c=60"). Is it because the assignments are considered as 'full-statements' and thus introduce a sequence point? Shouldn't that result in left-to-right argument/statement evaluation? Or, is it just manifestation of undefined behavior? –
Henson @Henson That printf statement involves undefined behaviour, for the same reason explained above. You are writing
b and c in 3rd & 4th arguments respectively and reading in 2nd argument. But there's no sequence between these expressions (2nd, 3rd, & 4th args). gcc/clang has an option -Wsequence-point which can help find these, too. –
Heligoland While it is unlikely that any compilers and processors would actually do so, it would be legal, under the C standard, for the compiler to implement "i++" with the sequence:
In a single operation, read `i` and lock it to prevent access until further notice
Compute (1+read_value)
In a single operation, unlock `i` and store the computed value
While I don't think any processors support the hardware to allow such a thing to be done efficiently, one can easily imagine situations where such behavior would make multi-threaded code easier (e.g. it would guarantee that if two threads try to perform the above sequence simultaneously, i would get incremented by two) and it's not totally inconceivable that some future processor might provide a feature something like that.
If the compiler were to write i++ as indicated above (legal under the standard) and were to intersperse the above instructions throughout the evaluation of the overall expression (also legal), and if it didn't happen to notice that one of the other instructions happened to access i, it would be possible (and legal) for the compiler to generate a sequence of instructions that would deadlock. To be sure, a compiler would almost certainly detect the problem in the case where the same variable i is used in both places, but if a routine accepts references to two pointers p and q, and uses (*p) and (*q) in the above expression (rather than using i twice) the compiler would not be required to recognize or avoid the deadlock that would occur if the same object's address were passed for both p and q.
While the syntax of the expressions like a = a++ or a++ + a++ is legal, the behaviour of these constructs is undefined because a shall in C standard is not obeyed. C99 6.5p2:
- Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. [72] Furthermore, the prior value shall be read only to determine the value to be stored [73]
With footnote 73 further clarifying that
This paragraph renders undefined statement expressions such as
i = ++i + 1; a[i++] = i;while allowing
i = i + 1; a[i] = i;
The various sequence points are listed in Annex C of C11 (and C99):
The following are the sequence points described in 5.1.2.3:
- Between the evaluations of the function designator and actual arguments in a function call and the actual call. (6.5.2.2).
- Between the evaluations of the first and second operands of the following operators: logical AND && (6.5.13); logical OR || (6.5.14); comma , (6.5.17).
- Between the evaluations of the first operand of the conditional ? : operator and whichever of the second and third operands is evaluated (6.5.15).
- The end of a full declarator: declarators (6.7.6);
- Between the evaluation of a full expression and the next full expression to be evaluated. The following are full expressions: an initializer that is not part of a compound literal (6.7.9); the expression in an expression statement (6.8.3); the controlling expression of a selection statement (if or switch) (6.8.4); the controlling expression of a while or do statement (6.8.5); each of the (optional) expressions of a for statement (6.8.5.3); the (optional) expression in a return statement (6.8.6.4).
- Immediately before a library function returns (7.1.4).
- After the actions associated with each formatted input/output function conversion specifier (7.21.6, 7.29.2).
- Immediately before and immediately after each call to a comparison function, and also between any call to a comparison function and any movement of the objects passed as arguments to that call (7.22.5).
The wording of the same paragraph in C11 is:
- If a side effect on a scalar object is unsequenced relative to either a different side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined. If there are multiple allowable orderings of the subexpressions of an expression, the behavior is undefined if such an unsequenced side effect occurs in any of the orderings.84)
You can detect such errors in a program by for example using a recent version of GCC with -Wall and -Werror, and then GCC will outright refuse to compile your program. The following is the output of gcc (Ubuntu 6.2.0-5ubuntu12) 6.2.0 20161005:
% gcc plusplus.c -Wall -Werror -pedantic
plusplus.c: In function ‘main’:
plusplus.c:6:6: error: operation on ‘i’ may be undefined [-Werror=sequence-point]
i = i++ + ++i;
~~^~~~~~~~~~~
plusplus.c:6:6: error: operation on ‘i’ may be undefined [-Werror=sequence-point]
plusplus.c:10:6: error: operation on ‘i’ may be undefined [-Werror=sequence-point]
i = (i++);
~~^~~~~~~
plusplus.c:14:6: error: operation on ‘u’ may be undefined [-Werror=sequence-point]
u = u++ + ++u;
~~^~~~~~~~~~~
plusplus.c:14:6: error: operation on ‘u’ may be undefined [-Werror=sequence-point]
plusplus.c:18:6: error: operation on ‘u’ may be undefined [-Werror=sequence-point]
u = (u++);
~~^~~~~~~
plusplus.c:22:6: error: operation on ‘v’ may be undefined [-Werror=sequence-point]
v = v++ + ++v;
~~^~~~~~~~~~~
plusplus.c:22:6: error: operation on ‘v’ may be undefined [-Werror=sequence-point]
cc1: all warnings being treated as errors
The important part is to know what a sequence point is -- and what is a sequence point and what isn't. For example the comma operator is a sequence point, so
j = (i ++, ++ i);
is well-defined, and will increment i by one, yielding the old value, discard that value; then at comma operator, settle the side effects; and then increment i by one, and the resulting value becomes the value of the expression - i.e. this is just a contrived way to write j = (i += 2) which is yet again a "clever" way to write
i += 2;
j = i;
However, the , in function argument lists is not a comma operator, and there is no sequence point between evaluations of distinct arguments; instead their evaluations are unsequenced with regard to each other; so the function call
int i = 0;
printf("%d %d\n", i++, ++i, i);
has undefined behaviour because there is no sequence point between the evaluations of i++ and ++i in function arguments, and the value of i is therefore modified twice, by both i++ and ++i, between the previous and the next sequence point.
The C standard says that a variable should only be assigned at most once between two sequence points. A semi-colon for instance is a sequence point.
So every statement of the form:
i = i++;
i = i++ + ++i;
and so on violate that rule. The standard also says that behavior is undefined and not unspecified. Some compilers do detect these and produce some result but this is not per standard.
However, two different variables can be incremented between two sequence points.
while(*src++ = *dst++);
The above is a common coding practice while copying/analysing strings.
Of course it doesn't apply to different variables within one expression. It would be a total design failure if it did! All you need in the 2nd example is for both to be incremented between the statement ending and the next one beginning, and that's guaranteed, precisely because of the concept of sequence points at the centre of all this. –
Mercorr
Your question was probably not, "Why are these constructs undefined behavior in C?". Your question was probably, "Why did this code (using ++) not give me the value I expected?", and someone marked your question as a duplicate, and sent you here.
This answer tries to answer that question: why did your code not give you the answer you expected, and how can you learn to recognize (and avoid) expressions that will not work as expected.
I assume you've heard the basic definition of C's ++ and -- operators by now, and how the prefix form ++x differs from the postfix form x++. But these operators are hard to think about, so to make sure you understood, perhaps you wrote a tiny little test program involving something like
int x = 5;
printf("%d %d %d\n", x, ++x, x++);
But, to your surprise, this program did not help you understand — it printed some strange, inexplicable output, suggesting that maybe ++ does something completely different, not at all what you thought it did.
Or, perhaps you're looking at a hard-to-understand expression like
int x = 5;
x = x++ + ++x;
printf("%d\n", x);
Perhaps someone gave you that code as a puzzle. This code also makes no sense, especially if you run it — and if you compile and run it under two different compilers, you're likely to get two different answers! What's up with that? Which answer is correct? (And the answer is that both of them are, or neither of them are.)
As you've heard by now, these expressions are undefined, which means that the C language makes no guarantee about what they'll do. This is a strange and unsettling result, because you probably thought that any program you could write, as long as it compiled and ran, would generate a unique, well-defined output. But in the case of undefined behavior, that's not so.
What makes an expression undefined? Are expressions involving ++ and -- always undefined? Of course not: these are useful operators, and if you use them properly, they're perfectly well-defined.
For the expressions we're talking about, what makes them undefined is when there's too much going on at once, when we can't tell what order things will happen in, but when the order matters to the result we'll get.
Let's go back to the two examples I've used in this answer. When I wrote
printf("%d %d %d\n", x, ++x, x++);
the question is, before actually calling printf, does the compiler compute the value of x first, or x++, or maybe ++x? But it turns out we don't know. There's no rule in C which says that the arguments to a function get evaluated left-to-right, or right-to-left, or in some other order. So we can't say whether the compiler will do x first, then ++x, then x++, or x++ then ++x then x, or some other order. But the order clearly matters, because depending on which order the compiler uses, we'll clearly get a different series of numbers printed out.
What about this crazy expression?
x = x++ + ++x;
The problem with this expression is that it contains three different attempts to modify the value of x: (1) the x++ part tries to take x's value, add 1, store the new value in x, and return the old value; (2) the ++x part tries to take x's value, add 1, store the new value in x, and return the new value; and (3) the x = part tries to assign the sum of the other two back to x. Which of those three attempted assignments will "win"? Which of the three values will actually determine the final value of x? Again, and perhaps surprisingly, there's no rule in C to tell us.
You might imagine that precedence or associativity or left-to-right evaluation tells you what order things happen in, but they do not. You may not believe me, but please take my word for it, and I'll say it again: precedence and associativity do not determine every aspect of the evaluation order of an expression in C. In particular, if within one expression there are multiple different spots where we try to assign a new value to something like x, precedence and associativity do not tell us which of those attempts happens first, or last, or anything.
So with all that background and introduction out of the way, if you want to make sure that all your programs are well-defined, which expressions can you write, and which ones can you not write?
These expressions are all fine:
y = x++;
z = x++ + y++;
x = x + 1;
x = a[i++];
x = a[i++] + b[j++];
x[i++] = a[j++] + b[k++];
x = *p++;
x = *p++ + *q++;
These expressions are all undefined:
x = x++;
x = x++ + ++x;
y = x + x++;
a[i] = i++;
a[i++] = i;
printf("%d %d %d\n", x, ++x, x++);
And the last question is, how can you tell which expressions are well-defined, and which expressions are undefined?
As I said earlier, the undefined expressions are the ones where there's too much going at once, where you can't be sure what order things happen in, and where the order matters:
- If there's one variable that's getting modified (assigned to) in two or more different places, how do you know which modification happens first?
- If there's a variable that's getting modified in one place, and having its value used in another place, how do you know whether it uses the old value or the new value?
As an example of #1, in the expression
x = x++ + ++x;
there are three attempts to modify x.
As an example of #2, in the expression
y = x + x++;
we both use the value of x, and modify it.
So that's the answer: make sure that in any expression you write, each variable is modified at most once, and if a variable is modified, you don't also attempt to use the value of that variable somewhere else.
One more thing. You might be wondering how to "fix" the undefined expressions I started this answer by presenting.
In the case of printf("%d %d %d\n", x, ++x, x++);, it's easy — just write it as three separate printf calls:
printf("%d ", x);
printf("%d ", ++x);
printf("%d\n", x++);
Now the behavior is perfectly well defined, and you'll get sensible results.
In the case of x = x++ + ++x, on the other hand, there's no way to fix it. There's no way to write it so that it has guaranteed behavior matching your expectations — but that's okay, because you would never write an expression like x = x++ + ++x in a real program anyway.
In https://mcmap.net/q/23902/-why-are-these-constructs-using-pre-and-post-increment-undefined-behavior someone asked about a statement like:
int k[] = {0,1,2,3,4,5,6,7,8,9,10};
int i = 0;
int num;
num = k[++i+k[++i]] + k[++i];
printf("%d", num);
which prints 7... the OP expected it to print 6.
The ++i increments aren't guaranteed to all complete before the rest of the calculations. In fact, different compilers will get different results here. In the example you provided, the first 2 ++i executed, then the values of k[] were read, then the last ++i then k[].
num = k[i+1]+k[i+2] + k[i+3];
i += 3
Modern compilers will optimize this very well. In fact, possibly better than the code you originally wrote (assuming it had worked the way you had hoped).
A good explanation about what happens in this kind of computation is provided in the document n1188 from the ISO W14 site.
I explain the ideas.
The main rule from the standard ISO 9899 that applies in this situation is 6.5p2.
Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.
The sequence points in an expression like i=i++ are before i= and after i++.
In the paper that I quoted above it is explained that you can figure out the program as being formed by small boxes, each box containing the instructions between 2 consecutive sequence points. The sequence points are defined in annex C of the standard, in the case of i=i++ there are 2 sequence points that delimit a full-expression. Such an expression is syntactically equivalent with an entry of expression-statement in the Backus-Naur form of the grammar (a grammar is provided in annex A of the Standard).
So the order of instructions inside a box has no clear order.
i=i++
can be interpreted as
tmp = i
i=i+1
i = tmp
or as
tmp = i
i = tmp
i=i+1
because both all these forms to interpret the code i=i++ are valid and because both generate different answers, the behavior is undefined.
So a sequence point can be seen by the beginning and the end of each box that composes the program [the boxes are atomic units in C] and inside a box the order of instructions is not defined in all cases. Changing that order one can change the result sometimes.
EDIT:
Other good source for explaining such ambiguities are the entries from c-faq site (also published as a book) , namely here and here and here .
How this answer added new to the existing answers? Also the explanations for
i=i++ is very similar to this answer. –
Inflect @Inflect I did not read the other answers. I wanted to explain in my own language what I learned from the mentioned document from the official site of ISO 9899 open-std.org/jtc1/sc22/wg14/www/docs/n1188.pdf –
Tamikatamiko
@Inflect this answer is ok besides the fact it's a copy of your answer, but I would ask instead, what all other answers are doing here and why they have so much rep while missing the main point of question, that's explaining the details of the UB in examples. –
Goyette
@SoupEndless There are many answers because this is a canonical question for a number of similar (but not direct) duplicates. Without the overhead of creating different canonical posts for minor variants of the same question, often others post answers much later (often years later!) to make a question an ideal candidate for dup hammers. That's what happened here. It's pointless to repeat the same answer (especially after a few years, it's been answered!). So the latter answerers didn't really "miss the point". This is just how SO works. –
Heligoland
The reason is that the program is running undefined behavior. The problem lies in the evaluation order, because there is no sequence points required according to C++98 standard ( no operations is sequenced before or after another according to C++11 terminology).
However if you stick to one compiler, you will find the behavior persistent, as long as you don't add function calls or pointers, which would make the behavior more messy.
Using Nuwen MinGW 15 GCC 7.1 you will get:
#include<stdio.h>
int main(int argc, char ** argv)
{
int i = 0;
i = i++ + ++i;
printf("%d\n", i); // 2
i = 1;
i = (i++);
printf("%d\n", i); //1
volatile int u = 0;
u = u++ + ++u;
printf("%d\n", u); // 2
u = 1;
u = (u++);
printf("%d\n", u); //1
register int v = 0;
v = v++ + ++v;
printf("%d\n", v); //2
}
How does GCC work? it evaluates sub expressions at a left to right order for the right hand side (RHS) , then assigns the value to the left hand side (LHS) . This is exactly how Java and C# behave and define their standards. (Yes, the equivalent software in Java and C# has defined behaviors). It evaluate each sub expression one by one in the RHS Statement in a left to right order; for each sub expression: the ++c (pre-increment) is evaluated first then the value c is used for the operation, then the post increment c++).
according to GCC C++: Operators
In GCC C++, the precedence of the operators controls the order in which the individual operators are evaluated
the equivalent code in defined behavior C++ as GCC understands:
#include<stdio.h>
int main(int argc, char ** argv)
{
int i = 0;
//i = i++ + ++i;
int r;
r=i;
i++;
++i;
r+=i;
i=r;
printf("%d\n", i); // 2
i = 1;
//i = (i++);
r=i;
i++;
i=r;
printf("%d\n", i); // 1
volatile int u = 0;
//u = u++ + ++u;
r=u;
u++;
++u;
r+=u;
u=r;
printf("%d\n", u); // 2
u = 1;
//u = (u++);
r=u;
u++;
u=r;
printf("%d\n", u); // 1
register int v = 0;
//v = v++ + ++v;
r=v;
v++;
++v;
r+=v;
v=r;
printf("%d\n", v); //2
}
Then we go to Visual Studio. Visual Studio 2015, you get:
#include<stdio.h>
int main(int argc, char ** argv)
{
int i = 0;
i = i++ + ++i;
printf("%d\n", i); // 3
i = 1;
i = (i++);
printf("%d\n", i); // 2
volatile int u = 0;
u = u++ + ++u;
printf("%d\n", u); // 3
u = 1;
u = (u++);
printf("%d\n", u); // 2
register int v = 0;
v = v++ + ++v;
printf("%d\n", v); // 3
}
How does Visual Studio work, it takes another approach, it evaluates all pre-increments expressions in first pass, then uses variables values in the operations in second pass, assign from RHS to LHS in third pass, then at last pass it evaluates all the post-increment expressions in one pass.
So the equivalent in defined behavior C++ as Visual C++ understands:
#include<stdio.h>
int main(int argc, char ** argv)
{
int r;
int i = 0;
//i = i++ + ++i;
++i;
r = i + i;
i = r;
i++;
printf("%d\n", i); // 3
i = 1;
//i = (i++);
r = i;
i = r;
i++;
printf("%d\n", i); // 2
volatile int u = 0;
//u = u++ + ++u;
++u;
r = u + u;
u = r;
u++;
printf("%d\n", u); // 3
u = 1;
//u = (u++);
r = u;
u = r;
u++;
printf("%d\n", u); // 2
register int v = 0;
//v = v++ + ++v;
++v;
r = v + v;
v = r;
v++;
printf("%d\n", v); // 3
}
as Visual Studio documentation states at Precedence and Order of Evaluation:
Where several operators appear together, they have equal precedence and are evaluated according to their associativity. The operators in the table are described in the sections beginning with Postfix Operators.
I've edited the question to add the UB in evaluation of function arguments, as this question is often used as a duplicate for that. (The last example) –
Hesperidium
Also the question is about c now, not C++ –
Hesperidium
However if you stick to one compiler, you will find the behavior persistent. Well, no, not necessarily. If you, for example, change optimization flags, the compiler may quite easily end up emitting code which makes the undefined behavior behave differently. Also if you make seemingly-unrelated changes to nearby code. –
Wedgwood
The key to understanding this is that the value of the expression i++ is i and it's effect is to add 1 to i (i.e. store the value i+1 in the variable i) but that does not mean that the store will take place when the value is determined.
In an expression like i++ + ++i the value of the left-hand-side of the addition is i and right-hand-side is `i+1'.
But it's undefined when the effect of either side takes place so undefined what the value of the whole expression (i++ + ++i).
It's perfectly reasonable (and possibly efficient) for a compilter to notice that the net effect is to increment i by 2 and evaluate (what amounts to i+i+1 and later store i+2 in i, or not do that.
What you should not do is try and work out what your compiler does and play to it.
Changes to the compiler optimisation settings, apparently (to you!) unrelated changes to the surrounding code or new releases of the compiler could all change the behaviour.
You lay yourself open to one of the most time consuming kinds of bug that suddenly arise in apparently unchanged code.
Write the code you need (e.g. 2*i+1; i+=2;) and realise that all modern commercial compilers will (when optimisation is on) translate that into the most efficient code for your platform.
I would even suggest never using ++ in any other expression than standalone and then only because it's easy to read. Don't imagine it's somehow more efficient than i=i+1 because all modern commercial compilers will emit the same code for both. They ain't daft.
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(i++)still evaluates to 1, regardless of parentheses – Balkini = (i++);was intended to do, there is certainly a clearer way to write it. That would be true even if it were well defined. Even in Java, which defines the behavior ofi = (i++);, it's still bad code. Just writei++;– AlburC, it may be of interest to some aspects related to this is going to change with the next version ofC++, with the voted in C++17 evaluation order guarantees (P0145R2) More: #38502087 – Npint i=5; System.out.printf(",%d,%d,%d,%d,%d",i++,i--,++i,--i,i);gcc 5.3.0: Output: 4,5,5,5,5 Java1.8 Output: 5,6,6,5,5 – Acinusmemcpyin cases where the source and destination might occasionally be equal, e.g. because the cost of an occasional redundant copy would be less than... – Fashionx=(*p)++ + (*q)++; b=*p; c=*p;, for example, it may be reasonable to say that with some optimizations enabled a compiler could at its option independently treatbandcas either holding the one plus the value that was read before the increment of*p, or as holding a value which is read from*pat any time between the increment and the assignment toborc. – Fashion