Short circuit evaluation of a statement with ++ operator in C

Question

I have executed the following code in Code::Blocks 10.05 on Windows 7.

int a=0,b=0,c;
c=a++&&b++;
printf("\na=%d\nb=%d\nc=%d\n\n",a,b,c);

The output I obtained is given below,

a=1
b=0
c=0

This makes perfect sense because of short circuit evaluation.

The expression a++ is post increment and 0 is returned to the logical and (&&). Hence the part b++ is not evaluated since both 0 && 0 and 0 && 1 evaluates to 0.

But here arises my doubt. The precedence value of operators clearly states that ++ is having higher precedence over &&. So my understanding was like this, both a++ and b++ are evaluated and then && only checks the result of expression a++ to come to a decision. But this has not happened only a++ is evaluated here.

What is the reason for this behavior? Does && being a sequence point has something to do with this behavior? If so why we say that && is having lower precedence than ++?

Fennel

there is one more rule that is above them, a++ will only increment after a statement has executed. So, whatever its precedence is, its going to be incremenmted after statment is finished. – Sandusky

Wow, this is a nasty trap. – Arnelle

Apart from the interesting language aspects of this question, I would recommend making your code immediately legible and obvious. At present this is not the case. – Leong

I think it's sad that such code gets ANY upvotes. – Bobwhite

@saurabhagarwal a++ will increment at some indeterminate time between when a is evaluated and when the statement finishes. That's why you can't use a again within the same expression (before or after the a++). – Judges

To be honest, I think it's sad that we've had two people complaining about the code instead of editing the post. (To hold off obvious responses: I'll leave prettifying to people who actually work with the language.) – Tap

Answer 1

You are confused about precedence and order of evaluation.

Precedence defines how the operators are grouped, i.e

c = a++ && b++;

is equivalent to:

c = ((a++) && (b++));

Order of evaluation defines how the expression is evaluated, the short circuit of && means a++ is evaluated first, if it's zero, the end; if it's not zero, b++ is then evaluated.

---

As another example:

c = (a++) + (b++);

Is a++ evaluated before b++? The answer is we don't know. Most operators don't define the order of evaluation. && is one of the few operators that do define. (The rest are ||, , and ?:)

Blood

Another way to say this is that && defines a sequence point after the evaluation of its left argument and before that of its second argument. There is no sequence point defined for +. Wikipedia has a good article about sequence points. – Calices

Answer 2

There are two concepts here - order of precedence and order of evaluation. Order of precedence will have an impact only if an expression (or sub-expression) is evaluated.

In general, the order of evaluation is not sequenced. Given an operator, its operands can be evaluated in any order. The arguments of a function can be evaluated in any order.

From the C++ Standard:

1.9 Program execution

15 Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced.

and

8.3.6 Default arguments

9 Default arguments are evaluated each time the function is called. The order of evaluation of function arguments is unspecified.

For the logical AND operator, &&, the C++11 standard says:

5.14 Logical AND operator

1 The && operator groups left-to-right. The operands are both contextually converted to type bool (Clause 4). The result is true if both operands are true and false otherwise. Unlike &, && guarantees left-to-right evaluation: the second operand is not evaluated if the first operand is false.

Similar exception is specified for the logical OR operator, ||.

Since b++ is not evaluated due to short circuiting of the expression because of && operator, the order of precedence of the operators has no significance in this particular case.

Shoal

saying "order of precedence" confuses the issue, "order of evaluation" and "precedence" need to be kept separate – Soul