order of evaluation of operands [duplicate]

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In the expression a + b, is a guaranteed to be evaluated before b, or is the order of evaluation unspecified? I think it is the latter, but I struggle to find a definite answer in the standard.

Since I don't know whether C handles this different from C++, or if evaluation order rules were simplified in C++11, I'm gonna tag the question as all three.

Tetrad answered 18/8, 2011 at 18:28 Comment(2)

+1 Because I don't see any reason for this to be downvoted. Infact this might serve as a very good faq, where we have relevant quotes from c99, c++03 and c++11 standards at one place. – Individualism 18/8, 2011 at 19:0

@Als: didn't downvote it yet, but some research (#4176828) would have been good. – Sitdown 18/8, 2011 at 19:6

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In C++, for user-defined types a + b is a function call, and the standard says:

§5.2.2.8 - [...] The order of evaluation of function arguments is unspecified. [...]

For normal operators, the standard says:

§5.4 - Except where noted, the order of evaluation of operands of individual operators and subexpressions of individual expressions, and the order in which side effects take place, is unspecified. [...]

These haven't been changed for C++11. However, the wording changes in the second one to say that the order is "unsequenced" rather than unspecified, but it is essentially the same.

I don't have a copy of the C standard, but I imagine that it is the same there as well.

Mckinley answered 18/8, 2011 at 18:38 Comment(1)

CodeMonkey, already poosted the quote from the c99 standard as an comment under my answer. – Individualism 18/8, 2011 at 18:50

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It is Unspecified.

Reference - C++03 Standard:

Section 5: Expressions, Para 4:

except where noted [e.g. special rules for && and ||], the order of evaluation of operands of individual operators and subexpressions of individual expressions, and the order in which side effects take place, is Unspecified.

Individualism answered 18/8, 2011 at 18:32 Comment(3)

5.2 does not seem like the right section... I am looking at it now - edit oops I am looking at C99 spec it is 6.5 – Livvy 18/8, 2011 at 18:34

@Code Monkey: Oops, Added that quote belongs to C++03. – Individualism 18/8, 2011 at 18:35

In C99 it is at 6.5. "The grouping of operators and operands is indicated by the syntax.72) 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." – Livvy 18/8, 2011 at 18:35

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It is unspecified. C and C++ follow the same logic in selecting sequence points:

https://en.wikipedia.org/wiki/Sequence_point

Immortalize answered 18/8, 2011 at 18:34 Comment(1)

Its not undefined. Its unspecified. The Standard makes a difference between these two terminologies. – Monkhmer 18/8, 2011 at 18:42

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C++0x FDIS section 1.9 "Program Execution" §15 is similar to the corresponding paragraph in C++03, just reworded to accommodate the conceptual change from "sequence points" to "being sequenced":

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

Tetrad answered 18/8, 2011 at 18:46 Comment(0)

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For C: "Order of operations is not defined by the language. The compiler is free to evaluate such expressions in any order, if the compiler can guarantee a consistent result." [...] "Only the sequential-evaluation (,), logical-AND (&&), logical-OR (||), conditional-expression (? :), and function-call operators constitute sequence points and therefore guarantee a particular order of evaluation for their operands."

Source: http://msdn.microsoft.com/en-us/library/2bxt6kc4.aspx

The way this is organized in that article, it seems to indicate this applies to C++ too, which is confirmed by the answer to this question: Operator Precedence vs Order of Evaluation.

Handrail answered 18/8, 2011 at 18:31 Comment(0)

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According to the current C standard, C11, it also specifies that the order of evaluation of subexpressions (a and b in this case) is indeterminate. In fact, this order doesn't even have to be the same if the same expression is evaluated multiple times.

From section 6.5:

The grouping of operators and operands is indicated by the syntax.⁸⁵⁾ Except as specified later, side effects and value computations of subexpressions are unsequenced.⁸⁶⁾

86) In an expression that is evaluated more than once during the execution of a program, unsequenced and indeterminately sequenced evaluations of its subexpressions need not be performed consistently in different evaluations.

Boeke answered 3/12, 2016 at 22:31 Comment(0)

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