will right hand side of an expression always evaluated first

Question

Will right side always evaluated to ahead of left side? And then the result of right side will be passed on to left side. I am not talking about the exception such as A[i]=i++

I am talking about the normal cases:

A[i] = (j+32+43 & K); 
A[j] != (A[j] + A[k]);

will the right part of all these expression evaluated first and then the result is compared to the left side? (Always)

In C, some items, like plain expressions evaluate right to left, however, there are several statements in C that evaluate left to right. — user3629249, Oct 15 '14 at 05:33

score 7 · Accepted Answer · edited May 23 '17 at 12:01

In general the order of evaluation of sub-expressions is unspecified, there are a few exceptions such as logical and, logical or, comma operator, etc...

Since you comment stated you are interested in the general rule:

any operstor @YuHao if there is any general rule

that would be covered by the draft C99 standard section 6.5 Expressions paragraph 3 which says (emphasis mine going forward):

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.

this is basically the same in the draft C11 standard expect C11 does not list the exceptions so quoting C99 is more convenient. Paragraph 3 in C11 says:

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)

Specifically for assignment operators C99 says:

The order of evaluation of the operands is unspecified [...]

and C11 says:

[...] The evaluations of the operands are unsequenced.

this line of yours:`the order in which side effects take place are both unspecified` made everything clear. — codejam.tk, Oct 14 '14 at 02:48
@codejam.tk I added links to the draft C99 and C11 standards. I get all my draft standard links from [Where do I find the current C or C++ standard documents?](http://stackoverflow.com/q/81656) which as far as I can tell links to every public draft standard. — Shafik Yaghmour, Oct 14 '14 at 03:29

score 5 · Answer 2 · answered Oct 14 '14 at 01:08

5

No, there is no such guarantee, N1570 §6.5.16/p3 (emphasis mine):

An assignment operator stores a value in the object designated by the left operand. An assignment expression has the value of the left operand after the assignment,111) but is not an lvalue. The type of an assignment expression is the type the left operand would have after lvalue conversion. The side effect of updating the stored value of the left operand is sequenced after the value computations of the left and right operands. The evaluations of the operands are unsequenced.

Note that assignment operator "consumes" two operands and has side-effect of modifying a lvalue.

answered Oct 14 '14 at 01:08

Grzegorz Szpetkowski

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everytime i ask questions on stackoverflow people copy and explain the content from original standard. I have K&R book. I don't think it has that documentation for standards. Can you tell me where can i read those standards book or pdf something. All of them. – codejam.tk Oct 14 '14 at 02:52
@codejam.tk: Drafts are freely available from [WG14](http://www.open-std.org/jtc1/sc22/wg14/www/standards) website. – Grzegorz Szpetkowski Oct 14 '14 at 09:06

score 1 · Answer 3 · answered Jul 29 '17 at 10:54

I just encountered this today, and spent 1 hour debugging code like this:

int a[1], b=0;

a[b++] = b;

I expected a[0] to contain 0 after this, but the compiler actually decided to evaluate b++ first, then right side of assignment, and store result in a[0] (so b++ on the left side worked as it should). So this effectively became:

b++;
a[0] = b; // 1

score 0 · Answer 4 · answered Oct 14 '14 at 01:07

0

This will depend on the precedence and associativity of the operators involved.

A full list can be found here

answered Oct 14 '14 at 01:07

Ed Heal

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This will depend mostly on the compiler implementation. Precedence and associativity are factors in the compiler's decision, but they don't define the order of evaluation. – Idan Arye Oct 14 '14 at 01:20
@IdanArye - Have you any evidence for this? Are you saying that compilers can ignore precedence/associativity of the operators? – Ed Heal Oct 14 '14 at 01:22
I'm not saying that compilers can ignore precedence and associativity - I'm saying these things are different from order of evaluation. If you have `a * b + c * d` the compiler can choose to evaluate `a * b` first then `c * d` then add them, or it can choose to evaluate `c * d` first then `a * b` then add them - and both choices respect precedence and associativity. – Idan Arye Oct 14 '14 at 01:36
@IdanArye yaa i read that point in ANSI C. It depends (mostly) on the compiler implementation. But I am not sure if this word was mostly or sometimes. – codejam.tk Oct 14 '14 at 02:55

will right hand side of an expression always evaluated first

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