Order vs. Associativity

Question

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cout << order of call to functions it prints?

What is the difference between order and associativity when evaluating a compound expression?

In the following example, I don't see the effect of order on the result of expression. The result is always 3 like the functions would have been called from left to right as arithmetic operators being left associative.

#include <iostream>
using std::cout;
using std::endl;

int Func1(int &i)
{
    return i;
}

int Func2(int &i)
{
    return i++;
}

int main()
{
    for (int index = 0; index < 999999999; index++)
    {
        int i = 0;

        int result = (Func2(i) + Func1(i) + Func1(i) + Func2(i));

        cout << result << endl;
    }
}

@Charles: I think, it's not a duplicate. the order in case of `cout` is well-defined! — Nawaz, Feb 25 '11 at 10:07
@Nawaz: It's exactly the same issue. The associativity of `+` is well-defined too. — CB Bailey, Feb 25 '11 at 10:24
@Nawaz The order of evaluation isn’t well-defined in the case of `cout` either. — Konrad Rudolph, Feb 25 '11 at 10:24
Is there any way to know if the order of evaluation isn't well-defined in my case? — user963241, Feb 25 '11 at 10:28
@cpx: Yes. See my answer and the quotation from the Standard itself. — Nawaz, Feb 25 '11 at 10:38
I think I'll just have to go with it except that I don't see any change in `result` evaluated from my expression. — user963241, Feb 25 '11 at 10:42

Nawaz · Answer 1 · 2011-02-25T10:19:35.703

4

 int result = (Func2(i) + Func1(i) + Func1(i) + Func2(i));

The order in which these functions are called is unspecified by the language!

The section $5/4 from the C++ Standard (2003) reads,

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.

So the free advice is : avoid writing such code. They're non-portable!

edited Feb 25 '11 at 10:19

answered Feb 25 '11 at 09:59

Nawaz

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What do you mean by "the result is predictable"? – user963241 Feb 25 '11 at 10:08
@cpx: Seeing the function definitions, you can calculate the value! – Nawaz Feb 25 '11 at 10:09
But if they were called in different order, the result won't be the same. – user963241 Feb 25 '11 at 10:11
@cpx: You were right. I didn't notice that argument is passed by reference! – Nawaz Feb 25 '11 at 10:14
Please note the value has been passed by reference to each function thus which function will change it first, is undefined. – user963241 Feb 25 '11 at 10:15
@cpx: I've quoted a reference from the standard. I think, that is helpful. – Nawaz Feb 25 '11 at 10:38

score 0 · Answer 2 · answered Feb 25 '11 at 16:36

The result is very likely to be the same if you run the code multiple times, using the same compiler with the same compile options. However, if you change the options or try another compiler, you can get a different result.

As you use only simple functions on ints, there is nothing much to be gained from calling the functions in a different order, so either left-to-right or right-to-left would be the obvious choice. And your test code can't tell the difference! :-)

Order vs. Associativity

2 Answers2