comma operator in c++ doesn't evaluate second expression

Question

I wrote the following code:

#include <iostream>
using namespace std;

int f()
{
    cout << "f()" << endl;
    return 3;
}

int v()
{
    cout << "v()" << endl;
    return 4;
}

int main()
{
    int m = f(),v();
    cout << m << endl;
    return 0;
}

I expected it to print:

f()
v()
3

compiling with g++ -O0 test.cpp -o test.out and running results:

f()
3

Why the call to v is omitted? (this can't be do to optimization, because I added the flag -O0)

possible duplicate of [How does the Comma Operator work](http://stackoverflow.com/questions/54142/how-does-the-comma-operator-work) — Mohit Jain, Jul 09 '15 at 13:10

score 21 · Accepted Answer · edited Nov 10 '13 at 07:41

21

int m = f(),v();

This statement executes f() and assign return value to m, then declare function v() which return int type. int v(); also known as most vexing parse.

To achieve your comma operator test, try:

int m;
m = f(),v();
cout << m << endl;

see live sample.

edited Nov 10 '13 at 07:41

user207421

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answered Nov 10 '13 at 07:34

billz

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Also note that the comma separator has a lower precedence than the assignment operator. – JustSid Nov 10 '13 at 07:35
@Nawaz see my live sample. – billz Nov 10 '13 at 07:38
@Nawaz I tried int m; m= f(), v(); and it produced the expected f() v() 3 – LeeNeverGup Nov 10 '13 at 07:38
2

Also, you can try int m = (f(),v()); – LeeNeverGup Nov 10 '13 at 07:39
2

@billz: I misunderstood the question. +1. – Nawaz Nov 10 '13 at 07:40
3

@billz I wish we could scotch the idea that this is the most vexing parse. If you look at the article you linked to you'll see that the most vexing parse is more complicated `T1 x(T2());` not `T x();`. – john Nov 10 '13 at 07:51
3

The ambiguity that is present in this declaration has no relation to "most vexing parse" at all. – AnT Nov 10 '13 at 08:18

kfsone · Answer 2 · 2013-11-10T18:18:47.853

The following variation of your code demonstrates what this is doing:

http://ideone.com/GBae0p

#include <iostream>

int f() { std::cout << "f()" << std::endl; return 1; }

int main() {
   int t = f(), v();
   std::cout << t << std::endl;
   return 0;
}

This compiles without error, even though we don't have a "v()".

The comma operator is splitting this line

   int t = f(), v();

into

   int t = f();
   int v();

The second line is a function prototype which declares that there will be a function, int v(), which takes no parameters and returns an int.

It doesn't call it, it just pre-declares it - incase the compiler encounters a call it to it before it is actually defined - because the body is an empty statement (;). The compiler assumes we will implement it later or in a different compilation unit.

Because this commonly confuses experience and new programmers alike, because it is introduced by C++ allowing you to put function prototypes inside function bodies, this is called the most vexing parse.

comma operator in c++ doesn't evaluate second expression

2 Answers2