What does this warning mean in orwell dev C++ IDE?

Question

I recently wrote below simple program but compiler shows warning.

#include <iostream>
int main()
{
    int a();
    std::cout<<a;
    return 0;
}

[Warning] the address of 'int a()' will always evaluate as 'true' [-Waddress]

What is the meaning of the above warning? Why value of a is 1 not 0?

This is the [most vexing parse](http://stackoverflow.com/questions/20529434/member-must-have-class-struct-union) ... there is an ambiguity in the C++ grammar and so anything that can be interpreted as a function declaration will be. — Shafik Yaghmour, Sep 16 '14 at 13:40
@Shafik Technically this is not the most vexing parse, though it is related. — Lightness Races in Orbit, Sep 16 '14 at 13:47
@LightnessRacesinOrbit I asked Scott myself and he said he thinks of all the ambiguities as the most vexing parse. I had this discussion several times and so I just ended up emailing him and asking since he coined the term. — Shafik Yaghmour, Sep 16 '14 at 13:47
The question I linked above is not a perfect duplicate but it is close, I can not find a better one. — Shafik Yaghmour, Sep 16 '14 at 13:55
@LightnessRacesinOrbit: Is the behaviour of above program undefined? If yes then why? — Destructor, Nov 25 '15 at 08:51
@PravasiMeet: Why, over a year later, do you suddenly think it would be undefined? The warning literally tells you that you can _always_ expect a certain value here. — Lightness Races in Orbit, Nov 25 '15 at 09:45
@LightnessRacesinOrbit: It's an ODR violation for which no diagnostic is required, meaning that my code has UB. got my answer. — Destructor, Nov 25 '15 at 09:46
@PravasiMeet: Oh, you mean because of the missing definition and _odr-use_? — Lightness Races in Orbit, Nov 25 '15 at 11:27

Paul Evans · Answer 1 · 2014-09-16T13:46:21.423

3

It might look like a definition of a as an int, but:

int a();

declares a function a taking no parameters and return int.

Use:

int a{};

instead.

std::cout<<a;

calls operator<<() with bool which is always nonzero, hence true.

edited Sep 16 '14 at 13:46

answered Sep 16 '14 at 13:42

Paul Evans

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score 2 · Accepted Answer · answered Sep 16 '14 at 13:43

int a(); declares a function, not a variable. If you want a to be a zero-initialised variable, then you'll need one of

int a{};  // C++11 or later
int a = int();
int a(0);
int a = 0;

<< doesn't have an overload that can directly take a function; so it looks for a suitable conversion sequence to a type that it is overloaded for, and finds:

int() -> int(*)() -> bool

that is, using the standard function-to-pointer and pointer-to-boolean conversions. The function pointer won't be null, since a declared function must exist and have an address; so the boolean value will be true.

What does this warning mean in orwell dev C++ IDE?

2 Answers2