CODEWITHSUNDEEP
c++compilationparentheses
samkpo
samkpo

Reputation: 143

Why doesn't the C++ compiler complain when I use functions without parentheses?

I was looking at some code a friend sent me, and he said: "It compiles, but doesn't work". I saw that he used the functions without the parentheses, something like this:

void foo(){
  cout<< "Hello world\n";
}

int main(){
  foo; //function without parentheses
  return 0;
}

The first I said was "use parentheses, you have to". And then I tested that code - it does compile, but when executed doesn't work (no "Hello world" shown).

So, why does it compile (no warning at all from the compiler GCC 4.7), but doesn't work?

Upvotes: 7

Views: 3409

Answers (3)

Edmund
Edmund

Reputation: 10809

foo;

You're not actually 'using' the function here. You're just using the address of it. In this case, you're taking it but not really using it.

Addresses of functions (i.e. their names, without any parenthesis) are useful when you want to pass that function as a callback to some other function.

Upvotes: 3

CB Bailey
CB Bailey

Reputation: 791709

You need to increase the warning level that you use. foo; is a valid expression statement (the name of a function converts to a pointer to the named function) but it has no effect.

I usually use -std=c++98 -Wall -Wextra -pedantic which gives:

<stdin>: In function 'void foo()':
<stdin>:2: error: 'cout' was not declared in this scope
<stdin>: In function 'int main()':
<stdin>:6: warning: statement is a reference, not call, to function 'foo'
<stdin>:6: warning: statement has no effect

Upvotes: 11

Bo Persson
Bo Persson

Reputation: 92241

It surely warns if you set the warning level high enough.

A function name evaluates to the address of the function, and is a legal expression. Usually it is saved in a function pointer, 

void (*fptr)() = foo;

but that is not required.

Upvotes: 12

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