function declaration without definition [duplicate]

Question

The following code will compile and is deterministic according to cppquiz.org (Question #30)

#include <iostream>
struct X {
  X() { std::cout << "X"; }
};

int main() { X x(); }

The output of the program is nothing, as

X x();

is a function declaration.

But still I wonder why this compiles though this declaration is never defined anywhere?

It would compile even if you did call x. It wouldn't link, though. — Qaz
– Qaz, Commented Jun 25, 2014 at 17:39

Paul Evans · Accepted Answer · 2014-06-25 17:27:58Z

6

Since x() is never called, there's nothing to link so no error from linker that it's not defined. It's only declared as a function taking no arguments and returning an X: X x();.

answered Jun 25, 2014 at 17:27

Paul Evans

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Comments

haccks · Accepted Answer · 2014-06-25 17:28:25Z

3

X x(); itself a declaration (prototype), not a function call. If a function call is made prior to seeing its declaration then it would not compile.

answered Jun 25, 2014 at 17:28

haccks

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