default constructor for int [duplicate]

Question

Possible Duplicate:
Why is it an error to use an empty set of brackets to call a constructor with no arguments?

In an answer to this question it's said that

ints are default-constructed as 0, as if you initialized them with int(). Other primitive types are initialized similarly (e.g., double(), long(), bool(), etc.).

Just while I was explaining this to a colleague of mine I made up the following code, compiled (gcc-4.3.4) and ran, and observed unexpected behavior.

#include <iostream>  int main() {   int i();    std::cout << i << std::endl; // output is 1 } 

Why is the output 1 and not 0 ?

Answer 1

Most vexing parse comes into play here. You're actually declaring a function i, not an int variable. It shouldn't even compile (unless you actually have a function i defined somewhere... do you?).

To value-initialize the int, you need:

int i = int();