Question

After learning from this question that in C++ It is okay to put the name of the variable in parenthesis
I tried this program:

#include <iostream>
int main()
{
    int (a)();
    std::cout << "if this works then deafult value of int should be " << a << std::endl;
    return 0;
}

And got output of ‘if this works then deafult value of int should be 1’
So, is this true?

EDIT::
After reading @james-mcnellis answer when i tried to assign a value to a, it gives an error as assignment of function ‘int a()’.

so now it is clear that here a is a function not the variable.

Answer 1

a is not an int: it is a function that has no parameters and returns an int. Because it is a function declaration, a is also not a local variable and it does not have a “default value.”

The program is ill-formed because you never define the function a but you attempt to use it (by taking its address in the insertion expression). It therefore violates the one definition rule.

If you define a in the program, 1 will be printed because the address of the function a will be converted to bool: the operator<< overload that has a bool parameter is the best match for the function pointer argument type.

[Note: if you define a and compile with Visual C++, it will print the address of the function, not 1. This is (I think) because Visual C++ allows a function pointer to be implicitly converted to void*, and then the operator<< overload that has a void const* parameter is the best match for the function pointer argument type. If you compile with language extensions disabled (/Za), the overload with a bool parameter will be selected as expected.]