Re: Function pointers and default arguments
James Kanze wrote:
On Aug 5, 2:13 am, Gianni Mariani <gi3nos...@mariani.ws> wrote:
James Kanze wrote:
[Note that as far as I can tell, this comment has nothing to
do with what preceding in the thread.]
Interesting; this code compiles in gcc but not on comeau or VC++ 2005.
int f( int p = 5 )
{
return p;
}
template <typename T>
T F( T fp )
{
return fp;
}
#include <iostream>
int main()
{
std::cout << F( f )() << "\n";
}
Ok, so who is wrong, GCC or Comeau + VC++ ?
What operator<< is being called?
int ... I think.
... The type of F(f) is "int (*)(
int )" (a "function" type is treated as a pointer to function
type when it is a function parameter). There's no such
operator<< in the standard, and there's no implicit conversion
from a function pointer to anything which has an operator<<, so
a priori, it shouldn't compile.
What does the output look like?
5
... Could it be that g++ is
(illegally) converting the function pointer to void*?
No. I don't think so.