How to print function pointers with cout?

Question

I want to print out a function pointer using cout, and found it did not work. But it worked after I converting the function pointer to (void *), so does printf with %p, such

Answer 1

maybe (in one time I stay intersecting about the address of function) one of decision )))

#include      
#include 

void alexf();

int main()
{ 
    int int_output;

    printf("printf(\"%%p\", pf) is %p\n", alexf);

    asm( "movl %[input], %%eax\n"                
         "movl %%eax, %[output]\n" 
         : [output] "+m" (int_output)
         : [input] "r" (&alexf)
         : "eax", "ebx"
    );

        std::cout<<"" <

passing the pointer to function (&alexf) or other pointer using & use constraint r. Let gcc to use register for input argument)).