Address of function pointers in C++ -

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• how print function pointers cout? 6 answers

i'm not clear on values being returning calling:

&next, fp, *fp, &return_func_ptr, fp_ptr, &fp_ptr, *fp_ptr 

they seem give me value 1. mean?

also, how declare

int (*return_f())(char) 

to receive parameter without using typedef?

#include <iostream>  int next(int n){     return n+99;     }  // returns pointer function typedef int (*fptr)(int);               // using typdef fptr return_func_ptr(){     return next;     }  int f(char){     return 0; } int (*return_f())(char){                // how pass parameter here?      // std::cout << "do " << param << std::endl;     return f; }    int main() {      int x = 5;      // p points x     int *p = &x;     std::cout << "x=" << x << std::endl;        // 5,               value of x     std::cout << "&x=" << &x << std::endl;      // 0x7fff6447a82c,  address of x     std::cout << "p=" << p << std::endl;        // 0x7fff6447a82c,  value of p address of x     std::cout << "*p=" << *p << std::endl;      // 5,               value of x (p dereferenced)     std::cout << "&p=" << &p << std::endl;      // 0x7fff6447a820,  address of p pointer      // change value of x thru p     // p = 6;                                   // error,           can't set int* int     *p = 6;                                          std::cout << "x=" << x << std::endl;        // 6       int y = 2;     // int *q = y;                              // error can't initiate type int, needs int*       // pointer function     int (*fp)(int);     std::cout << "&fp=" << &fp << std::endl;        // 0x7fff66da6810,  address of pointer fp     std::cout << "fp=" << fp << std::endl;          // 0,               value of pointer fp      fp = &next;                                     // fp points function next(int)     fp = next;                                           std::cout << "&next=" << &next << std::endl;    // 1,               address of function?     std::cout << "fp=" << fp << std::endl;          // 1,               value address of function?     std::cout << "&fp=" << &fp << std::endl;        // 0x7fff66da6810,  address of pointer fp?     std::cout << "*fp=" << *fp << std::endl;        // 1,               address of function?      // calling function thru pointer     int = 0;     = (*fp)(i);     std::cout << "i=" << << std::endl;            // 99     = fp(i);     std::cout << "i=" << << std::endl;            // 198        // function returns pointer function     fptr fp_ptr = return_func_ptr();     std::cout << "&return_func_ptr=" << &return_func_ptr << std::endl;      // 1     std::cout << "fp_ptr=" << *fp_ptr << std::endl;                         // 1     std::cout << "&fp_ptr=" << *fp_ptr << std::endl;                        // 1     std::cout << "*fp_ptr=" << *fp_ptr << std::endl;                        // 1      int j = fp_ptr(1);                                   std::cout << "j=" << j << std::endl;                                    // 100  } 

there pointer here seems not clear :

// pointer function int (*fp)(int); std::cout << "&fp=" << &fp << std::endl;        // 0x7fff66da6810,  address of pointer fp std::cout << "fp=" << fp << std::endl;          // 0,               value of pointer fp 

here fp undefined. lines have undefined behaviour.

after :

// function returns pointer function fptr fp_ptr = return_func_ptr(); std::cout << "&return_func_ptr=" << &return_func_ptr << std::endl;      // 1 std::cout << "fp_ptr=" << *fp_ptr << std::endl;                         // 1 std::cout << "&fp_ptr=" << *fp_ptr << std::endl;                        // 1 //           ^^^^^^^^^^    ^^^^^^^ std::cout << "*fp_ptr=" << *fp_ptr << std::endl;                        // 1 

there 2 things here :

• on line pointed, i'm not sure wanted test.

• also, cout doesn't have overload take function pointer, take bool instead. should :

  std::cout << "fn_ptr=" << reinterpret_cast<void*>( fn_ptr ) << std::endl; 

i suggest read article function pointer, explains need know : http://www.learncpp.com/cpp-tutorial/78-function-pointers/