c++11 lambda expression acting on traditional C callback function

--std=c++11#include <functional> #include  <cstdio>typedef bool  (*is_x_num) ( int); Void show_x_num (Int * array, int len, is_x_num is_x_num) {     for (int i = 0; i < len; i++)  {         if (Is_x_num (array[i))  {             printf ("%d ",  array[i]);        }     }    printf ("\ n");} VOID&NBSP;SHOW_X_NUM2 (int * array, int len, std::function<bool (int) > is_x_ num) {    for (int i = 0; i < len; i++)  {         if (Is_x_num (array[i))  {             printf ("%d ",  array[i]); &NBSP;&Nbsp;      }    }    printf ("\ n");} Int main ()  {    int list[] = { 2, 3, 5, 8,  19, 20, 21 };    printf ("show all:\t");     Show_x_num (list, 7, [] (int) {return true;});     printf ("show even num:\t");     show_x_num (List, 7,  [] (int a) {return ! ( A&NBSP;%&NBSP;2);});     int two = 2;    printf ("Show even num:\t" );     //show_x_num (list, 7, [=] (int a) {return ! ( A % two);});     //error: cannot convert  ' main ():: __lambda2 '  to  ' is_x_num  {aka bool  (*) (int)} '  for argument  ' 3 '  to  ' Void show_x_num (int*,  int, is_x_num) '     show_x_num2 (list, 7, [=] (int a) {return ! ( A % two);});     return 0;}

In the above example, the 3rd parameter of Show_x_num is a traditional function pointer, and we can use a lambda expression without capture ([Captrue]) as the 3rd parameter of the Show_x_num ( section Lines 31 and 34), but using a lambda with capture will compile an error (line 39 ).

If you want to use a captured lambda, declare the function parameter as std::function<> (line 17th), and line 42nd uses the captured lambda to succeed.

This is the case (a lambda expression with a capture cannot be converted to a traditional function pointer), and I understand that a lambda with a capture actually adds parameters.

c++11 lambda expression acting on traditional C callback function