定义函数的一种奇怪类型

我想这是关于返回一个函数指针

FunctionReturnType func(TheWeirdArea);

以下是一个示例:

#include <stdio.h>

int bar1(int b)
{
    printf("bar1\n");
    return 42 + b;
}

int bar2(int b)
{
    printf("bar2\n");
    return 100 + b;
}

int (*foo(int a))(int b)  // Returns a function pointer
{
    if (a == 1) return bar1;
    return bar2;
}

int main(void) {
    printf("%d\n", foo(1)(100));
    printf("%d\n", foo(2)(200));
    return 0;
}

输出:

bar1
142
bar2
300

希望这个虚拟示例对您有所帮助

int add(int x,int y)
{
return x+y;   //very dummy function return x+y
}


typedef int (*pointer_to_add)(int,int);
/* this is a declaration of  function pointer works 

with any function has an int return type and

 accept two int arguments not only add */

int (*my_function(int arg1, double arg2))(int, int)
{
   /*This a function accept an int argument 

arg1 and double arg2 and return pointer to a function or address of function

which has a two int type arguments and return an int */

pointer_to_add  ptr=add; //ptr is a pointer now to add function.

return  ptr;
} 

更简单的形式

pointer_to_add my_function(int arg1,double arg2)
{

 pointer_to_add ptr=add;

 return ptr;

 }

现在,您可以在主代码中使用返回的地址,如:

 int main(void)
  {
    pointer_to_add P=my_function(25,3.5);
    printf("%i",p(22,33));
    return 0;
    }