如果未初始化，以函数指针作为参数的函数如何接受它？

Question

any_function声明中的函数指针如何接收任何值？在 main 函数中，any_function 从未被赋予任何函数指针，只有函数本身。这是否意味着函数指针 any_function 正在递归调用自身？

#include <stdio.h>

int sum(int, int);
int product(int, int);
int difference(int, int);
int any_function(int(*pfun)(int, int), int x, int y);

int main(void)
{
  int a = 10; /* Initial value for a */
  int b = 5; /* Initial value for b */
  int result = 0; /* Storage for results */
  int (*pf)(int, int) = sum; /* Pointer to function */

  /* Passing a pointer to a function */
  result = any_function(pf, a, b);
  printf("\nresult = %d", result );

  /* Passing the address of a function */
  result = any_function(product,a, b);
  printf("\nresult = %d", result );
  printf("\nresult = %d\n", any_function(difference, a, b));

  return 0;
}

/* Definition of a function to call a function */
int any_function(int(*pfun)(int, int), int x, int y)
{
  return pfun(x, y);
}
/* Definition of the function sum */
int sum(int x, int y)
{
  return x + y;
}

/* Definition of the function product */
int product(int x, int y)
{
  return x * y;
}

/*Defintion of the function product*/
int difference(int x, int y)
{
  return x - y;
}

Answer 1

How does the function pointer in the declaration of any_function receive any value?

因为函数调用指定了一个值。与任何其他参数相同。

例如当你写：

int my_function(int x) {
    return x + 1;
}

// in main
printf("%d\n", my_function(5));

它打印出 6。但是 my_function 的声明中的 x 是如何接收到任何值的呢？嗯，写my_function(5).

的时候就指定了

In the main function, any_function is never given any function pointers, only functions themselves.

如果您使用这样的函数，它会自动转换为指针。换句话说 pf = sum; 是 pf = ∑

的缩写

Does this mean the function pointer any_function is recursively calling itself?

不是，这个想法是从哪里来的？

如果未初始化，以函数指针作为参数的函数如何接受它？

How do functions with a function pointer as an argument accept it if it is uninitialized?

c

function-pointers