Question

我正在尝试实现一个简单的函数指针程序，但是我收到了这个警告：

警告：在将函数地址分配给函数指针
时从不兼容的指针类型进行赋值

我的计划如下：

#include <stdio.h>
#include <stdlib.h>

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int add(int a, int b);
int sub(int a, int b);
int Add3Num(int a, int b, int c);

int main(int argc, char *argv[]) {
  int (*func)(int , ...);
  int a = 20, b = 10, c = 5, result;

  func = add;
  result = func(a, b);  // Warning over here
  printf("Result of 20 + 10 = %d\n", result);

  func = sub;
  result = func(a, b);  // Warning over here
  printf("Result of 20 - 10 = %d\n", result);

  func = Add3Num;
  result = func(a, b, c);  // Warning over here
  printf("Result of 20 + 10 + 5 = %d\n", result);

  return 0;
}

int add(int a, int b){
  return a+b;
}
int sub(int a, int b){
    return a-b;
}
int Add3Num(int a, int b, int c){
  return a+b+c;
}

Answer 1

将指针声明为指向无原型函数的指针（一个函数采用未指定数量的参数）：

  int (*func)();

然后你的程序应该工作，without any need for casts（只要每次调用继续匹配当前指向的函数）。

#include <stdio.h>
int add(int a, int b);
int sub(int a, int b);
int Add3Num(int a, int b, int c);
int main(){
  int (*func)(); /*unspecified number of arguments*/
  int a = 20, b = 10, c = 5, result;

  /*Casts not needed for these function pointer assignments
    because: https://stackoverflow.com/questions/49847582/implicit-function-pointer-conversions */
  func = add;
  result = func(a, b);
  printf("Result of 20 + 10 = %d\n", result);

  func = sub;
  result = func(a, b);
  printf("Result of 20 - 10 = %d\n", result);

  func = Add3Num;
  result = func(a, b, c);
  printf("Result of 20 + 10 + 5 = %d\n", result);
  return 0;
}
/*...*/

使用原型函数，函数指针类型之间的匹配需要或多或少精确（有一些注意事项，如顶级限定符无关紧要或事物可以拼写不同），否则标准将事物定义为不确定

或者，也许最好假设无原型函数和函数指针是一个过时的特性，你可以使用强类型指针（第一个int (*)(int,int)然后int (*)(int,int,int)）并使用强制转换强制事物。

#include <stdio.h>
int add(int a, int b);
int sub(int a, int b);
int Add3Num(int a, int b, int c);
int main(){
  int (*func)(int , int);
  int a = 20, b = 10, c = 5, result;

  func = add;
  result = func(a, b);
  printf("Result of 20 + 10 = %d\n", result);

  func = sub;
  result = func(a, b);
  printf("Result of 20 - 10 = %d\n", result);

  /*cast it so it can be stored in func*/
  func = (int (*)(int,int))Add3Num; 
  /*cast func back so the call is defined (and compiles)*/
  result = ((int (*)(int,int,int))func)(a, b, c); 
  printf("Result of 20 + 10 + 5 = %d\n", result);
  return 0;
}
/*...*/

如何创建一个指向具有不同参数数量的函数的函数指针？

1 个答案: