如何創建一個指向具有不同參數數量的函數的函數指針？

Question

我正在嘗試實現一個簡單的函數指針程序，但我收到此警告：

警告：在將函數地址分配給函數指針時從不兼容的指針類型進行賦值

我的計划如下：

#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

將指針聲明為指向無原型函數的指針（一個函數采用未指定數量的參數）：

  int (*func)();

然后你的程序應該工作， 而不需要演員表 （只要每個調用繼續匹配當前指向的函數）。

#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;
}
/*...*/

對於原型函數，函數指針類型之間的匹配需要或多或少精確（有一些注意事項，如頂級限定符無關緊要或事物可以拼寫不同），否則標准會保留未定義的內容。

或者，也許最好假設無原型函數和函數指針是一個過時的特性，你可以使用強類型指針（第一個int (*)(int,int)然后int (*)(int,int,int) ）並使用強迫事情。

#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;
}
/*...*/