回调函数详解

2.实例讲解

? 先让我们引入例子：

? 这是一个我们自制模拟计算器计算两个整数加减乘除的代码：

//使?回调函数改造前 
#include <stdio.h>
int add(int a, int b)
{
 return a + b;
}
int sub(int a, int b)
{
 return a - b;
}
int mul(int a, int b)
{
 return a * b;
}
int div(int a, int b)
{
 return a / b;
}
int main()
{
 int x, y;
 int input = 1;
 int ret = 0;
 do
 {
 printf("******************
 printf(" 1:add 
 printf(" 3:mul 
 printf("******************
 printf("请选择：");
 scanf("%d", &input);
 switch (input)
 {
 case 1:
 printf("输?操作数：");
 scanf("%d %d", &x, &y)
 ret = add(x, y);
 printf("ret = %d\n", r
 break;
 case 2:
 printf("输?操作数：");
 scanf("%d %d", &x, &y)
 ret = sub(x, y);
 printf("ret = %d\n", r
 break;
 case 3:
 printf("输?操作数：");
 scanf("%d %d", &x, &y)
 ret = mul(x, y);
 printf("ret = %d\n", r
 break;
 case 4:
 printf("输?操作数：");
 scanf("%d %d", &x, &y)
 ret = div(x, y);
 printf("ret = %d\n", r
 break;
 case 0:
 printf("退出程序\n");
 break;
 default:
 printf("选择错误\n");
 break;
 }
 } while (input);
 return 0; }

? 我们发现在每一个case的后面都有着一堆相似重复的代码，这些代码会增加计算机的负担，对此，我们可以通过回调函数来对代码进行优化。

//使?回到函数改造后 #include <stdio.h>
int add(int a, int b)
{
 return a + b;
}
int sub(int a, int b)
{
 return a - b;
}
int mul(int a, int b)
{
 return a * b;
}
int div(int a, int b)
{
 return a / b;
}
void calc(int(*pf)(int, int))
{
 int ret = 0;
 int x, y;
 printf("输?操作数：");
 scanf("%d %d", &x, &y);
 ret = pf(x, y);
 printf("ret = %d\n", ret);
}
int main()
{
 int input = 1;
 do
 {
 printf("******************
 printf(" 1:add 
 printf(" 3:mul 
 printf("******************
 printf("请选择：");
 scanf("%d", &input);
 switch (input)
 {
 case 1:
 calc(add);
 break;
 case 2:
 calc(sub);
 break;
 case 3:
 calc(mul);
 break;
 case 4:
 calc(div);
 break;
 case 0:
 printf("退出程序\n");
 break;
 default:
 printf("选择错误\n");
 break;
 }
 } while (input);
return 0;}