#include<stdio.h>
void menu()
{
printf("****************************\n");
printf("****0.exit 1.add************\n");
printf("****2.sub 3.mul************\n");
printf("****4.div ************\n");
printf("****************************\n");
}
int add(int x, int y)
{
return x + y;
}
int sub(int x, int y)
{
return x - y;
}
int mul(int x, int y)
{
return x * y;
}
int div(int x, int y)
{
return 0;
}
int main()
{
int input = 0;
int x = 0;
int y = 0;
int ret = 0;
do
{
menu();
printf("请选择\n");
scanf("%d ", &input);
switch (input)
{
case 1:
scanf("%d %d", &x, &y);
ret = add(x, y);
printf("%d\n", ret);
break;
case 2:
scanf("%d %d", &x, &y);
ret = sub(x, y);
printf("%d\n", ret);
break;
case 3:
scanf("%d %d", &x, &y);
ret = mul(x, y);
printf("%d\n", ret);
break;
case 4:
scanf("%d %d", &x, &y);
ret = div(x, y);
printf("%d\n", ret);
break;
case 0:
printf("退出计算器\n");
default:
printf("选择错误 请重新选择\n");
}
} while (input);
return 0;
}
可以看到再switch的case语句中有大量的冗余 如何改进他就需要用到回调函数了
下面给出改进后的代码
#include<stdio.h>
void menu()
{
printf("****************************\n");
printf("****0.exit 1.add************\n");
printf("****2.sub 3.mul************\n");
printf("****4.div ************\n");
printf("****************************\n");
}
int add(int x, int y)
{
return x + y;
}
int sub(int x, int y)
{
return x - y;
}
int mul(int x, int y)
{
return x * y;
}
int div(int x, int y)
{
return 0;
}
void calc(int (*pa)(int , int))
{
int x = 0;
int y = 0;
int ret = 0;
scanf("%d %d", &x, &y);
ret = pa(x, y);
printf("%d\n", ret);
}
int main()
{
int input = 0;
do
{
menu();
printf("请选择\n");
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("退出计算器\n");
default:
printf("选择错误 请重新选择\n");
}
} while (input);
return 0;
}