实现一个计算器的功能：

一般的形式：

#include<stdio.h>
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 x / y;
}
void menu()
{printf("************************\n");printf("***  1.Add   2.Sub   ***\n");printf("***  3.Mul   4.Div   ***\n");printf("***  0.exit          ***\n");printf("************************\n");
}
int main()
{int ret = 0;int x, y;int input = 0;do{menu();printf("请输入你的选择：");scanf("%d", &input);switch (input){case 1:printf("请输入两个数字：");scanf("%d %d", &x, &y);ret = Add(x, y);printf("%d\n", ret);break;case 2:printf("请输入两个数字：");scanf("%d %d", &x, &y);ret = Sub(x, y);printf("%d\n", ret);break;case 3:printf("请输入两个数字：");scanf("%d %d", &x, &y);ret = Mul(x, y);printf("%d\n", ret);break;case 4:printf("请输入两个数字：");scanf("%d %d", &x, &y);ret = Div(x, y);printf("%d\n", ret);break;case 0:printf("退出！\n");break;default:printf("请输入合适的数字！\n");break;}} while (input);return 0;
}

这种一般的形式，比较冗余，有很多重复的代码，所以我们考虑到了函数指针数组的形式。

---

函数指针数组的形式

//利用函数指针数组的形式完成计算器
#include<stdio.h>
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 x / y;
}
void menu()
{printf("************************\n");printf("***  1.Add   2.Sub   ***\n");printf("***  3.Mul   4.Div   ***\n");printf("***  0.exit          ***\n");printf("************************\n");
} 
int main()
{int ret = 0;int x, y;int input = 0;int (*pf[5])(int, int) = { 0,Add,Sub,Mul,Div };do {menu();printf("请输入你的选择：");scanf("%d", &input);if (input >= 1 && input <= 4){printf("请输入两个整数：");scanf("%d %d", &x, &y);ret=pf[input](x,y);printf("%d\n", ret);}else if (input == 0){printf("退出！\n");}else{printf("请重新输入一个值\n");}} while (input);
}

除了函数指针数组可以减少冗余，还有一种回调函数，也可以减少冗余度。

---

回调函数的形式

//利用函数回调的方法
#include<stdio.h>
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 x / y;
}
void menu()
{printf("************************\n");printf("***  1.Add   2.Sub   ***\n");printf("***  3.Mul   4.Div   ***\n");printf("***  0.exit          ***\n");printf("************************\n");
}
void cals(int(*pf)(int, int))
{int ret = 0;int x, y;printf("请输入两个数字：");scanf("%d %d", &x, &y);ret = pf(x, y);printf("%d\n", ret);
}
int main()
{int ret = 0;int x, y;int input = 0;int (*pf[5])(int, int) = { 0,Add,Sub,Mul,Div };do {menu();printf("请输入你的选择：");scanf("%d", &input);switch (input){case 1:cals(Add);break;case 2:cals(Sub);break;case 3:cals(Mul);break;case 4:cals(Div);break;case 0:printf("退出！\n");break;default:printf("请输入合适的数字！\n");break;}} while (input);
}

输出结果：