简单版

​
//模拟计算器：
void menu()
{
	printf("*****************************************\n");
	printf("************  1.add    2.sub  ***********\n");
	printf("************  3.mul    4.div  ***********\n");
	printf("************      0.exit      ***********\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 x / y;
}

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:
			printf("请输入两个操作数\n");
			scanf("%d %d",&x, &y);
			ret = Add(x, y);
			printf("ret = %d\n", ret);
			break;
		case 2:
			printf("请输入两个操作数\n");
			scanf("%d %d", &x, &y);
			ret = Sub(x, y);
			printf("ret = %d\n", ret);
			break;
		case 3:
			printf("请输入两个操作数\n");
			scanf("%d %d", &x, &y);
			ret = Mul(x, y);
			printf("ret = %d\n", ret);
			break;
		case 4:
			printf("请输入两个操作数\n");
			scanf("%d %d", &x, &y);
			ret = Div(x, y);
			printf("ret = %d\n", ret);
			break;
		case 0:
			printf("退出计算器\n");
			break;
		default:
			break;
		}
	} while (input); 
	return 0;
}

​

运行结果： 

//模拟计算器简单改良版
void menu()
{
	printf("*****************************************\n");
	printf("************  1.add    2.sub  ***********\n");
	printf("************  3.mul    4.div  ***********\n");
	printf("************      0.exit      ***********\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 x / y;
}

int main()
{
	int input = 0;
	int x = 0;
	int y = 0;
	int ret = 0;
	do
	{
		menu();
		printf("请选择:>\n");
		scanf("%d", &input);
		printf("请输入两个操作数\n");
		scanf("%d %d", &x, &y);
		switch (input)
		{
		case 1:
			ret = Add(x, y);
			printf("ret = %d\n", ret);
			break;
		case 2:
			ret = Sub(x, y);
			printf("ret = %d\n", ret);
			break;
		case 3:
			ret = Mul(x, y);
			printf("ret = %d\n", ret);
			break;
		case 4:
			ret = Div(x, y);
			printf("ret = %d\n", ret);
			break;
		case 0:
			printf("退出计算器\n");
			break;
		default:
			break;
		}
	} while (input);
	return 0;
}

//函数指针数组的应用
//case语句越来越长的增强：
void menu()
{
	printf("*****************************************\n");
	printf("************  1.add    2.sub  ***********\n");
	printf("************  3.mul    4.div  ***********\n");
	printf("************      0.exit      ***********\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 x / y;
}

int main()
{
	int input = 0;
	int x = 0;
	int y = 0;
	int ret = 0;
	do
	{
		menu();
		printf("请选择:>\n");
		scanf("%d", &input);
		//创建一个函数指针数组
      //函数指针数组  转移表
		int (*pfArr[])(int, int) = { NULL,Add,Sub,Mul,Div };
		//                            0    1   2   3   4
		if (0 == input)
		{
			printf("退出计算器\n");
		}
		else if (input >= 1 && input <= 4)
		{
			printf("请输入两个操作数\n");
			scanf("%d %d", &x, &y);
			ret = pfArr[input](x, y);
			printf("ret=%d\n", ret);
		}
		else
		{
			printf("选择错误，重新选择!\n");
		}
	} while (input);
	return 0;
}

最终版 回调函数 使得函数更加通用，通过参数不同输入，相应不同的函数

void menu()
{
	printf("*****************************************\n");
	printf("************  1.add    2.sub  ***********\n");
	printf("************  3.mul    4.div  ***********\n");
	printf("************      0.exit      ***********\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 x / y;
}

void calc(int (*pf)(int, int))//函数指针
{
	int x = 0;
	int y = 0;
	int ret = 0;
	printf("请输入两个操作数\n");//没有直接调用函数，将地址传给这个函数，在这个函数内部调用
	scanf("%d %d", &x, &y);//这里的加减乘除就是回调函数
	ret = pf(x, y);//调用函数并返回
	printf("ret = %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");
			break;
		default:
			break;
		}
	} while (input);
	return 0;
}