計算器的實現：
方法一：

#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;
while (input)
{
printf( "*************************\n"
 
 );
printf( " 1:add 2:sub \n" );
printf( " 3:mul 4:div \n" );
printf( "*************************\n" );
printf( "請選擇：" );
scanf( "%d", &input);
switch (input)
{
case 1:
printf( "輸入運算元：" );
scanf( "%d %d", &x, &y);
ret = add(x, y);
break;
case 2:
printf( "輸入運算元：" );
scanf( "%d %d", &x, &y
 
);
ret = sub(x, y);
break;
case 3:
printf( "輸入運算元：" );
scanf( "%d %d", &x, &y);
ret = mul(x, y);
break;
case 4:
printf( "輸入運算元：" );
scanf( "%d %d", &x, &y);
ret = div(x, y);
break;
default:
printf( "選擇錯誤\n" );
break;
}
printf( "ret = %d\n", ret);
}
return 0;
}

方法二：

#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;
int(*p[5])(int x, int y) = { 0, add, sub, mul, div }; //轉移
表
while (input)
{
printf( "*************************\n" );
printf( " 1:add 2:sub \n" );
printf( " 3:mul 4:div \n" );
printf( "*************************\n" );
printf( "請選擇：" );
scanf( "%d", &input);
if ((input < 4 && input > 1))
{
printf( "輸入運算元：" );
scanf( "%d %d", &x, &y);
ret = (*p[input])(x, y);
}
else
printf( "輸入有誤\n" );
printf( "ret = %d\n", ret);
}
return 0;
}