Write a program that does the following:

Displays your name on the screen

Calls a function

- to ask for 2 numbers and

- a character input to say whether they want to add or multiply the 2 numbers

- a Switch to determine which function  to execute

                        - do the add

                        Or

                        - do the multiply

Back in main a printf that will show the values of both numbers, the action taken (+ or *) and the result.
不求程序，只求思路。

这应该很简单吧，我的思路是这样的：1.printf()输出你的名字

2.定义三个变量，两个是你要输入的数字，一个是字符来判断是加是乘，，用scanf()输入

3.用switch判断字符变量，进行两种运算

老师要求要做函数的调用（function），下面是我编写的程序，但是怎么也compile不了，请大侠给我看看，谢谢。
#include<stdio.h>
int add(int, int);
int multiply(int, int);
main()
{
char choice;

printf("A for add\n");
printf("M for multiply\n");
printf("\nPlease input your choice:");
scanf("%c", &choice);

switch (choice)
{
case 'A':
     int a, b, total1;
     printf("Enter the first integer: ");
     scanf("%d", &a);
     printf("Enter the second integer: ");
     scanf("%d", &b);
     total1=add(a, b);
     printf("\nThe total1 is%d", total1);
     break;
case 'M':
     int c, d, total2;
     printf("Enter the first integer: ");
     scanf("%d", &c);
     printf("Enter the second integer: ");
     scanf("%d", &d);
     total2=multiply(c, d);
     printf("\nThe total is%d", total2);
     break;
default: printf("Invalid choice, please try again.\n\n");

}
}
int add(int x, int y)
     {
     int z;
     z=x+y;
     return z;
     }
int multiply(int g, int h)
     {
     int i;
     i=g*h;
     return i;
     }

vs2008环境下正常
#include<stdio.h>
#include "stdafx.h"
int add(int, int);
int multiply(int, int);
void main()
{
    char choice;

    printf("A for add\n");
    printf("M for multiply\n");
    printf("\nPlease input your choice:");
    scanf("%c", &choice);

    switch (choice)
    {
    case 'A':
        int a, b, total1;
        printf("Enter the first integer: ");
        scanf("%d", &a);
        printf("Enter the second integer: ");
        scanf("%d", &b);
        total1=add(a, b);
        printf("\nThe total1 is %d", total1);
        getchar();
        break;
    case 'M':
        int c, d, total2;
        printf("Enter the first integer: ");
        scanf("%d", &c);
        printf("Enter the second integer: ");
        scanf("%d", &d);
        total2=multiply(c, d);
        printf("\nThe total is %d", total2);
        getchar();
        break;
    default: printf("Invalid choice, please try again.\n\n");

    }
    getchar();
}
int add(int x, int y)
{
    int z;
    z=x+y;
    return z;
}
int multiply(int g, int h)
{
    int i;
    i=g*h;
    return i;
}

谢谢楼上的相助。让我烦恼的是，如何让case A break出来就运行
int add(int x, int y)
     {
     int z;
     z=x+y;
     return z;
     }
而case M出来就运行
     int multiply(int g, int h)
     {
     int i;
     i=g*h;
     return i;
     }
呢？

你的意思是数据在一开始就输入，然后根据A or M来调用函数吗？

（choce)，改为(‘choce’)是不

还有break作用域到哪，case后面的语句太多，直接那些都扔涵数里得了，朱程序，case掉用下就好，看起了简洁。

是的。

没问题呀，VC6.0运行正常
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