Question

http://tinypic.com/r/1pcky1/9 这是关于我的练习中期问题的图片。

这是我的代码：

#include <stdio.h>

int FindProd(int memkey1, int memkey2, int cd1, int cd2);
int FindSum(int prod1, int prod2);

void main() {

    int memkey1, cd1, memkey2, cd2;
    int tProd, jProd, tjSum;

    printf("How many memory keys did Tom buy? \n");
    scanf("%d", &memkey1);
    printf("How many memory keys did Jane buy? \n");
    scanf("%d", &memkey2);
    printf("How many CDs did Tom and Jane buy? (In such order) \n");
    scanf("%d %d", &cd1, &cd2);

    FindProd(memkey1, memkey2, cd1, cd2);
    FindSum(tProd, jProd);

    printf("Tom spent:\t %d \n", tProd);
    printf("Jane spent:\t %d \n", jProd);
    printf("Together they spent:\t %d \n", tjSum);

}

int FindProd(int memkey1, int memkey2, int cd1, int cd2)
{


  int  prod1 = (memkey1 * 20) + (cd1 * 10);
  int  prod2 = (memkey2 * 20) + (cd2 * 10);


}

int FindSum(int prod1, int prod2)
{

    int prodSum = prod1 + prod2;
    return prodSum;


}

据说汤姆花了50，简花了8，而他们共花了58。输入在这里完全无关紧要。我明白，为了提高效率，我的程序函数FindProd可能很糟糕，但是这里有什么？

Answer 1

好的，我终于明白了。这是我的新计划：

#include <stdio.h>

int FindProd(int memkey, int cd);
int FindSum(int prod1, int prod2);

void main() {

    int memkey1, cd1, memkey2, cd2;
    int tProd, jProd, tjSum;

    printf("How many memory keys did Tom buy? \n");
    scanf("%d", &memkey1);
    printf("How many memory keys did Jane buy? \n");
    scanf("%d", &memkey2);
    printf("How many CDs did Tom and Jane buy? (In such order) \n");
    scanf("%d %d", &cd1, &cd2);

    printf("Tom spent:\t %d \n", FindProd(memkey1, cd1));
    printf("Jane spent:\t %d \n", FindProd(memkey2, cd2));

    tProd = FindProd(memkey1, cd1);
    jProd = FindProd(memkey2, cd2);
    FindSum(tProd, jProd);
    printf("Together they spent:\t %d \n",  FindSum(tProd, jProd));

}

int FindProd(int memkey, int cd)
{


 int   prod = (memkey * 20) + (cd * 10);
    return prod;

}

int FindSum(int prod1, int prod2)
{

    int prodSum = prod1 + prod2;
    return prodSum;


}

我的程序返回相同的输出，无论输入我输入（C程序）

1 个答案: