Question

我有一个奇怪的问题。我开始阅读让我们C 并遇到一种情况。当我给出以下代码percentage变量时，我得到0作为输出，但是当我修改公式时，它不会这样做。有人可以解释原因吗？

int m1,m2,m3,m4,m5,aggregate;
float percentage;

 printf("Please enter marks of the student in 5 subjects : \n");
 scanf("%d %d %d %d %d",&m1,&m2,&m3,&m4,&m5);
 aggregate = m1+m2+m3+m4+m5;
 percentage= (aggregate/500)*100;

上面的输出如下：

Please enter marks of the student in 5 subjects : 
50 50 50 50 50
The aggregate of marks obtained by the student is : 250 
The percentage of marks obtained by the student is : 0.000000*

但如果我修改百分比公式如下：百分比=聚合/ 5;

我得到了正确的输出。

Please enter marks of the student in 5 subjects : 
50 50 50 50 50
The aggregate of marks obtained by the student is : 250 
The percentage of marks obtained by the student is : 50.000000
RUN SUCCESSFUL (total time: 6s)*

有人可以解释为什么会发生这种情况，即使两者都使用相同的公式吗？

Answer 1

由于aggregate是一个int，(aggregate/500)*100将产生一个int。更改为(aggregate/500.0)*100或aggregate/5.0

percentage= aggregate/5“工作原因”是纯粹的运气。它不会包含所有数字。当聚合为5的倍数时，它将给出正确的结果。

用于计算C

1 个答案: