C语言 除法结果始终为零
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Division result is always zero
提问by VaioIsBorn
I got this C code.
我得到了这个 C 代码。
#include <stdio.h>
int main(void)
{
int n, d, i;
double t=0, k;
scanf("%d %d", &n, &d);
t = (1/100) * d;
k = n / 3;
printf("%.2lf\t%.2lf\n", t, k);
return 0;
}
I want to know why my variable 't' is always zero (in the printf function) ?
我想知道为什么我的变量 't' 始终为零（在 printf 函数中）？
回答by John Knoeller
because in this expression
因为在这个表达式中
t = (1/100) * d;
1 and 100 are integer values, integer division truncates, so this It's the same as this
1 和 100 是整数值，整数除法截断，所以 this 和 this 一样
t = (0) * d;
you need make that a float constant like this
你需要把它变成一个像这样的浮动常量
t = (1.0/100.0) * d;
you may also want to do the same with this
你可能也想用这个做同样的事情
k = n / 3.0;
回答by MikeP
You are using integer division, and 1/100 is always going to round down to zero in integer division.
您正在使用整数除法，并且 1/100 总是会在整数除法中向下舍入为零。
If you wanted to do floating point division and simply truncate the result, you can ensure that you are using floating pointer literals instead, and d will be implicitly converted for you:
如果您想进行浮点除法并简单地截断结果，您可以确保您使用的是浮点字面量，并且 d 将为您隐式转换：
t = (int)((1.0 / 100.0) * d);
回答by Eric
I think its because of
我认为是因为
t = (1/100) * d;
1/100 as integer division = 0
1/100 作为整数除法 = 0
then 0 * d always equals 0
那么 0 * d 总是等于 0
if you do 1.0/100.0 i think it will work correctly
如果你做 1.0/100.0 我认为它会正常工作
回答by Alex
t = (1/100) * d;
That is always equals 0,you can do this
那总是等于0，你可以这样做
t=(1%100)*d
and add it to 0
并将其添加到 0
