C语言 如何在整数除法中获得分数？
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原文地址: http://stackoverflow.com/questions/2976011/
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How to get fractions in an integer division?
提问by Chad Carisch
How do you divide two integers and get a double or float answer in C?
你如何在 C 中除以两个整数并得到双精度或浮点数的答案？
回答by Mark Rushakoff
You need to cast one or the other to a floator double.
您需要将一个或另一个转换为 a floator double。
int x = 1;
int y = 3;
// Before
x / y; // (0!)
// After
((double)x) / y; // (0.33333...)
x / ((double)y); // (0.33333...)
Of course, make sure that you are store the resultof the division in a doubleor float! It doesn't do you any good if you store the result in another int.
当然，请确保您将除法结果存储在 a doubleor 中float！如果将结果存储在另一个int.
Regarding @Chad's comment ("[tailsPerField setIntValue:tailsPer]"):
关于@Chad 的评论（“ [tailsPerField setIntValue:tailsPer]”）：
Don't pass a double or float to setIntValuewhen you have setDoubleValue, etc. available. That's probably the same issue as I mentioned in the comment, where you aren't using an explicit cast, and you're getting an invalid value because a double is being read as an int.
setIntValue当您有setDoubleValue等 可用时，不要传递双精度或浮点数。这可能与我在评论中提到的问题相同，在那里您没有使用显式强制转换，并且您得到一个无效值，因为 double 被读取为 int。
For example, on my system, the file:
例如，在我的系统上，文件：
#include <stdio.h>
int main()
{
double x = 3.14;
printf("%d", x);
return 0;
}
outputs:
输出：
1374389535
because the double was attempted to be read as an int.
因为试图将 double 读为 int。
回答by Pranav Totla
Use type-casting. For example,
使用类型转换。例如，
main()
{
float a;
int b = 2, c = 3;
a = (float) b / (float) c; // This is type-casting
printf("%f", a);
}
