C ++浮点数和精度 [英] C++ Float Division and Precision

问题描述

我知道511除以512实际上等于0.998046875。我也知道浮点的精度是7位数。我的问题是，当我做这个数学在C + +（GCC），我得到的结果是0.998047，这是一个舍入的值。我希望只是得到截断值0.998046，我该怎么办？

I know that 511 divided by 512 actually equals 0.998046875. I also know that the precision of floats is 7 digits. My question is, when I do this math in C++ (GCC) the result I get is 0.998047, which is a rounded value. I'd prefer to just get the truncated value of 0.998046, how can I do that?

  float a = 511.0f;
  float b = 512.0f;
  float c = a / b;

推荐答案

作为 float 的 511/512 的值是精确的。不进行舍入。您可以通过要求超过七位数字来检查：

Well, here's one problem. The value of 511/512, as a float, is exact. No rounding is done. You can check this by asking for more than seven digits:

#include <stdio.h>
int main(int argc, char *argv[])
{
    float x = 511.0f, y = 512.0f;
    printf("%.15f\n", x/y);
    return 0;
}

输出：

0.998046875000000

A code>不是作为十进制数存储，而是二进制。如果你用一个数字除以2的幂，例如512，结果几乎总是准确的。发生的是 float 的精度不是7位数，而是精确的23位。

A float is stored not as a decimal number, but binary. If you divide a number by a power of 2, such as 512, the result will almost always be exact. What's going on is the precision of a float is not simply 7 digits, it is really 23 bits of precision.

请参见每个计算机科学家应该知道的浮动 - 点算术。

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