如何在C中生成随机浮点数 [英] How to generate random float number in C

问题描述

我找不到任何解决方案来生成 [0,a] 范围内的随机浮点数，其中 a 是用户定义的某个浮点数.

I can't find any solution to generate a random float number in the range of [0,a], where a is some float defined by a user.

我尝试了以下方法，但似乎无法正常工作.

I have tried the following, but it doesn't seem to work correctly.

float x=(float)rand()/((float)RAND_MAX/a)

推荐答案

尝试:

float x = (float)rand()/(float)(RAND_MAX/a);

要了解其工作原理，请考虑以下内容.

To understand how this works consider the following.

N = a random value in [0..RAND_MAX] inclusively.

上面的等式(为了清晰起见去掉了演员表)变成:

The above equation (removing the casts for clarity) becomes:

N/(RAND_MAX/a)

但除以一个分数相当于乘以该分数的倒数，所以这相当于:

But division by a fraction is the equivalent to multiplying by said fraction's reciprocal, so this is equivalent to:

N * (a/RAND_MAX)

可以改写为:

a * (N/RAND_MAX)

考虑到 N/RAND_MAX 总是一个介于 0.0 和 1.0 之间的浮点值，这将生成一个介于 0.0 和 a 之间的值.

Considering N/RAND_MAX is always a floating point value between 0.0 and 1.0, this will generate a value between 0.0 and a.

或者，您可以使用以下内容，它可以有效地完成我上面显示的细分.我实际上更喜欢这个，因为它更清楚实际发生的事情(无论如何对我来说):

Alternatively, you can use the following, which effectively does the breakdown I showed above. I actually prefer this simply because it is clearer what is actually going on (to me, anyway):

float x = ((float)rand()/(float)(RAND_MAX)) * a;

注意:a 的浮点表示必须是 exact 否则这永远不会达到 a 的绝对边缘情况(它会得到关闭).有关原因的详细信息，请参阅这篇文章.

Note: the floating point representation of a must be exact or this will never hit your absolute edge case of a (it will get close). See this article for the gritty details about why.

示例

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main(int argc, char *argv[])
{
    srand((unsigned int)time(NULL));

    float a = 5.0;
    for (int i=0;i<20;i++)
        printf("%f
", ((float)rand()/(float)(RAND_MAX)) * a);
    return 0;
}

输出

1.625741
3.832026
4.853078
0.687247
0.568085
2.810053
3.561830
3.674827
2.814782
3.047727
3.154944
0.141873
4.464814
0.124696
0.766487
2.349450
2.201889
2.148071
2.624953
2.578719

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