C语言 如何在C中生成随机浮点数
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How to generate random float number in C
提问by fragon
I can't find any solution to generate a random float number in the range of [0,a], where ais some float defined by a user.
我找不到任何解决方案来生成范围内的随机浮点数[0,a]，其中a是用户定义的一些浮点数。
I have tried the following, but it doesn't seem to work correctly.
我尝试了以下方法，但似乎无法正常工作。
float x=(float)rand()/((float)RAND_MAX/a)
回答by WhozCraig
Try:
尝试：
float x = (float)rand()/(float)(RAND_MAX/a);
To understand how this works consider the following.
要了解这是如何工作的，请考虑以下内容。
N = a random value in [0..RAND_MAX] inclusively.
The above equation (removing the casts for clarity) becomes:
上面的等式（为了清楚起见去掉了演员表）变成：
N/(RAND_MAX/a)
But division by a fraction is the equivalent to multiplying by said fraction's reciprocal, so this is equivalent to:
但是除以一个分数相当于乘以该分数的倒数，所以这相当于：
N * (a/RAND_MAX)
which can be rewritten as:
可以改写为：
a * (N/RAND_MAX)
Considering N/RAND_MAXis always a floating point value between 0.0 and 1.0, this will generate a value between 0.0 and a.
考虑到N/RAND_MAX总是介于 0.0 和 1.0 之间的浮点值，这将生成介于 0.0 和a.
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):
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）：
float x = ((float)rand()/(float)(RAND_MAX)) * a;
Note: the floating point representation of amust be exactor this will never hit your absolute edge case of a(it will get close). See this articlefor the gritty details about why.
注意： 的浮点表示a必须是精确的，否则这永远不会达到你的绝对边缘情况a（它会接近）。有关原因的详细信息，请参阅本文。
Sample
样本
#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\n", ((float)rand()/(float)(RAND_MAX)) * a);
return 0;
}
Output
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
回答by baz
You can also generate in a range [min, max] with something like
您还可以在 [min, max] 范围内生成类似的东西
float float_rand( float min, float max )
{
float scale = rand() / (float) RAND_MAX; /* [0, 1.0] */
return min + scale * ( max - min ); /* [min, max] */
}
回答by PADYMKO
If you want to generate a random float in a range, try a next solution.
如果要在范围内生成随机浮点数，请尝试下一个解决方案。
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
float
random_float(const float min, const float max)
{
if (max == min) return min;
else if (min < max) return (max - min) * ((float)rand() / RAND_MAX) + min;
// return 0 if min > max
return 0;
}
int
main (const int argc, const char *argv[])
{
srand(time(NULL));
char line[] = "-------------------------------------------";
float data[10][2] = {
{-10, 10},
{-5., 5},
{-1, 1},
{-0.25, -0.15},
{1.5, 1.52},
{-1700, 8000},
{-0.1, 0.1},
{-1, 0},
{-1, -2},
{1.2, 1.1}
};
puts(line);
puts(" From | Result | To");
puts(line);
int i;
for (i = 0; i < 10; ++i) {
printf("%12f | %12f | %12f\n", data[i][0], random_float(data[i][0], data[i][1]), data[i][1]);
}
puts(line);
return 0;
}
A result (values is fickle)
结果（值变化无常）
-------------------------------------------
From | Result | To
-------------------------------------------
-10.000000 | 2.330828 | 10.000000
-5.000000 | -4.945523 | 5.000000
-1.000000 | 0.004242 | 1.000000
-0.250000 | -0.203197 | -0.150000
1.500000 | 1.513431 | 1.520000
-1700.000000 | 3292.941895 | 8000.000000
-0.100000 | -0.021541 | 0.100000
-1.000000 | -0.148299 | 0.000000
-1.000000 | 0.000000 | -2.000000
1.200000 | 0.000000 | 1.100000
-------------------------------------------
