c - C 语言中的 Intel RdRand 工作示例。如何生成 -100.001 到 +100.001 范围内的 float

Question

有一个英特尔 DRNG 库允许您使用基于处理器的晶体熵效应的随机数生成器。

库本身及其使用说明:https://software.intel.com/en-us/articles/intel-digital-random-number-generator-drng-library-implementation-and-uses

库中有一个示例，它只打印随机生成的数组的内容。

请分享 C 中的工作示例，它允许使用此库生成 -100.001 到 +100.001 范围内的浮点型数字

我只能找到一个基于伪随机数生成器的代码，但这不是我需要的:

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

float randoms(float min, float max)
{
    return (float)(rand())/RAND_MAX*(max - min) + min;
}

int main()
{
    srand((unsigned int)time(0));
    printf("%f\n",randoms(-100.001, 100.001));
    return 0;
}

提前致谢。

Answer 1

答案已发布on the Intel's DRNG page不久前。我想在这里引用它:

You can almost use that same algorithm. You just need a way to check for the (highly unlikely) chance the RDRAND instruction will not return a value.

Here's how I would modify your code snippet for Linux (you'll need to supply the -mrdrnd option to gcc to compile this):

#include <stdio.h>
#include <limits.h>

char randoms(float *randf, float min, float max)
{
    int retries= 10;
    unsigned long long rand64;

    while(retries--) {
        if ( __builtin_ia32_rdrand64_step(&rand64) ) {
            *randf= (float)rand64/ULONG_MAX*(max - min) + min;
            return 1;
        }
    }
    return 0;
}

int main()
{
    float randf;

    if ( randoms(&randf, -100.001, 100.001) ) printf("%f\n", randf);
    else printf("Failed to get a random value\n");
    return 0;
}

See section 4.2.1 in the above document:

4.2.1 Retry Recommendations

It is recommended that applications attempt 10 retries in a tight loop in the unlikely event that the RDRAND instruction does not return a random number. This number is based on a binomial probability argument: given the design margins of the DRNG, the odds of ten failures in a row are astronomically small and would in fact be an indication of a larger CPU issue.