使用getrandom在C中随机浮动 [英] Random float in C using getrandom
问题描述
我正在尝试生成一个介于0和1之间的随机浮点数(对我来说，不管是[0,1]还是[0,1)).在线上有关此问题的每一个问题似乎都涉及rand()调用，并以time(NULL)作为种子，但是我希望能够每秒多次调用我的程序，并且每次都获得不同的随机数.这导致我进入Linux中的getrandom系统调用，该系统调用来自/dev/urandom.我想到了这个:
I'm trying to generate a random floating point number in between 0 and 1 (whether it's on [0,1] or [0,1) shouldn't matter for me). Every question online about this seems to involves the rand() call, seeded with time(NULL), but I want to be able to invoke my program more than once a second and get different random numbers every time. This lead me to the getrandom syscall in Linux, which pulls from /dev/urandom. I came up with this:
#include <stdio.h>
#include <sys/syscall.h>
#include <unistd.h>
#include <stdint.h>
int main() {
uint32_t r = 0;
for (int i = 0; i < 20; i++) {
syscall(SYS_getrandom, &r, sizeof(uint32_t), 0);
printf("%f\n", ((double)r)/UINT32_MAX);
}
return 0;
}
我的问题很简单，就是我是否正确执行了此操作.它似乎可以正常工作，但是我担心自己滥用了某些东西，并且几乎没有在线使用getrandom()的示例.
My question is simply whether or not I'm doing this correctly. It appears to work, but I'm worried that I'm misusing something, and there are next to no examples using getrandom() online.
推荐答案
OP有2个问题:
-
如何非常随机地开始序列.
How to started the sequence very randomly.
如何在[0 ... 1)范围内生成double.
How to generate a double on the [0...1) range.
通常的方法是获取一个非常随机的来源，例如/dev/urandom或syscall()甚至是seed = time() ^ process_id;的结果，并通过srand()进行播种.然后根据需要调用rand().
The usual method is to take a very random source like /dev/urandom or the result from the syscall() or maybe even seed = time() ^ process_id; and seed via srand(). Then call rand() as needed.
以下包括一种快速转换的方法来生成均匀的[0.0 to 1.0)(线性分布).但是，像所有随机生成函数一样，真正好的函数是建立在广泛研究基础之上的.这个只是根据DBL_MANT_DIG和RAND_MAX调用rand()几次，
Below includes a quickly turned method to generate a uniform [0.0 to 1.0) (linear distribution). But like all random generating functions, really good ones are base on extensive study. This one simply calls rand() a few times based on DBL_MANT_DIG and RAND_MAX,
原始double rand_01(void)的一个弱点在于，它仅生成2 ^ 52个不同的double而不是2 ^ 53.它已被修改.另一种选择:远远低于rand_01_ld(void)的double版本.
Original double rand_01(void) has a weakness in that it only generates a 2^52 different doubles rather than 2^53. It has been amended. Alternative: a double version of rand_01_ld(void) far below.
#include <assert.h>
#include <float.h>
#include <limits.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
double rand_01(void) {
assert(FLT_RADIX == 2); // needed for DBL_MANT_DIG
unsigned long long limit = (1ull << DBL_MANT_DIG) - 1;
double r = 0.0;
do {
r += rand();
// Assume RAND_MAX is a power-of-2 - 1
r /= (RAND_MAX/2 + 1)*2.0;
limit = limit / (RAND_MAX/2 + 1) / 2;
} while (limit);
// Use only DBL_MANT_DIG (53) bits of precision.
if (r < 0.5) {
volatile double sum = 0.5 + r;
r = sum - 0.5;
}
return r;
}
int main(void) {
FILE *istream = fopen("/dev/urandom", "rb");
assert(istream);
unsigned long seed = 0;
for (unsigned i = 0; i < sizeof seed; i++) {
seed *= (UCHAR_MAX + 1);
int ch = fgetc(istream);
assert(ch != EOF);
seed += (unsigned) ch;
}
fclose(istream);
srand(seed);
for (int i=0; i<20; i++) {
printf("%f\n", rand_01());
}
return 0;
}
如果要扩展到更宽的FP，则无符号的宽整数类型可能不足.下面是一种不受此限制的可移植方法.
If one wanted to extend to an even wider FP, unsigned wide integer types may be insufficient. Below is a portable method that does not have that limitation.
long double rand_01_ld(void) {
// These should be calculated once rather than each function call
// Leave that as a separate implementation problem
// Assume RAND_MAX is power-of-2 - 1
assert((RAND_MAX & (RAND_MAX + 1U)) == 0);
double rand_max_p1 = (RAND_MAX/2 + 1)*2.0;
unsigned BitsPerRand = (unsigned) round(log2(rand_max_p1));
assert(FLT_RADIX != 10);
unsigned BitsPerFP = (unsigned) round(log2(FLT_RADIX)*LDBL_MANT_DIG);
long double r = 0.0;
unsigned i;
for (i = BitsPerFP; i >= BitsPerRand; i -= BitsPerRand) {
r += rand();
r /= rand_max_p1;
}
if (i) {
r += rand() % (1 << i);
r /= 1 << i;
}
return r;
}
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