有没有一个函数生成一个随机int数在C?或者我必须使用第三方库?
当前回答
FWIW,答案是肯定的,有一个stdlib.h函数叫rand;此函数主要针对速度和分布进行调优,而不是针对不可预测性。几乎所有语言和框架的内置随机函数都默认使用这个函数。还有“加密”随机数生成器,它们的可预测性要低得多,但运行速度要慢得多。在任何类型的与安全相关的应用程序中都应该使用它们。
其他回答
注意:为了安全性,不要使用rand()。如果您需要加密安全的号码,请参阅此答案。
#include <time.h>
#include <stdlib.h>
srand(time(NULL)); // Initialization, should only be called once.
int r = rand(); // Returns a pseudo-random integer between 0 and RAND_MAX.
在Linux上,您可能更喜欢使用random和srandom。
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
//generate number in range [min,max)
int random(int min, int max){
int number = min + rand() % (max - min);
return number;
}
//Driver code
int main(){
srand(time(NULL));
for(int i = 1; i <= 10; i++){
printf("%d\t", random(10, 100));
}
return 0;
}
如果您需要128个安全随机位,符合RFC 1750的解决方案是读取已知可以生成可用熵位的硬件源(例如旋转磁盘)。更好的是,好的实现应该使用混合函数组合多个源,并最终通过重新映射或删除输出来消除输出分布的倾斜。
如果你需要更多的比特,你需要做的就是从128个安全随机比特的序列开始,并将其拉伸到所需的长度,将其映射到人类可读的文本等等。
如果你想在C中生成一个安全的随机数,我将遵循这里的源代码:
https://wiki.sei.cmu.edu/confluence/display/c/MSC30-C.+Do+not+use+the+rand%28%29+function+for+generating+pseudorandom+numbers
注意,对于Windows bbcryptgenrandom是使用的,而不是CryptGenRandom,在过去的20年里已经变得不安全。您可以亲自确认BCryptGenRandom符合RFC 1750。
For POSIX-compliant operating systems, e.g. Ubuntu (a flavor of Linux), you can simply read from /dev/urandom or /dev/random, which is a file-like interface to a device that generates bits of entropy by combining multiple sources in an RFC 1750 compliant fashion. You can read a desired number of bytes from these "files" with read or fread just like you would any other file, but note that reads from /dev/random will block until a enough new bits of entropy are available, whereas /dev/urandom will not, which can be a security issue. You can get around that by checking the size of the available entropy pool, either my reading from entropy_avail, or by using ioctl.
<stdlib.h>中的rand()函数返回一个介于0和RAND_MAX之间的伪随机整数。你可以使用srand(unsigned int seed)来设置种子。
通常的做法是将%操作符与rand()结合使用以获得不同的范围(但请记住,这在一定程度上破坏了一致性)。例如:
/* random int between 0 and 19 */
int r = rand() % 20;
如果你真的在乎一致性,你可以这样做:
/* Returns an integer in the range [0, n).
*
* Uses rand(), and so is affected-by/affects the same seed.
*/
int randint(int n) {
if ((n - 1) == RAND_MAX) {
return rand();
} else {
// Supporting larger values for n would requires an even more
// elaborate implementation that combines multiple calls to rand()
assert (n <= RAND_MAX)
// Chop off all of the values that would cause skew...
int end = RAND_MAX / n; // truncate skew
assert (end > 0);
end *= n;
// ... and ignore results from rand() that fall above that limit.
// (Worst case the loop condition should succeed 50% of the time,
// so we can expect to bail out of this loop pretty quickly.)
int r;
while ((r = rand()) >= end);
return r % n;
}
}
下面是我的方法(围绕rand()的包装器):
我还扩展到允许min为INT_MIN而max为INT_MAX的情况,这通常不可能单独使用rand(),因为它返回从0到RAND_MAX的值,包括(1/2范围)。
像这样使用它:
const int MIN = 1;
const int MAX = 1024;
// Get a pseudo-random number between MIN and MAX, **inclusive**.
// Seeding of the pseudo-random number generator automatically occurs
// the very first time you call it.
int random_num = utils_rand(MIN, MAX);
定义和氧描述:
#include <assert.h>
#include <stdbool.h>
#include <stdlib.h>
/// \brief Use linear interpolation to rescale, or "map" value `val` from range
/// `in_min` to `in_max`, inclusive, to range `out_min` to `out_max`, inclusive.
/// \details Similar to Arduino's ingenious `map()` function:
/// https://www.arduino.cc/reference/en/language/functions/math/map/
///
/// TODO(gabriel): turn this into a gcc statement expression instead to prevent the potential for
/// the "double evaluation" bug. See `MIN()` and `MAX()` above.
#define UTILS_MAP(val, in_min, in_max, out_min, out_max) \
(((val) - (in_min)) * ((out_max) - (out_min)) / ((in_max) - (in_min)) + (out_min))
/// \brief Obtain a pseudo-random integer value between `min` and `max`, **inclusive**.
/// \details 1. If `(max - min + 1) > RAND_MAX`, then the range of values returned will be
/// **scaled** to the range `max - min + 1`, and centered over the center of the
/// range at `(min + max)/2`. Scaling the numbers means that in the case of scaling,
/// not all numbers can even be reached. However, you will still be assured to have
/// a random distribution of numbers across the full range.
/// 2. Also, the first time per program run that you call this function, it will
/// automatically seed the pseudo-random number generator with your system's
/// current time in seconds.
/// \param[in] min The minimum pseudo-random number you'd like, inclusive. Can be positive
/// OR negative.
/// \param[in] max The maximum pseudo-random number you'd like, inclusive. Can be positive
/// OR negative.
/// \return A pseudo-random integer value between `min` and `max`, **inclusive**.
int utils_rand(int min, int max)
{
static bool first_run = true;
if (first_run)
{
// seed the pseudo-random number generator with the seconds time the very first run
time_t time_now_sec = time(NULL);
srand(time_now_sec);
first_run = false;
}
int range = max - min + 1;
int random_num = rand(); // random num from 0 to RAND_MAX, inclusive
if (range > RAND_MAX)
{
static_assert(
sizeof(long int) > sizeof(int),
"This must be true or else the below mapping/scaling may have undefined overflow "
"and not work properly. In such a case, try casting to `long long int` instead of "
"just `long int`, and update this static_assert accordingly.");
random_num = UTILS_MAP((long int)random_num, (long int)0, (long int)RAND_MAX, (long int)min,
(long int)max);
return random_num;
}
// This is presumably a faster approach than the map/scaling function above, so do this faster
// approach below whenever you don't **have** to do the more-complicated approach above.
random_num %= range;
random_num += min;
return random_num;
}
参见:
[我在写下上面的答案后发现了这个问答,但它显然非常相关,他们对非缩放范围的情况做了同样的事情]我如何从rand()中获得特定的数字范围? [我需要进一步研究和阅读这个答案-似乎有一些好的观点,保持良好的随机性不使用模量]我如何从rand()得到一个特定的数字范围? http://c-faq.com/lib/randrange.html