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
STL是c++,不是C,所以我不知道你想要什么。然而,如果你想使用C语言,则有rand()和srand()函数:
int rand(void);
void srand(unsigned seed);
它们都是ANSI c的一部分。还有random()函数:
long random(void);
但据我所知,random()不是标准的ANSI c。第三方库可能不是一个坏主意,但这完全取决于您真正需要生成的数字的随机程度。
其他回答
对于c, STL不存在,你必须调用rand,或者更好的是,随机。它们在标准库头文件stdlib.h中声明。rand是POSIX, random是BSD规范函数。
rand和random之间的区别是random返回一个更有用的32位随机数,而rand通常返回一个16位数。BSD手册显示rand的较低位是循环的和可预测的,因此rand对于较小的数字可能是无用的。
STL是c++,不是C,所以我不知道你想要什么。然而,如果你想使用C语言,则有rand()和srand()函数:
int rand(void);
void srand(unsigned seed);
它们都是ANSI c的一部分。还有random()函数:
long random(void);
但据我所知,random()不是标准的ANSI c。第三方库可能不是一个坏主意,但这完全取决于您真正需要生成的数字的随机程度。
C程序生成9到50之间的随机数
#include <time.h>
#include <stdlib.h>
int main()
{
srand(time(NULL));
int lowerLimit = 10, upperLimit = 50;
int r = lowerLimit + rand() % (upperLimit - lowerLimit);
printf("%d", r);
}
通常我们可以生成一个介于lowerLimit和upperLimit-1之间的随机数
例如,lowerLimit包含或说r∈[lowerLimit, upperLimit)
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
我的极简解决方案应该适用于范围内的随机数[min, max)。在调用函数之前使用srand(time(NULL))。
int range_rand(int min_num, int max_num) {
if (min_num >= max_num) {
fprintf(stderr, "min_num is greater or equal than max_num!\n");
}
return min_num + (rand() % (max_num - min_num));
}