某些范围内的随机数c ++

Question

可能重复:
在整个范围内均匀生成随机数

我想在c ++中生成随机数,在某些范围内,假设我希望数字在25到63之间.

我怎么能这样.

谢谢

Answer 1

既然没人发布现代C++方法,

#include <iostream>
#include <random>
int main()
{
    std::random_device rd; // obtain a random number from hardware
    std::mt19937 eng(rd()); // seed the generator
    std::uniform_int_distribution<> distr(25, 63); // define the range

    for(int n=0; n<40; ++n)
        std::cout << distr(eng) << ' '; // generate numbers
}

Answer 2

您可以使用标准库(TR1)添加中包含的随机功能.或者您可以使用在纯C中工作的相同旧技术:

25 + ( std::rand() % ( 63 - 25 + 1 ) )

Answer 3

int random(int min, int max) //range : [min, max)
{
   static bool first = true;
   if (first) 
   {  
      srand( time(NULL) ); //seeding for the first time only!
      first = false;
   }
   return min + rand() % (( max + 1 ) - min);
}

Answer 4

int range = max - min + 1;
int num = rand() % range + min;

Answer 5

使用rand函数：

http://www.cplusplus.com/reference/clibrary/cstdlib/rand/

引用：

A typical way to generate pseudo-random numbers in a determined range using rand is to use the modulo of the returned value by the range span and add the initial value of the range:

( value % 100 ) is in the range 0 to 99
( value % 100 + 1 ) is in the range 1 to 100
( value % 30 + 1985 ) is in the range 1985 to 2014

Answer 6

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