模数和rand（）如何工作？ [英] How does modulus and rand() work?

问题描述

rand（）％6总是产生一个0-5之间的结果。

$ b $

我应该有rand（）％6 + 6

  0 + 6 = 6. 
 1 + 6 = 7. 
 ... 
 5 + 6 = 11。 
  

那么我需要+ 7如果我想要间隔6-12？但是，0 + 7 = 7。什么时候会随机化6？

我在这里缺少什么？哪一个是正确的方法有一个随机的数字在6和12之间？为什么？

解决方案

如果 C ++ 11 是一个选项，您应该使用随机标头和 uniform_int_distrubution 。正如James在使用 rand 和％的评论中指出的，有很多问题，包括有偏差的分布：

  #include< iostream> 
 #include< random> 
 
 int main（）
 {
 std :: random_device rd; 
 
 std :: mt19937 e2（rd（））; 
 
 std :: uniform_int_distribution< int> dist（6，12）; 
 
 for（int n = 0; n <10; ++ n）{
 std :: cout< dist（e2） ，; 
} 
 std :: cout<< std :: endl; 
} 
  

如果您必须使用 rand 那么应该这样做：

  rand（）％7 + 6 
  

更新 p>

使用 rand 的更好方法如下：

  6 + rand（）/（RAND_MAX /（12  -  6 + 1）+ 1）
  

$ b b

我从 C常见问题中获得了这一点，如何在一定范围内获取随机整数？问题。

更新2

Boost 也是一个选项：

  #include< iostream> 
 #include< boost / random / mersenne_twister.hpp> 
 #include< boost / random / uniform_int_distribution.hpp> 
 
 int main（）
 {
 boost :: random :: mt19937 gen; 
 boost :: random :: uniform_int_distribution<> dist（6，12）; 
 
 for（int n = 0; n <10; ++ n）{
 std :: cout< dist（gen）< ，; 
} 
 std :: cout<< std :: endl; 
} 
  

So, I've been nuts on this.

rand() % 6 will always produce a result between 0-5.

However when I need between, let's say 6-12.

Should I have rand() % 6 + 6

0+6 = 6.
1+6 = 7.
...
5+6 = 11. ???

So do I need to + 7 If I want the interval 6-12? But then, 0+7 =7. When will it randomize 6?

What am I missing here? Which one is the correct way to have a randomized number between 6 and 12? And why? It seems like I am missing something here.

解决方案

If C++11 is an option then you should use the random header and uniform_int_distrubution. As James pointed out in the comments using rand and % has a lot of issues including a biased distribution:

#include <iostream>
#include <random>

int main()
{
    std::random_device rd;

    std::mt19937 e2(rd());

    std::uniform_int_distribution<int> dist(6, 12);

    for (int n = 0; n < 10; ++n) {
            std::cout << dist(e2) << ", " ;
    }
    std::cout << std::endl ;
}

if you have to use rand then this should do:

rand() % 7 + 6

Update

A better method using rand would be as follows:

6 + rand() / (RAND_MAX / (12 - 6 + 1) + 1)

I obtained this from the C FAQ and it is explained How can I get random integers in a certain range? question.

Update 2

Boost is also an option:

#include <iostream>
#include <boost/random/mersenne_twister.hpp>
#include <boost/random/uniform_int_distribution.hpp>

int main()
{
  boost::random::mt19937 gen;
  boost::random::uniform_int_distribution<> dist(6, 12);

  for (int n = 0; n < 10; ++n) {
    std::cout << dist(gen) << ", ";
  }
  std::cout << std::endl ;
}

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