問題描述

所以，我一直對這個很著迷.

So, I've been nuts on this.

rand() % 6 將始終產生 0-5 之間的結果.

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

但是，當我需要介于兩者之間時，可以說是 6-12.

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

我應該有 rand() % 6 + 6

Should I have rand() % 6 + 6

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

如果我想要間隔 6-12，那么我需要 +7 嗎?但是，0+7 = 7.什么時候會隨機6個?

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

我在這里錯過了什么?哪個是在 6 到 12 之間隨機數的正確方法?為什么?我好像在這里遺漏了什么.

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.

推薦答案

如果 C++11 是一個選項，那么您應該使用 隨機標頭 和 uniform_int_distrubution.正如 James 在評論中指出的那樣，使用 rand 和 % 有很多問題，包括偏差分布:

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 ;
}

如果您必須使用 rand 那么應該這樣做:

if you have to use rand then this should do:

rand() % 7 + 6

更新

使用 rand 的更好方法如下:

A better method using rand would be as follows:

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

我從 C 常見問題解答 中獲得了這個，并解釋了 如何我得到一定范圍內的隨機整數? 問題.

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

更新 2

Boost 也是一種選擇:

#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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