如何實現一個自己的發生器與std分佈一起使用

Question

我想用一個隨機數生成器來實現一些來自C++ 11 <random>庫的發行版。

是否有我的生成器類必須遵守的接口，以便它的一個實例可以傳遞給分配的operator()方法？

還是有基礎班，我必須從中派生我的班？

Answer 1

你的接口有相匹配的UniformRandomBitGenerator概念：

+-----------------+--------------+-------------------------------------------------------------------------------------------------------+ 
| Expression  | Return type | Requirements                       | 
+-----------------+--------------+-------------------------------------------------------------------------------------------------------+ 
| G::result_type | T   | T is an unsigned integer type                   | 
| G::min()  | T   | Returns the smallest value that G's operator() may return. The value is strictly less than G::max(). | 
| G::max()  | T   | Returns the largest value that G's operator() may return. The value is strictly greater than G::min() | 
| g()    | T   | Returns a value in the closed interval [G::min(), G::max()]. Has amortized constant complexity.  | 
+-----------------+--------------+-------------------------------------------------------------------------------------------------------+ 

您不必從基類派生的綁定發生在編譯時。

這裏是發電機的接口相匹配的玩具例子：

#include <random> 
#include <iostream> 

class MyGen 
{ 
public: 
    using result_type = int; 
    static constexpr result_type min() { return 0; } 
    static constexpr result_type max() { return 99; } 
    result_type operator()() { return rand() % 100; } 
}; 

int main() 
{ 
    std::uniform_int_distribution<int> dist{1, 6}; 
    MyGen gen; 

    for(int i = 0; i < 10; i++) 
    std::cout << dist(gen) << '\n'; 

    return 0; 
}