在C ++ 17/C ++ 2a中的编译时散列类型 [英] Hashing types at compile-time in C++17/C++2a

问题描述

考虑以下代码:

#include <iostream>
#include <type_traits>

template <class T>
constexpr std::size_t type_hash(T) noexcept 
{
    // Compute a hash for the type
    // DO SOMETHING SMART HERE
}

int main(int argc, char* argv[])
{
    auto x = []{};
    auto y = []{};
    auto z = x;
    std::cout << std::is_same_v<decltype(x), decltype(y)> << std::endl; // 0
    std::cout << std::is_same_v<decltype(x), decltype(z)> << std::endl; // 1
    constexpr std::size_t xhash = type_hash(x);
    constexpr std::size_t yhash = type_hash(y);
    constexpr std::size_t zhash = type_hash(z);
    std::cout << (xhash == yhash) << std::endl; // should be 0
    std::cout << (yhash == zhash) << std::endl; // should be 1
    return 0;
}

我希望type_hash函数在编译时返回该类型唯一的哈希键.是否可以在C ++ 17或C ++ 2a中做到这一点(理想情况下仅依赖于标准，而无需依赖编译器内部函数)?

I would like the type_hash function to return a hash key unique to the type, at compile-time. Is there a way to do that in C++17, or in C++2a (ideally only relying on the standard and without relying compiler intrinsics)?

推荐答案

我不知道一种为哈希获取std::size_t的方法.

I don't know a way to obtain a std::size_t for the hash.

但是，如果您接受指向某个东西的指针，也许可以在模板类中获取静态成员的地址.

But if you accept a pointer to something, maybe you can take the address of a static member in a template class.

我的意思是……如下

#include <iostream>
#include <type_traits>

template <typename>
struct type_hash
 {
   static constexpr int          i     { };
   static constexpr int const *  value { &i };
 };

template <typename T>
static constexpr auto type_hash_v = type_hash<T>::value;


int main ()
 {
   auto x = []{};
   auto y = []{};
   auto z = x;
   std::cout << std::is_same_v<decltype(x), decltype(y)> << std::endl; // 0
   std::cout << std::is_same_v<decltype(x), decltype(z)> << std::endl; // 1
   constexpr auto xhash = type_hash_v<decltype(x)>;
   constexpr auto yhash = type_hash_v<decltype(y)>;
   constexpr auto zhash = type_hash_v<decltype(z)>;
   std::cout << (xhash == yhash) << std::endl; // should be 0
   std::cout << (xhash == zhash) << std::endl; // should be 1
 } // ...........^^^^^  xhash, not yhash

如果您真的想将type_hash作为函数，我想您可以简单地创建一个返回所接收类型的type_hash_v<T>的函数.

If you really want type_hash as a function, I suppose you could simply create a function that return the type_hash_v<T> of the type received.

这篇关于在C ++ 17/C ++ 2a中的编译时散列类型的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！