问题与std :: reference_wrapper [英] Issue with std::reference_wrapper

问题描述

使用以下代码清楚地显示问题：

The issue is clear with the following code:

#include <functional>
#include <iostream>
#include <vector>

int main() {
  //std::vector<int> a, b;
  int a = 0, b = 0;
  auto refa = std::ref(a);
  auto refb = std::ref(b);
  std::cout << (refa < refb) << '\n';
  return 0;
}

如果我使用注释的 std :: vector< int> a，b; 而不是 int a = 0，b = 0; ，那么代码不能在GCC 5.1，clang 3.6 ，或MSVC'13。在我看来， std :: reference_wrapper< std :: vector< int>> 可隐式转换为 std :: vector< int& 这是LessThanComparable，因此它应该是LessThanComparable本身。

If I use the commented std::vector<int> a, b; instead of int a = 0, b = 0;, then the code does not compile on any of GCC 5.1, clang 3.6, or MSVC'13. In my opinion, std::reference_wrapper<std::vector<int>> is implicitly convertible to std::vector<int>& which is LessThanComparable, and thus it should be LessThanComparable itself. Could someone explain this to me?

推荐答案

问题是非成员 operator std :: vector 是一个函数模板：

The issue is that the non-member operator< for std::vector is a function template:

template< class T, class Alloc >
bool operator<( const vector<T,Alloc>& lhs,
                const vector<T,Alloc>& rhs );

在执行模板类型扣除时，隐式转换不 .arg.explicit]强调if：

Implicit conversions are not considered when doing template type deduction here, [temp.arg.explicit] emphasis on if:

隐式转换（第4条）将对函数参数执行，以将其转换为类型
对应的函数参数 if ，参数类型不包含参与模板参数扣除的
的模板参数。

Implicit conversions (Clause 4) will be performed on a function argument to convert it to the type of the corresponding function parameter if the parameter type contains no template-parameters that participate in template argument deduction.

但在这种情况下，参数类型确实参与了扣除。这就是为什么它不能找到。如果我们写我们自己的非 -template 运算符：

But in this case, the parameter type does participate in deduction. That's why it can't be found. Had we written our own non-template operator<:

bool operator<(const std::vector<int>& lhs, const std::vector<int>& rhs)
{
    return true;
}

您的代码将按预期工作。要使用通用的，你必须明确地拉出引用：

Your code would work as expected. To use the generic one though, you will have to explicitly pull out the reference:

std::cout << (refa.get() < refb.get()) << '\n';

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