问题描述

我正在编写一个自定义迭代器，它在取消引用时会返回一个引用元组.由于元组本身是短暂的，我认为我不能从 operator*() 返回引用.我认为我的迭代器在语义上是有意义的，因为它具有引用语义，即使 operator* 返回一个值.

I'm writing a custom iterator that, when dereferenced returns a tuple of references. Since the tuple itself is ephemeral, I don't think I can return a reference from operator*(). I think my iterator makes sense semantically, since it has reference semantics, even though operator* returns a value.

问题是，当我尝试调用 std::swap 时(或者更确切地说，当 std::sort 调用时)，如下所示，我收到错误，因为交换需要 l 值.这个问题有简单的解决方法吗?

The issue is, when I try to call std::swap (or rather, when std::sort does), like below, I get errors because the swap expects l-values. Is there an easy fix to this problem?

#include <vector>

class test {
  public:
  test()
    :v1(10), v2(10)
  {}

  class iterator {
    public:
    iterator(std::vector<int>& _v1,
             std::vector<int>& _v2)
      :v1(_v1), v2(_v2){}

    std::tuple<int&, int&> operator*(){
      return std::tuple<int&, int&>{v1[5], v2[5]};
    }
    std::vector<int>& v1;
    std::vector<int>& v2;
  };

  std::vector<int> v1, v2;
};



int main(){
  test t;
  //error, Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/../lib/c++/v1/type_traits:3003:1: note: candidate function [with _Tp = std::__1::tuple<int &, int &>] not viable: expects an l-value for 1st argument

  //deep within the bowels of std::sort ...
  std::swap(*(test::iterator(t.v1, t.v2)),
            *(test::iterator(t.v1, t.v2)));
}

推荐答案

在极少数情况上可能希望得到对临时的左值引用.这很容易通过与 std::move:

On rare occasions it may be desirable to get an lvalue reference to temporary. This easily achieved with a cast opposite to std::move:

template <typename T>
T & stay(T && t) { return t; }

用法:

std::swap(stay(foo()), stay(bar()));

<小时>

正如您已经说过的，如果您无法更改调用站点，您最好的选择可能是编写自己的引用包装器并为此使用 ADL:

As you already said, if you can't change the call site, your best option may be to write your own reference wrapper and use ADL for that:

namespace detail
{
    struct IntRefPair
    {
        int & a, & b;
        IntRefPair(int & x, int & y) : a(x), b(y) {}
    };

    void swap(IntRefPair && lhs, IntRefPair && rhs)
    { 
        std::swap(lhs.a, rhs.a);
        std::swap(lhs.b, rhs.b);
    }
}

// ...

IntRefPair operator*() { return IntRefPair(v1[5], v2[5]); } }

这篇关于交换引用的临时元组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编程学习网！