std::make_tuple 不做引用 [英] std::make_tuple doesn't make references

问题描述

我一直在尝试将 std::tuple 与参考文献结合使用:

I've been experimenting with std::tuple in combination with references:

#include <iostream>
#include <tuple>

int main() {
  int a,b;
  std::tuple<int&,int&> test(a,b);
  std::get<0>(test) = 1;
  std::get<1>(test) = 2;
  std::cout << a << ":" << b << std::endl;

  // doesn't make ref, not expected
  auto test2 = std::make_tuple(a,b);
  std::get<0>(test2) = -1;
  std::get<1>(test2) = -2;
  std::cout << a << ":" << b << std::endl;

  int &ar=a;
  int &br=b;
  // why does this not make a tuple of int& references? can we force it to notice?
  auto test3 = std::make_tuple(ar,br);
  std::get<0>(test3) = -1;
  std::get<1>(test3) = -2;
  std::cout << a << ":" << b << std::endl;
}

在这里的三个示例中，前两个按预期工作.然而第三个没有.我期望 auto 类型 (test3) 与 test 的类型相同(即 std::tuple<int&,int&>).

Of the three examples here the first two work as expected. The third one however does not. I was expecting the auto type (test3) to be the same as the type of test (i.e. std::tuple<int&,int&>).

似乎 std::make_tuple 不能自动生成引用的元组.为什么不?除了自己明确构建这种类型的东西之外，我还能做些什么来实现这一点?

It seems that std::make_tuple can't automatically make tuples of references. Why not? What can I do to make this the case, other than explicitly constructing something of that type myself?

(编译器是 g++ 4.4.5，使用 4.5 不会改变它)

(Compiler was g++ 4.4.5, using 4.5 doesn't change it)

推荐答案

试试forward_as_tuple:

auto test3 = std::forward_as_tuple(ar,br);

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