我们可以用一个c ++ 1y std :: tie（）类函数做深关系吗？ [英] can we do deep tie with a c++1y std::tie() -like function?

问题描述

有一种方法可以在c ++ 11 / 1y中编写一个 std :: tie 的变体，它深深地融入了一个元组。也就是说， tie（（x，y），z）= make_tuple（make_tuple（1,2），3）绑定 y，z 到 1,2和3 ，如以下示例所示。这将是很好。感谢。

Is there a way to write a variant of std::tie in c++11/1y that ties deeply into a tuple. That is, one in which tie((x,y),z) = make_tuple(make_tuple(1,2),3) binds x, y, z to 1, 2 and 3, respectively as in the following example. It would be nice. Thanks.

#include <tuple>
#include <iostream>
using namespace std;

int main() {
  int x, y ,z;
  auto t = make_tuple(1,2);
  std::tie(y,x)= t;
  //std::tie((x,y),z) = make_tuple(t,3); //not working
  cout << x << y << z << endl;
  return 0;
}

推荐答案

std :: tuple_cat ：

std::tie(x,y,z) = std::tuple_cat(t, make_tuple(3)); 

您可以将 tuples 元组，以避免处理嵌套元组。我认为使嵌套元组变平的解决方案会更复杂。

You can string together tuples as one long tuple, to avoid dealing with nested tuples. I think the solution to flattening nested tuples would be more complex.

只是为了澄清 std :: tie 工作（我想）。 std :: tie 从其参数构造一个包含左值引用的元组。使用赋值运算符时，将执行复制分配。 std :: tie（（x，y），z）不会做你想的。您将（x，y）赋给逗号运算符，其中x被丢弃。没有魔法继续，其中嵌套由括号确定。如果 std :: tie 的参数之一是元组，那么相应的参数也应该是元组。即： std :: tie（tuple，3）= std :: make_tuple（std :: make_tuple（1,2），3）。但是这不是你想要的，这是我的建议来自于，因为它似乎不是你的意图是扁平嵌套的元组。

Just to make a clarification on how std::tie works (I think). std::tie constructs a tuple of lvalue references from its arguments. When you use the assignment operator, copy assignments are performed. std::tie((x,y),z) doesn't do what you think. You're subjecting (x,y) to the comma operator, where x is discarded. There is no magic going on, where nesting is determined by parentheses. If one of the arguments to std::tie is a tuple, then the corresponding argument should be a tuple as well. i.e.: std::tie(tuple, 3) = std::make_tuple(std::make_tuple(1, 2), 3). However this is not what you want, which is where my suggestion comes from, because it doesn't seem like your intention is to flatten a nested tuple.

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