can we do deep tie with a c++1y std::tie() -like function?

Question

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;
}

Answer 1

You can go:

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

std::forward_as_tuple works very similarly to std::tie, except that unlike it it will not reject std::tie(x, y) as an argument.

Answer 2

Maybe you're looking for std::tuple_cat:

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

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.

---

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.