What is the difference between assigning to std::tie and tuple of references?

Question

I am a bit puzzled by the following tuple business:

int testint = 1;
float testfloat = .1f;
std::tie( testint, testfloat ) = std::make_tuple( testint, testfloat );
std::tuple<int&, float&> test = std::make_tuple( testint, testfloat );

With std::tie it works, but assigning directly to the tuple of references doesn't compile, giving

"error: conversion from ‘std::tuple<int, float>’ to non-scalar type ‘std::tuple<int&, float&>’ requested"

or

"no suitable user-defined conversion from std::tuple<int, float> to std::tuple<int&, float&>"

Why? I double checked with the compiler if it's really the same type that is being assigned to by doing this:

static_assert( std::is_same<decltype( std::tie( testint, testfloat ) ), std::tuple<int&, float&>>::value, "??" );

Which evaluates as true.

I also checked online to see if it maybe was the fault of msvc, but all compilers give the same result.

Answer 1

Both make_tuple and tie will deduce the returned type by arguments. But tie will make a lvalue reference type according to deduced type and make_tuple will make an actual tuple.

std::tuple<int&, float&> a = std::tie( testint, testfloat );

std::tuple<int , float > b = std::make_tuple( testint, testfloat );

 

The goal of tie is making a temporary tuple to avoid temporary copies of tied objects, the bad effect is, you can not return a tie if entry objects are local temporary.