Return std::tuple containing const-reference in C 11-CodePudding

I have something like this (C 11)

std::tuple<const MyType&, bool> func()
{
   return std::make_tuple(some_internal_reference, true);
}

the problem is that in the caller I cannot declare:

const MyType& obj; // this does not compile of course
bool b;

std::tie(obj, b) = func();

An idea is to return the boolean as an output param and drop the tuple, but is there a better solution?

CodePudding user response：

Use std::get. It returns reference to stored element.

#include <iostream>
#include <tuple>

using MyType = int;
MyType some_internal_reference = 42;

std::tuple<const MyType&, bool> func()
{
    return { some_internal_reference, true };
}

int main()
{   
    auto ret = func();
    const MyType& obj = std::get<0>(ret);
    
    std::cout << "before change: " << obj << '\n';
    
    some_internal_reference = 7;

    std::cout << "after change: " << obj << '\n';
}

It prints

before change: 42
after change: 7

Note, as per @StoryTeller-UnslanderMonica comment, don't use std::make_tuple in this case. It stores a decay (naked type) copy.