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Home/ Questions/Q 8180667
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Editorial Team
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Editorial Team
Asked: June 7, 20262026-06-07T00:15:43+00:00 2026-06-07T00:15:43+00:00

This function works fine: std::string get_str() { return std::get<0>( make_tuple(std::string(hi)) ); } But if

  • 0

This function works fine:

std::string get_str() {
  return std::get<0>( make_tuple(std::string("hi")) );
}

But if you try to do the same thing with a trailing return type defined by decltype, the function returns a dangling rvalue reference:

auto get_str() -> decltype( std::get<0>( make_tuple(std::string("hi")) ) ) {
  return std::get<0>( make_tuple(std::string("hi")) );
}

I have a cool application of trailing return type where I’d like to use std::get. Unfortunately, the return type of std::get is an rvalue reference in this case, so decltype is just doing its job…

Do you know of a way to use the trailing return type and decltype but avoid the dangling rvalue reference?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-07T00:15:44+00:00Added an answer on June 7, 2026 at 12:15 am

    You could remove the reference using the standard remove_reference type trait:

    #include <string>
    #include <tuple>
    #include <type_traits>
    
    auto get_str() ->
        std::remove_reference<
            decltype( std::get<0>( make_tuple( std::string("hi") ) ) )
        >::type
    {
        return std::get<0>( make_tuple( std::string("hi") ) );
    }
    
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