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c++c++17autodecltypetype-deduction

Type of variables in structured binding

#include <type_traits>

int main()
{
    int arr[1] = { 6 };

    auto& ref1 = arr[0];  
    static_assert( std::is_same_v<decltype( ref1 ), int&> ); //ok

    auto& [ ref2 ] = arr;
    static_assert( std::is_same_v<decltype( ref2 ), int> ); //ok
    static_assert( std::is_same_v<decltype( ref2 ), int&> ); //error
}

What is the consequential difference between identifiers ref1 and ref2 in that example? As I understand, ref2 in structure binding also has a reference type, but why does decltype indicate a non-referenced type for it?

Solution

  • decltype(e) behaves differently depending on what e is given as the argument. For the structured binding, decltype yields what follows, [dcl.type.simple]:

    For an expression e, the type denoted by decltype(e) is defined as follows:

    • if e is an unparenthesized id-expression naming a structured binding, decltype(e) is the referenced type as given in the specification of the structured binding declaration

    The referenced type for a structured binding declaration with an array type expression as the initializer is the type of the element [dcl.struct.bind]:

    If E is an array type with element type T, the number of elements in the identifier-list shall be equal to the number of elements of E. Each vi is the name of an lvalue that refers to the element i of the array and whose type is T; the referenced type is T. [ Note: The top-level cv-qualifiers of T are cv. — end note ]