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c++c++11decltype

Why decltype(a, b) is evaluated to a reference?

According to this answer, ref should be an int.

But for some reason it evaluated to int&, both in gcc and MSVC2015, while decltype(b) is correctly evaluated to just int. Why so?

int a = 1, b = 2;
decltype(a, b) ref; // ref is int&
decltype(b) var;    // var is int
Solution

  • a, b is an expression. According to decltype rules for expressions, if result of the expression is an lvalue, type is going to be deduced as T&

    7.1.6.2/4 Simple type specifiers [dcl.type.simple]
    For an expression e, the type denoted by decltype(e) is defined as follows:

    • if e is an unparenthesized id-expression or an unparenthesized class member access (5.2.5), decltype(e) is the type of the entity named by e. If there is no such entity, or if e names a set of overloaded functions, the program is ill-formed;
    • otherwise, if e is an xvalue, decltype(e) is T&&, where T is the type of e;
    • otherwise, if e is an lvalue, decltype(e) is T&, where T is the type of e;
    • otherwise, decltype(e) is the type of e.

    The confusing part about difference between "type of the entity named by e" and "type of e" is easy to understand with example:

    If some entity e is declared as int& e = x;, then later, in expression e, type of e is int, and type of the entity named by e is int&. In short, type of e drops reference qualifiers.