Question

typedef decltype(true ? (long&&)0 : (long&&)0) T;

What should T be?

According to gcc (4.7), it’s long. According to clang (trunk), it’s long&&. This difference is causing clang to fail to compile code that uses gcc 4.7’s libstdc++. Who is right?

UPDATE: As ildjarn points out, Clang is right, and as Richard Smith points out, the error libstdc++ is due to an error in the Standard. Here is the relevant GCC bug, and the relevant Defect Report.

Answer 1

Clang is right. N3337 §7.1.6.2/4:

The type denoted by decltype(e) is defined as follows:

• if e is an unparenthesized id-expression or an unparenthesized class member access, 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 operand of the decltype specifier is an unevaluated operand.

§5/6:

[ Note: An expression is an xvalue if it is:

• the result of calling a function, whether implicitly or explicitly, whose return type is an rvalue reference to object type,
• a cast to an rvalue reference to object type,
• a class member access expression designating a non-static data member of non-reference type in which the object expression is an xvalue, or
• a .* pointer-to-member expression in which the first operand is an xvalue and the second operand is a pointer to data member.

In general, the effect of this rule is that named rvalue references are treated as lvalues and unnamed rvalue references to objects are treated as xvalues; rvalue references to functions are treated as lvalues whether named or not. —end note ]

I was wary earlier that a literal 0 may somehow prevent this from qualifying as an object type in this context, but §3.9/8 clarifies things:

An object type is a (possibly cv-qualified) type that is not a function type, not a reference type, and not a void type.

The conditional operator doesn’t affect anything here – §5.16/4:

If the second and third operands are glvalues of the same value category and have the same type, the result is of that type and value category and it is a bit-field if the second or the third operand is a bit-field, or if both are bit-fields.

In this case both are of the same value category (xvalue), and xvalues are glvalues.