Is declval() the same as (*(T*)nullptr)?

Question

Is declval() just a replacement for the old trick of (*(T*)NULL) to get an instance of T in a decltype without needing to worry about T\'s

Answer 1

No, declval() is not the same as (*(T*)nullptr). And decltype(expr.bar) is not the same as decltype((expr.bar)).

The former comparison compares expressions. The latter use of decltype inspects the expression and the former use of decltype inspects the declared type of expr.bar. So you have to paren your uses of the decltype operand to make a useful comparison of the types and you will find they are different.

struct A {};

struct B {
    A a;
};

// E1: B().a 
// E2: declval().a
// E3: (*(B*)0).a
// E4: ((B&&)(*(B*)0)).a

In these 4 expressions, all expressions have type A. E1 is a prvalue (in C++14 it is an xvalue. Some compilers probably will treat it as an xvalue even in their C++11 mode), E2 is an xvalue. E3 is an lvalue and E4 is an xvalue again.

// T1: decltype((*(B*)0).a)
// T2: decltype(((*(B*)0).a))

In these two types, the first decltype gives the type of the member named by the expression. The member has type A, so T1 is A. The second decltype yields the type of the expression, modified by & if the expression is an lvalue and modified by && if the expression is an xvalue. The expression is an lvalue, so T2 is A&.