C ++: Why decltype (* this) returns a link?

Question

template<typename T> struct foo{ void f(){ decltype(*this) a(*this); do_some_test(a); } T data; }; //compiler won't accept this 

In my interpretation, decltype should return a type a so that we can use it in the declaration. But google says in decltype(x) , if x is an lvalue, it will return T& , where T is a type of x.

But what did they plan to return the link? Also, what should I do to create an instance of the class with the same type as *this in the template?

Answer 1

decltype prints expression type if it does not apply to the variable, in which case it infers the type of this variable:

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

- if e is a non-spheronized identifier or disjoint access to a class member, decltype(e) is the type of object called e . If such an object is absent or if e calls a set of overloaded functions, the program is poorly formed;

- otherwise, if e is the value of x, decltype(e) is T&& , where T is the type of e ;

- otherwise, if e is an lvalue, decltype(e) is T& , where T is a type of e ;

- otherwise decltype(e) is type e .

§7.1.6.2 [dcl.type.simple]

Pointer dereferencing gives an lvalue, and therefore decltype prints a lvalue reference to the pointer type:

The unary operator * performs an indirect direction: the expression to which it is applied must be a pointer to the type of the object or a pointer to the type of the function, and the result is the value l that refers to the object or function of the expression.

§5.3.1 [expr.unary.op]

Therefore decltype(*p) , for some pointer p displays an lvalue reference to the type of the pointer.

If you want to get a pointer type from some pointer p , you can use:

 std::remove_pointer<decltype(p)>::type 

Or:

 std::remove_reference<decltype(*p)>::type 

Or, in your example, you can just say foo , an output type is not required.