How do I value-initialize a Type* pointer using Type()-like syntax?

Question

Variables of built-in types can be value-initialized like this:

int var = int();

this way I get the default value of int without hardcoding the zero in my code.

However if I try to do similar stuff for a pointer:

int* ptr = int*();

the compiler (Visual C++ 10) refuses to compile that (says type int unexpected).

How do I value-initialize a pointer in similar manner?

Answer 1

How do I value-initialize a Type* pointer using Type()-like syntax?

You cannot. The syntax T() is defined in 5.2.3/1,2 (C++03, slightly different wording in C++11 FDIS). In particular the second paragraph states:

The expression T(), where T is a simple-type-specifier (7.1.5.2) for a non-array complete object type or the (possibly cv-qualified) void type, creates an rvalue of the specified type, which is value-initialized (8.5);

That means that int(), will create an rvalue of type int and value-initialize it. Now the problem is that int* is not a simple-type-specifier, but rather an elaborated-type-specifier. The definition of simple-type-specifier in the grammar is:

  simple-type-specifier:
    ::opt nested-name-specifieropt type-name
    ::opt nested-name-specifier template template-id
    char
    wchar_t
    bool
    short
    int
    long
    signed
    unsigned
    float
    double
    void

With type-name being defined as:

  type-name:
    class-name
    enum-name
    typedef-name

This is what makes the proposed solutions work. The creation of the typedef (either directly or through the template) creates a type-name (third type) that can be used as a simple-type-specifier (first type).