Re: what does this mean ?

"Doug Harrison [MVP]" <dsh@xxxxxxxx> wrote in message
news:jo00euo0aw6e.13puhb98z01aw.dlg@xxxxxxxxxx

On Fri, 8 Jul 2005 02:37:56 +1000, John Carson wrote:

By section 5.3.4/15 of the standard, the second form does value
initialisation, the details of which are described in section 8.5/5.

To illustrate, consider the following as compiled by VC++ 8.0 Beta 2:

struct Test
{
     int x, y;
};

int main()
{
     Test *ptr1 = new Test;
     cout << ptr1->x << ',' << ptr1->y << endl;

     Test *ptr2 = new Test();
     cout << ptr2->x << ',' << ptr2->y << endl;

     delete ptr1;
     delete ptr2;
     return 0;
}

x and y are uninitialised in the first case and zero initialised in
the second case. This is a change that occurred between the 1998
version of the standard and the 2003 version.

Don't you mean the 1998 and 2003 versions of VC++? :)

I am not sure how to interpret the :). The standard itself changed. Thus the 1998 section 8.5/5 says nothing about value initialisation. The 2003 version of section 8.5/5 reads as follows:

<standard>
To zero-initialize an object of type T means:
- if T is a scalar type (3.9), the object is set to the value of 0 (zero) converted to T;
- if T is a non-union class type, each nonstatic data member and each base-class subobject is zero-initialized;
- if T is a union type, the object's first named data member is zero-initialized;
- if T is an array type, each element is zero-initialized;
- if T is a reference type, no initialization is performed.

To default-initialize an object of type T means:
- if T is a non-POD class type (clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
- if T is an array type, each element is default-initialized;
- otherwise, the object is zero-initialized.

To value-initialize an object of type T means:
- if T is a class type (clause 9) with a user-declared constructor (12.1), then the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
- if T is a non-union class type without a user-declared constructor, then every non-static data member and base-class component of T is value-initialized;
- if T is an array type, then each element is value-initialized;
- otherwise, the object is zero-initialized
A program that calls for default-initialization or value-initialization of an entity of reference type is ill-formed. If T is a cv-qualified type, the cv-unqualified version of T is used for these definitions of zero-initialization, default-initialization, and value-initialization.
</standard>

Also, the zero-initialization occurs in the Test() case only if Test
is a POD type, which means, to a fair approximation, a type you can
write in C. In particular, it would not be done were Test defined as:

struct Test
{
  int x;
  std::string s;
};

While there's no user-defined ctor, the non-static std::string member
makes it a non-POD type, and "new Test()" would leave x uninitialized.

On my reading, your statement is in accordance with the 1998 standard, which said that new Test() would be default initialized. It is also what VC++ 8.0 Beta 2 does.

Under the 2003 standard, new Test() gives value initialization. Given your declaration of struct Test, the applicable item is:

- if T is a non-union class type without a user-declared constructor, then every non-static data member and base-class component of T is value-initialized;

Thus x is value initialized, which means, by the last item in the list, that it is zero-initialized.


--
John Carson

.

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