(nessun oggetto)

[Graziano Servizi]

I'm wondering in reading the 14.7.1.4 example of the standard, about
template implicit instantiation: there it is found the following code

template<class T> struct Z {
void f();
void g();
};
void h() {
Z<int> a; // instantiation of class Z<int> required
Z<char>* p; // instantiation of class Z<char> not required
Z<double>* q; // instantiation of class Z<double> not required
a.f(); // instantiation of Z<int>::f() required
p->g(); // instantiation of class Z<char> required, and
// instantiation of Z<char>::g() required
}

followed by this comment:

"Nothing in this example requires class

Z<double>, Z<int>::g(), or Z<char>::f()

to be implicitly instantiated."


Almost all is clear enaugh to me, but my question is: does the line

p->g();

imply allocation of the pointer p to hold an address for the member
function g ONLY?

I didn't find any new operator before.

And what about this pointer at the end of the expression? Is it
implicitly released as well as it was "implicitly allocated"?

Or all I wrote up to now is meaningless? ....

Thanks in advance.

GS

[Jonathan Wakely]

This is not the right list to ask questions about the standard, this
list is for help using GCC.

Post by Graziano Servizi
I'm wondering in reading the 14.7.1.4 example of the standard, about
template implicit instantiation: there it is found the following code
template<class T> struct Z {
void f();
void g();
};
void h() {
Z<int> a; // instantiation of class Z<int> required
Z<char>* p; // instantiation of class Z<char> not required
Z<double>* q; // instantiation of class Z<double> not required
a.f(); // instantiation of Z<int>::f() required
p->g(); // instantiation of class Z<char> required, and
// instantiation of Z<char>::g() required
}
"Nothing in this example requires class
Z<double>, Z<int>::g(), or Z<char>::f()
to be implicitly instantiated."
Almost all is clear enaugh to me, but my question is: does the line
p->g();
imply allocation of the pointer p to hold an address for the member function
g ONLY?
I didn't find any new operator before.
And what about this pointer at the end of the expression? Is it implicitly
released as well as it was "implicitly allocated"?
Or all I wrote up to now is meaningless? ....

Yes, I think it's meaningless. There is no allocation or deallocation
in the example, it is only talking about template instantiation.

If the code was executed the expression p->g() would probably crash
because the pointer is uninitialized.