typename D<T>::E e; // this->E would not be valid syntax

hi all:
I reading "C++ Templates: The Complete Guide "
Part II: Templates in Depth
he write:

template<typename T>
class B {
public:
enumE{e1=6,e2=28,e3=496};
virtual void zero(E e = e1);
virtual void one(E&);
};

template<typename T>
class D : public B<T{
public:
void f() {
typename D<T>::E e; // this->E would not be valid syntax
this->zero(); // D<T>::zero() would inhibit virtuality
one(e); // one is dependent because its argument
} // is dependent
};

who can tell me ,"this->E would not be valid syntax "
why?

* lizhuo:

Quote:

>
[with E a type]
who can tell me ,"this->E would not be valid syntax "
why?

this->SomeType is never valid syntax, because an object can't contain a
type: a class can contain a type, an object (instance of the class) can't.

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

Alf P. Steinbach wrote:

Quote:

* lizhuo:

Quote:

>>
>[with E a type]
>who can tell me ,"this->E would not be valid syntax "
>why?

>
this->SomeType is never valid syntax, because an object can't contain
a type: a class can contain a type, an object (instance of the class)
can't.

struct SomeType { int a; };
struct OtherType { double a; };
struct blah : SomeType, OtherType { void foo(); };

void blah::foo() {
this->SomeType ::a;
// ^^^^^^^^^^^^^^
}

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

lizhuo wrote:

Quote:

hi all:
I reading "C++ Templates: The Complete Guide "
Part II: Templates in Depth
he write:
>
template<typename T>
class B {
public:
enumE{e1=6,e2=28,e3=496};
virtual void zero(E e = e1);
virtual void one(E&);
};
>
template<typename T>
class D : public B<T{
public:
void f() {
typename D<T>::E e; // this->E would not be valid syntax
this->zero(); // D<T>::zero() would inhibit virtuality
one(e); // one is dependent because its argument
} // is dependent
};
>
who can tell me ,"this->E would not be valid syntax "
why?

The operator-is called the member selection operator. It is used to
select a member (non-static method or variable) of an object. Thus
this->E is invalid, as E is not the name of a method or variable of
class D. As Victor pointed out, this->B::one would be legal, as B:: in
this expression is used to denote the name scope of class B for looking
up the member one (which would mean that you want to invoke the method
'one' as it is defined for class B).

Regards,
Stuart

* Victor Bazarov:

Quote:

Alf P. Steinbach wrote:

Quote:

>* lizhuo:

Quote:

>>[with E a type]
>>who can tell me ,"this->E would not be valid syntax "
>>why?

>this->SomeType is never valid syntax, because an object can't contain
>a type: a class can contain a type, an object (instance of the class)
>can't.

>
struct SomeType { int a; };
struct OtherType { double a; };
struct blah : SomeType, OtherType { void foo(); };
>
void blah::foo() {
this->SomeType ::a;
// ^^^^^^^^^^^^^^
}

:-) Yeah yeah, but, for other readers, that's "this->SomeType::a",
referring to a non-type.

Cheerio,

- Alf

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

typename D<T>::E e; // this->E would not be valid syntax

re: typename D<T>::E e; // this->E would not be valid syntax

re: typename D<T>::E e; // this->E would not be valid syntax

re: typename D<T>::E e; // this->E would not be valid syntax

re: typename D<T>::E e; // this->E would not be valid syntax

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