Re: class scope typedef through dot operator

From:
red floyd <no.spam.here@its.invalid>
Newsgroups:
comp.lang.c++
Date:
Wed, 27 Oct 2010 19:19:36 -0700
Message-ID:
<iaamjr$n8k$1@news.eternal-september.org>
On 10/27/2010 7:04 PM, Jonathan Lee wrote:

On Oct 27, 8:12 pm, Joshua Maurice<joshuamaur...@gmail.com> wrote:

On Oct 27, 3:49 pm, Jonathan Lee<jonathan.lee....@gmail.com> wrote:

So I think I've seen somewhere that you can access a class-scoped
typedef via the dot operator, like other members. For example,

   std::vector<double> v;
   /* do something with v.value_type */

The thing is, I can't seem to actually do anything with it. Can't take
the size of it, can't construct a variable, etc. But the compiler
errors suggest it's actually figuring out that I *mean*
std::vector<double>::value_type. What can you actually accomplish with
this notation? Or is it even valid?


value_type is a type name, not a data member or function member. The
dot operator is used to access data members and function members. The
scope resolution operator "::" is used for things like nested type
names. The compiler is telling you exactly that. The syntax to use
value_type is "std::vector<double>::value_type". Ex:
   #include<vector>
   std::vector<double>::value_type some_variable;

What exactly is your question?


Sorry, I misremembered the example. I thought it was for typedefs,
but it was actually for enums. i.e,

class A {
public:
   enum dir { left, right };
};

int main() {
   A a;
   A* b =&a;

   std::cout<< static_cast<int>(a.left)<< std::endl;
   std::cout<< static_cast<int>(b->left)<< std::endl;
}


left and right should still be accessed using the scope operator:

A::left, A::right

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