one more template question and some typedef

[Tony Johansson]

Hello Experts!

I have the following Array template class see below.
I execute these three statements
statement 1: Array<int> x(5);
statement 2: cin >>x[1];
statement 3: Array<int>::element_type y = x[1];
but I can't understand the last one which is
Array<int>::element_type y = x[1];
We have a typedef T element_type; in the template Array class see below
and in my example I use int as the type parameter for T so
that would mean that it says
Array<int>::int y = x[1];
this is even more confusing for me.
If the Array template class for example had a static variable with name
number then you would
use Array<int>::number; to access this static variable.



template <typename T>
class Array
{
public:
Array(size_t=100); // size_t is predefined
Array(const Array<T>&);
Array<T>& operator=(const Array<T>&);
virtual ~Array();
virtual const T& operator[](int) const;
virtual T& operator[](int);
int size() const;

typedef T element_type;
protected:
size_t size_;
T* array;
};

Many thanks

//Tony

 

[Victor Bazarov]

I have the following Array template class see below.
I execute these three statements
statement 1: Array<int> x(5);
statement 2: cin >>x[1];
statement 3: Array<int>::element_type y = x[1];
but I can't understand the last one which is
Array<int>::element_type y = x[1];
We have a typedef T element_type; in the template Array class see below
and in my example I use int as the type parameter for T so
that would mean that it says
Array<int>::int y = x[1];
this is even more confusing for me.

No, it doesn't mean it says "Array<int>::int y = x[1];".
It means what it says,

Array<int>::element_type y = x[1];

or you can rewrite it to say

int y = x[1];

'element_type' does not mean 'int' by itself. It only means 'int' if
used _along_ with the class in which it's defined. For example, in
Array<string>, 'element_type' would mean 'string'. So, if you write

Array<string>::element_type

it would be the same as writing

string

If the Array template class for example had a static variable with name
number then you would
use Array<int>::number; to access this static variable.

Yes.

V

 

[Tony Johansson]

I have the following Array template class see below.
I execute these three statements
statement 1: Array<int> x(5);
statement 2: cin >>x[1];
statement 3: Array<int>::element_type y = x[1];
but I can't understand the last one which is
Array<int>::element_type y = x[1];
We have a typedef T element_type; in the template Array class see below
and in my example I use int as the type parameter for T so
that would mean that it says
Array<int>::int y = x[1];
this is even more confusing for me.

No, it doesn't mean it says "Array<int>::int y = x[1];".
It means what it says,

Array<int>::element_type y = x[1];

or you can rewrite it to say

int y = x[1];

'element_type' does not mean 'int' by itself. It only means 'int' if
used _along_ with the class in which it's defined. For example, in
Array<string>, 'element_type' would mean 'string'. So, if you write

Array<string>::element_type

it would be the same as writing

string

If the Array template class for example had a static variable with name
number then you would
use Array<int>::number; to access this static variable.

Yes.

V

I can't understand how Array<int>::element_type
is the same as int in my example.
or that Array<string>::element_type is the same as string
Some question that might help me to understand this.
How does the scope operation be understood here?
What meaning have T in the typedef T element_type;?
I want to understand the scope operatorn in such way that there is some
variable
in the Array template class that has name element_type.
The fully name of the Array template class when you have int as the
parameter is
Array<int>
Can you give me some more explanation about Array<int>::element_type
that might help me to understand this.

//Tony

 

[Victor Bazarov]

I have the following Array template class see below.
I execute these three statements
statement 1: Array<int> x(5);
statement 2: cin >>x[1];
statement 3: Array<int>::element_type y = x[1];
but I can't understand the last one which is
Array<int>::element_type y = x[1];
We have a typedef T element_type; in the template Array class see below
and in my example I use int as the type parameter for T so
that would mean that it says
Array<int>::int y = x[1];
this is even more confusing for me.

No, it doesn't mean it says "Array<int>::int y = x[1];".
It means what it says,

Array<int>::element_type y = x[1];

or you can rewrite it to say

int y = x[1];

'element_type' does not mean 'int' by itself. It only means 'int' if
used _along_ with the class in which it's defined. For example, in
Array<string>, 'element_type' would mean 'string'. So, if you write

Array<string>::element_type

it would be the same as writing

string

If the Array template class for example had a static variable with name
number then you would
use Array<int>::number; to access this static variable.

Yes.

V

I can't understand how Array<int>::element_type
is the same as int in my example.

Try a simple substitution for this struct template:

template<class T> struct A { typedef T type; };

If I instantiate A<int>, what happens? It's as if I declare a type

struct A<int> /* incorrect syntax, but will do */ {
typedef int type;
};

(anywhere 'T' was used inside the 'A' definition, I put 'int'). Now, what
does 'A<int>::type' stand for? It's typedef'ed to be the same as 'int'.

or that Array<string>::element_type is the same as string
Some question that might help me to understand this.
How does the scope operation be understood here?

It's not "the scope operation". It's what you may call a "class-wide
member access operation". A typedef inside a class is a class-wide
member. It has a name which you can access using the '::'. The name
of that member in my example is 'type'. A way to access it is to supply
the name of the _class_ with the scope resolution operator:

A<int> :: type
^^^^^^ ^^ ^^^^
Name of the class Scope resolution Member name

What meaning have T in the typedef T element_type;?

Just like with any other typedef, 'element_type' is now the synonym for
whatever 'T' is. For any particular instantiation of the template, 'T'
means something specific. So, in any particular instantiation of your
'Array' template, 'element_type' will mean something in particular.

I want to understand the scope operatorn in such way that there is some
variable
in the Array template class that has name element_type.

I don't understand the sentence above.

The fully name of the Array template class when you have int as the
parameter is
Array<int>

It's called "instantiation", not "fully name of the Array template class".

Can you give me some more explanation about Array<int>::element_type
that might help me to understand this.

See above. And, yes, I can give you more if you ask more specific
questions.

V

 

[Tony Johansson]

Tony Johansson wrote:

I have the following Array template class see below.
I execute these three statements
statement 1: Array<int> x(5);
statement 2: cin >>x[1];
statement 3: Array<int>::element_type y = x[1];
but I can't understand the last one which is
Array<int>::element_type y = x[1];
We have a typedef T element_type; in the template Array class see below
and in my example I use int as the type parameter for T so
that would mean that it says
Array<int>::int y = x[1];
this is even more confusing for me.

No, it doesn't mean it says "Array<int>::int y = x[1];".
It means what it says,

Array<int>::element_type y = x[1];

or you can rewrite it to say

int y = x[1];

'element_type' does not mean 'int' by itself. It only means 'int' if
used _along_ with the class in which it's defined. For example, in
Array<string>, 'element_type' would mean 'string'. So, if you write

Array<string>::element_type

it would be the same as writing

string


If the Array template class for example had a static variable with name
number then you would
use Array<int>::number; to access this static variable.

Yes.

V

I can't understand how Array<int>::element_type
is the same as int in my example.

Try a simple substitution for this struct template:

template<class T> struct A { typedef T type; };

If I instantiate A<int>, what happens? It's as if I declare a type

struct A<int> /* incorrect syntax, but will do */ {
typedef int type;
};

(anywhere 'T' was used inside the 'A' definition, I put 'int'). Now, what
does 'A<int>::type' stand for? It's typedef'ed to be the same as 'int'.

or that Array<string>::element_type is the same as string
Some question that might help me to understand this.
How does the scope operation be understood here?

It's not "the scope operation". It's what you may call a "class-wide
member access operation". A typedef inside a class is a class-wide
member. It has a name which you can access using the '::'. The name
of that member in my example is 'type'. A way to access it is to supply
the name of the _class_ with the scope resolution operator:

A<int> :: type
^^^^^^ ^^ ^^^^
Name of the class Scope resolution Member name

What meaning have T in the typedef T element_type;?

Just like with any other typedef, 'element_type' is now the synonym for
whatever 'T' is. For any particular instantiation of the template, 'T'
means something specific. So, in any particular instantiation of your
'Array' template, 'element_type' will mean something in particular.

I want to understand the scope operatorn in such way that there is some
variable
in the Array template class that has name element_type.

I don't understand the sentence above.

The fully name of the Array template class when you have int as the
parameter is
Array<int>

It's called "instantiation", not "fully name of the Array template class".

Can you give me some more explanation about Array<int>::element_type
that might help me to understand this.

See above. And, yes, I can give you more if you ask more specific
questions.

V

I think I start to understand it now. I just check with you Victor hope you
read this.
Assume there exist a valid type x that I use to get a specialisation of the
template Array.
If you instantiate the template class Array and pass type x as parameter you
get the instantiation name Array<x>.
If I want to access a class-wide member in the Array template I use the ::
In my case I have a class-wide member with name element_type so I access
this element_type by giving statement
Array<x>::element_type
this whole expession can always be substituted with the type that this
template was called with in my example here it's x because we have this
typedef T element_type

Does my thinking seems to be about correct?

//Tony

 

[Victor Bazarov]

[..]
Assume there exist a valid type x that I use to get a specialisation of the
template Array.
If you instantiate the template class Array and pass type x as parameter you
get the instantiation name Array<x>.
If I want to access a class-wide member in the Array template I use the ::
In my case I have a class-wide member with name element_type so I access
this element_type by giving statement
Array<x>::element_type
this whole expession can always be substituted with the type that this
template was called with in my example here it's x because we have this
typedef T element_type

Does my thinking seems to be about correct?

Yes... I am trying to find some kind of hidden pothole in that or
some additional condition that has to be satisfied, but it all seems
to be fine.

I recommend looking at the standard library containers and how they
define 'value_type' members. That's a common requirement for all of
them, and you should be able to rely on that, for example.

V

 

[neerajk]

"Array<int>::element_type" is really "typename
Array<int>::element_type"

and in fact the newer compilers issue warnings asking you to write it
this way.

It is a hint to the compiler to look for a type definition rather than
a data/function member.

 

[Victor Bazarov]

"Array<int>::element_type" is really "typename
Array<int>::element_type"

and in fact the newer compilers issue warnings asking you to write it
this way.

Really? Which ones? Please post a complete compilable piece of code
that illustrates what you're talking about.

V