Accessing a variable using namespace and a template class

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I can't access the variable width.

The A.h file:

namespace AN{

    template <typename T> class A{

    public:

        unsigned int width; #The variable

        ...

    }

}

The B.cpp file:

#include "A.h"

using namespace AN;

namespace BN{

    bool something(){

        unsigned int * w = AN::&width;

    }

}

I tried also AN::A::&width but it doesn't work as well.

edited Jan 3 at 16:16

François Andrieux

16.5k32950

asked Jan 3 at 16:03

Giwrgos Kostopoulos

AN::&width makes no sense. width is not a member of AN, but of AN::A<SomeType>. You need an instance of class A<SomeType>, then you can refer to the data member of that instance.

– Igor Tandetnik
Jan 3 at 16:06

width is a non-static member variable. So you need an instance of A in order to access it's variable. AN::A<int> a; unsigned int* w = &a.width;

– john
Jan 3 at 16:06

unsigned int * w = AN::A<unsigned int>::&width; .. I tried it but it is stays that width is undefined..

– Giwrgos Kostopoulos
Jan 3 at 16:08

if i make static would it work?

– Giwrgos Kostopoulos
Jan 3 at 16:09

@GiwrgosKostopoulos No, you need a variable of type A, AN::A<unsigned int> a; unsigned int * w = &a.width;. I think really need you need to learn some basic C++ concepts.

– john
Jan 3 at 16:09

|
show 3 more comments

I can't access the variable width.

The A.h file:

namespace AN{

    template <typename T> class A{

    public:

        unsigned int width; #The variable

        ...

    }

}

The B.cpp file:

#include "A.h"

using namespace AN;

namespace BN{

    bool something(){

        unsigned int * w = AN::&width;

    }

}

I tried also AN::A::&width but it doesn't work as well.

edited Jan 3 at 16:16

François Andrieux

16.5k32950

asked Jan 3 at 16:03

Giwrgos Kostopoulos

AN::&width makes no sense. width is not a member of AN, but of AN::A<SomeType>. You need an instance of class A<SomeType>, then you can refer to the data member of that instance.

– Igor Tandetnik
Jan 3 at 16:06

width is a non-static member variable. So you need an instance of A in order to access it's variable. AN::A<int> a; unsigned int* w = &a.width;

– john
Jan 3 at 16:06

unsigned int * w = AN::A<unsigned int>::&width; .. I tried it but it is stays that width is undefined..

– Giwrgos Kostopoulos
Jan 3 at 16:08

if i make static would it work?

– Giwrgos Kostopoulos
Jan 3 at 16:09

@GiwrgosKostopoulos No, you need a variable of type A, AN::A<unsigned int> a; unsigned int * w = &a.width;. I think really need you need to learn some basic C++ concepts.

– john
Jan 3 at 16:09

|
show 3 more comments

I can't access the variable width.

The A.h file:

namespace AN{

    template <typename T> class A{

    public:

        unsigned int width; #The variable

        ...

    }

}

The B.cpp file:

#include "A.h"

using namespace AN;

namespace BN{

    bool something(){

        unsigned int * w = AN::&width;

    }

}

I tried also AN::A::&width but it doesn't work as well.

edited Jan 3 at 16:16

François Andrieux

16.5k32950

asked Jan 3 at 16:03

Giwrgos Kostopoulos

I can't access the variable width.

The A.h file:

namespace AN{

    template <typename T> class A{

    public:

        unsigned int width; #The variable

        ...

    }

}

The B.cpp file:

#include "A.h"

using namespace AN;

namespace BN{

    bool something(){

        unsigned int * w = AN::&width;

    }

}

I tried also AN::A::&width but it doesn't work as well.

c++ namespaces

edited Jan 3 at 16:16

François Andrieux

16.5k32950

asked Jan 3 at 16:03

Giwrgos Kostopoulos

edited Jan 3 at 16:16

François Andrieux

16.5k32950

asked Jan 3 at 16:03

Giwrgos Kostopoulos

edited Jan 3 at 16:16

François Andrieux

16.5k32950

edited Jan 3 at 16:16

François Andrieux

16.5k32950

edited Jan 3 at 16:16

François Andrieux

16.5k32950

asked Jan 3 at 16:03

Giwrgos Kostopoulos

asked Jan 3 at 16:03

Giwrgos Kostopoulos

asked Jan 3 at 16:03

Giwrgos Kostopoulos

AN::&width makes no sense. width is not a member of AN, but of AN::A<SomeType>. You need an instance of class A<SomeType>, then you can refer to the data member of that instance.

– Igor Tandetnik
Jan 3 at 16:06

width is a non-static member variable. So you need an instance of A in order to access it's variable. AN::A<int> a; unsigned int* w = &a.width;

– john
Jan 3 at 16:06

unsigned int * w = AN::A<unsigned int>::&width; .. I tried it but it is stays that width is undefined..

– Giwrgos Kostopoulos
Jan 3 at 16:08

if i make static would it work?

– Giwrgos Kostopoulos
Jan 3 at 16:09

@GiwrgosKostopoulos No, you need a variable of type A, AN::A<unsigned int> a; unsigned int * w = &a.width;. I think really need you need to learn some basic C++ concepts.

– john
Jan 3 at 16:09

|
show 3 more comments

AN::&width makes no sense. width is not a member of AN, but of AN::A<SomeType>. You need an instance of class A<SomeType>, then you can refer to the data member of that instance.

– Igor Tandetnik
Jan 3 at 16:06

width is a non-static member variable. So you need an instance of A in order to access it's variable. AN::A<int> a; unsigned int* w = &a.width;

– john
Jan 3 at 16:06

unsigned int * w = AN::A<unsigned int>::&width; .. I tried it but it is stays that width is undefined..

– Giwrgos Kostopoulos
Jan 3 at 16:08

if i make static would it work?

– Giwrgos Kostopoulos
Jan 3 at 16:09

@GiwrgosKostopoulos No, you need a variable of type A, AN::A<unsigned int> a; unsigned int * w = &a.width;. I think really need you need to learn some basic C++ concepts.

– john
Jan 3 at 16:09

AN::&width makes no sense. width is not a member of AN, but of AN::A<SomeType>. You need an instance of class A<SomeType>, then you can refer to the data member of that instance.

– Igor Tandetnik
Jan 3 at 16:06

width is a non-static member variable. So you need an instance of A in order to access it's variable. AN::A<int> a; unsigned int* w = &a.width;

– john
Jan 3 at 16:06

unsigned int * w = AN::A<unsigned int>::&width; .. I tried it but it is stays that width is undefined..

– Giwrgos Kostopoulos
Jan 3 at 16:08

if i make static would it work?

– Giwrgos Kostopoulos
Jan 3 at 16:09

@GiwrgosKostopoulos No, you need a variable of type A, AN::A<unsigned int> a; unsigned int * w = &a.width;. I think really need you need to learn some basic C++ concepts.

– john
Jan 3 at 16:09

|
show 3 more comments

1 Answer
1

active

oldest

votes

This has nothing to do with templates. It's about classes and objects. The address of width is determined by the object that it's a part of; without an object, there is no width.

However, without an object you can create a pointer-to-member; it's not an ordinary pointer (if it was, it would be called "pointer"). Like this:

class A {

public:

    int width;

};



int A::*w = &A::width;

You use it to access that variable when you create an object:

A a;

a.*w = 3;

A aa;

aa.*w = 4;

If you really only want one value of width for every object of your type, yes, you can make it a static member:

class A {

public:

    static int width;

};

int A::width;

Now you can create a pointer to that member as an ordinary pointer:

int* w = &A::width;

and you can use w as an ordinary pointer:

*w = 3;

answered Jan 3 at 16:15

Pete Becker

59k442122

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

This has nothing to do with templates. It's about classes and objects. The address of width is determined by the object that it's a part of; without an object, there is no width.

However, without an object you can create a pointer-to-member; it's not an ordinary pointer (if it was, it would be called "pointer"). Like this:

class A {

public:

    int width;

};



int A::*w = &A::width;

You use it to access that variable when you create an object:

A a;

a.*w = 3;

A aa;

aa.*w = 4;

If you really only want one value of width for every object of your type, yes, you can make it a static member:

class A {

public:

    static int width;

};

int A::width;

Now you can create a pointer to that member as an ordinary pointer:

int* w = &A::width;

and you can use w as an ordinary pointer:

*w = 3;

answered Jan 3 at 16:15

Pete Becker

59k442122

add a comment |

This has nothing to do with templates. It's about classes and objects. The address of width is determined by the object that it's a part of; without an object, there is no width.

However, without an object you can create a pointer-to-member; it's not an ordinary pointer (if it was, it would be called "pointer"). Like this:

class A {

public:

    int width;

};



int A::*w = &A::width;

You use it to access that variable when you create an object:

A a;

a.*w = 3;

A aa;

aa.*w = 4;

If you really only want one value of width for every object of your type, yes, you can make it a static member:

class A {

public:

    static int width;

};

int A::width;

Now you can create a pointer to that member as an ordinary pointer:

int* w = &A::width;

and you can use w as an ordinary pointer:

*w = 3;

answered Jan 3 at 16:15

Pete Becker

59k442122

add a comment |

This has nothing to do with templates. It's about classes and objects. The address of width is determined by the object that it's a part of; without an object, there is no width.

However, without an object you can create a pointer-to-member; it's not an ordinary pointer (if it was, it would be called "pointer"). Like this:

class A {

public:

    int width;

};



int A::*w = &A::width;

You use it to access that variable when you create an object:

A a;

a.*w = 3;

A aa;

aa.*w = 4;

If you really only want one value of width for every object of your type, yes, you can make it a static member:

class A {

public:

    static int width;

};

int A::width;

Now you can create a pointer to that member as an ordinary pointer:

int* w = &A::width;

and you can use w as an ordinary pointer:

*w = 3;

answered Jan 3 at 16:15

Pete Becker

59k442122

This has nothing to do with templates. It's about classes and objects. The address of width is determined by the object that it's a part of; without an object, there is no width.

However, without an object you can create a pointer-to-member; it's not an ordinary pointer (if it was, it would be called "pointer"). Like this:

class A {

public:

    int width;

};



int A::*w = &A::width;

You use it to access that variable when you create an object:

A a;

a.*w = 3;

A aa;

aa.*w = 4;

If you really only want one value of width for every object of your type, yes, you can make it a static member:

class A {

public:

    static int width;

};

int A::width;

Now you can create a pointer to that member as an ordinary pointer:

int* w = &A::width;

and you can use w as an ordinary pointer:

*w = 3;

answered Jan 3 at 16:15

Pete Becker

59k442122

answered Jan 3 at 16:15

Pete Becker

59k442122

answered Jan 3 at 16:15

Pete Becker

59k442122

answered Jan 3 at 16:15

Pete Becker

59k442122

add a comment |

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