Completely emulate a pointer using templates



 DEVELOP > c-Plus-Plus > Completely emulate a pointer using templates

LINK TO THIS PAGE  


rating :  0   |  0


  Page 1 of 1

1

 
Topic: DEVELOP > c-Plus-Plus
User: ""
Date: 18 Oct 2006 02:54:17 AM
Object: Completely emulate a pointer using templates
Hello!
I would like to write a pointer to type template that behaves exactly and
completely like normal pointers, which I will use as starting point for
other kinds of pointers that hold extra data (for example, one of my aims
is to have 64bit pointers on a 32bit machine, and ignore the highmost
32bits of the address, while still keeping it in structures and function
parameters, so I can start to migrate my 32bit OS to 64bit, although I
don't have a 64bit CPU, and make it behave like a 64bit system where only
the first 4GB space is really used (the highmost 32 bits are allocated but
simply always ignored)).
While I am fluent with various assembly languages, I'm not so much with C++,
but I'd like to learn. So far my efforts brought me to:
template<typename Type> struct Pt {
private:
void* Address; // on my machine, a void* is 32 bit
//void* Unused; // later I will add this member, to double the sizeof(Pt)
public:
// Default Constructor
Pt() {
}
// Copy Constructor
Pt(const Pt& Object) {
Address=Object.Address;
}
Pt& operator = (const Type* Address) {
this->Address=(void*)Address;
return(*this);
}
operator Type* () {
return((Type*)Address);
}
Type& operator * () const {
return(*(Type*)Address);
}
Type& operator [] (const int Subscript) const {
return(((Type*)Address)[Subscript]);
}
Type* operator -> () const {
return(Address);
}
Pt& operator + (const int Offset) {
*((Type*)Address)+=Offset;
return(*this);
}
Pt& operator - (const int Offset) {
*((Type*)Address)-=Offset;
return(*this);
}
// No need for a Destructor
//~Pt() {
//}
};
for each type, so far I typedef'ed the corresponding pointer this way:
typedef MyType* MyTypePt;
but from now on I will do it this other way instead:
typedef Pt<MyType> MyTypePt;
and expect identical functionality, but with twice as big pointers (when
I will uncomment that "Unused" member, of course).
Will this work ALWAYS? Have I missed the overloading of some important
operator? So far things seem to work, but I'm not sure that some subtle
bugs haven't appeared somewhere.
Could anybody, please, be so kind to fix eventual errors and address me
to the eventual extensions I need to make, in order to complete my pointer
emulation via template?
Thanks!
Mario
.

User: "Kai-Uwe Bux"

Title: Re: Completely emulate a pointer using templates 18 Oct 2006 03:29:35 AM
wrote:


Hello!
I would like to write a pointer to type template that behaves exactly and
completely like normal pointers, which I will use as starting point for
other kinds of pointers that hold extra data (for example, one of my aims
is to have 64bit pointers on a 32bit machine, and ignore the highmost
32bits of the address, while still keeping it in structures and function
parameters, so I can start to migrate my 32bit OS to 64bit, although I
don't have a 64bit CPU, and make it behave like a 64bit system where only
the first 4GB space is really used (the highmost 32 bits are allocated but
simply always ignored)).

While I am fluent with various assembly languages, I'm not so much with
C++, but I'd like to learn. So far my efforts brought me to:

template<typename Type> struct Pt {
private:
void* Address; // on my machine, a void* is 32 bit

Why not Type* ?

//void* Unused; // later I will add this member, to double the
sizeof(Pt)
public:
// Default Constructor
Pt() {
}
// Copy Constructor
Pt(const Pt& Object) {
Address=Object.Address;

What is Object.Address? Did you mean &Object or boost::addressof( Object ) ?

}
Pt& operator = (const Type* Address) {

Why the const? Do you define only pointers to const objects? Did you mean:
Type* const Address // const pointer to Type
as opposed to:
const Type* // pointer to const Type

this->Address=(void*)Address;
return(*this);
}
operator Type* () {
return((Type*)Address);
}
Type& operator * () const {
return(*(Type*)Address);
}
Type& operator [] (const int Subscript) const {
return(((Type*)Address)[Subscript]);
}
Type* operator -> () const {
return(Address);
}
Pt& operator + (const int Offset) {
*((Type*)Address)+=Offset;
return(*this);
}
Pt& operator - (const int Offset) {
*((Type*)Address)-=Offset;
return(*this);
}

// No need for a Destructor
//~Pt() {
//}
};

for each type, so far I typedef'ed the corresponding pointer this way:

typedef MyType* MyTypePt;

but from now on I will do it this other way instead:

typedef Pt<MyType> MyTypePt;

and expect identical functionality, but with twice as big pointers (when
I will uncomment that "Unused" member, of course).

Will this work ALWAYS? Have I missed the overloading of some important
operator? So far things seem to work, but I'm not sure that some subtle
bugs haven't appeared somewhere.

Could anybody, please, be so kind to fix eventual errors and address me
to the eventual extensions I need to make, in order to complete my pointer
emulation via template?

One omission: you should provide comparison operators ==, !=, <, >, <=, >=.
If you want to make those emulate pointer behavior, then you should also
specialize std::less< Pt<T> > to get a total ordering for the use with
standard containers.
Another omission is exemplified by this piece of code:
struct X {};
struct Y : public X {};
int main ( void ) {
Pt<Y> y_ptr ;
Pt<X> x_ptr ( y_ptr );
Y* y_raw;
X* x_raw ( y_raw );
}
Best
Kai-Uwe Bux
.

User: "David Harmon"

Title: Re: Completely emulate a pointer using templates 18 Oct 2006 09:33:22 AM
On 18 Oct 2006 07:54:17 GMT in comp.lang.c++,
mario.rossi@REMOVETHISTOMAILMENOT.com wrote,

Pt& operator + (const int Offset) {
*((Type*)Address)+=Offset;
return(*this);
}
Pt& operator - (const int Offset) {
*((Type*)Address)-=Offset;
return(*this);
}

That part is wrong, operator+ should not modify the object.
Typically you want to implement operator+= and operator-= as member
functions, and then write operator+ and operator- as NON-member
functions using the former. That way e.g. (1+pt) works.
.

User: "Roland Pibinger"

Title: Re: Completely emulate a pointer using templates 18 Oct 2006 04:31:56 AM
On 18 Oct 2006 07:54:17 GMT, mario.rossi wrote:

I would like to write a pointer to type template that behaves exactly and
completely like normal pointers, which I will use as starting point for
other kinds of pointers

In short, you cannot completely emulate a pointer. The relationship
between the type of the pointer and the type of the pointed-to object
is defined by the C and the C++ language for real pointes. It is
undefined for 'smart pointers'. You can emulate parts of the pointer
semantics if that's really what you want.
Best regards,
Roland Pibinger
.


  Page 1 of 1

1

 


Related Articles
 

NEWER

pg.1232     pg.940     pg.716     pg.544     pg.412     pg.311     pg.234     pg.175     pg.130     pg.96     pg.70     pg.50     pg.35     pg.24     pg.16     pg.10     pg.6     pg.3     pg.1

OLDER