Suppose I wanted to a class to contain a pointer, but I didn't want to
make the copy constructor or assignment operator reinitialize this
pointer. (Think of a class that contains an mmapped pointer - having to
duplicate the memory could be very expensive.) Instead, I wanted only
the non-copy construction to create the pointer, and have copies of the
object, achieved through either copy construction or assignment, refer
to the pointer through a reference; in other words, "proxy" the original
object. If an original object changes its pointer, all copy constructed
or reassigned objects will see the update with minimal overhead.
I developed this simple class below to demonstrate how this could be done:
class PtrClass
{
public:
PtrClass(char *p = NULL) : ptr_(p), cpref_(ptr_) { }
PtrClass(const PtrClass &co) : ptr_(NULL), cpref_(co.cpref _) { }
PtrClass &
operator =(const PtrClass &rhs)
{
if (this != &rhs)
{
memcpy(this, &rhs, sizeof(*this));
ptr_ = NULL;
}
return *this;
}
char *getPtr() const { return cpref_; }
~PtrClass() { ptr_ = NULL; }
private:
char *ptr_;
char * &cpref_;
};
My question now: Is there any way within the C++ language to implement
the assignment operator so that the reference is carried from the right
hand side to the destination? Since references can never be reassigned,
the only way I can see around it is by going around C++'s back with the
memcpy().
Thanks,
Jim 2 2447
Jim Campbell <ji*@rochester. rr.com> wrote in message
news:Lw******** **********@twis ter.nyroc.rr.co m... Suppose I wanted to a class to contain a pointer, but I didn't want to make the copy constructor or assignment operator reinitialize this pointer. (Think of a class that contains an mmapped pointer - having to duplicate the memory could be very expensive.) Instead, I wanted only the non-copy construction to create the pointer, and have copies of the object, achieved through either copy construction or assignment, refer to the pointer through a reference; in other words, "proxy" the original object. If an original object changes its pointer, all copy constructed or reassigned objects will see the update with minimal overhead.
I developed this simple class below to demonstrate how this could be done:
class PtrClass { public: PtrClass(char *p = NULL) : ptr_(p), cpref_(ptr_) { } PtrClass(const PtrClass &co) : ptr_(NULL), cpref_(co.cpref _) { }
PtrClass & operator =(const PtrClass &rhs) { if (this != &rhs) { memcpy(this, &rhs, sizeof(*this));
Very nasty.
ptr_ = NULL; } return *this; }
char *getPtr() const { return cpref_; }
~PtrClass() { ptr_ = NULL; }
private: char *ptr_; char * &cpref_; };
My question now: Is there any way within the C++ language to implement the assignment operator so that the reference is carried from the right hand side to the destination?
No. You can't make the reference refer to a pointer different from the one
it was initialized to refer to.
Since references can never be reassigned, the only way I can see around it is by going around C++'s back with the memcpy().
Why not a pointer to a pointer instead of a reference to a pointer?
DW
Yes, a pointer to a pointer would be the obvious choice for
implementing this "proxy". I just wanted to see if a reference could
get me what I needed.
The "memcpy()" approach is apparently unportable, in that I've only
gotten it to work under the GNU C++ compiler. On the Sun Workshop 6
compiler, it fails.
- Jim
"David White" <no@email.provi ded> wrote in message news:<hY******* ***********@nas al.pacific.net. au>... Jim Campbell <ji*@rochester. rr.com> wrote in message news:Lw******** **********@twis ter.nyroc.rr.co m... Suppose I wanted to a class to contain a pointer, but I didn't want to make the copy constructor or assignment operator reinitialize this pointer. (Think of a class that contains an mmapped pointer - having to duplicate the memory could be very expensive.) Instead, I wanted only the non-copy construction to create the pointer, and have copies of the object, achieved through either copy construction or assignment, refer to the pointer through a reference; in other words, "proxy" the original object. If an original object changes its pointer, all copy constructed or reassigned objects will see the update with minimal overhead.
I developed this simple class below to demonstrate how this could be done:
class PtrClass { public: PtrClass(char *p = NULL) : ptr_(p), cpref_(ptr_) { } PtrClass(const PtrClass &co) : ptr_(NULL), cpref_(co.cpref _) { }
PtrClass & operator =(const PtrClass &rhs) { if (this != &rhs) { memcpy(this, &rhs, sizeof(*this));
Very nasty.
ptr_ = NULL; } return *this; }
char *getPtr() const { return cpref_; }
~PtrClass() { ptr_ = NULL; }
private: char *ptr_; char * &cpref_; };
My question now: Is there any way within the C++ language to implement the assignment operator so that the reference is carried from the right hand side to the destination?
No. You can't make the reference refer to a pointer different from the one it was initialized to refer to.
Since references can never be reassigned, the only way I can see around it is by going around C++'s back with the memcpy().
Why not a pointer to a pointer instead of a reference to a pointer?
DW This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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Hello,
My question concerns as to how a pointer is passed by reference as a
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in include files:
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I mistakenly set this to the comp.std.c++ a few days back. I don't believe
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ISO/IEC 14882:2003(E) §8.5 says:
To zero-initialize an object of type T means:
5
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