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February 10, 2007
&this pointer
Hi,

I'm a beginner with D, I've got some basis with C/C++ but I mainly 
program with PHP. Ad far as I'm concerned, D is a very interesting 
language, hence I started to develop a personal project in D to get a 
clear idea of its features ; nonetheless, during my research, I've 
noticed a behavior I considered very weird about the 'this' reference. 
According to what I understood from the documentation, it is a reference 
to the object itself, so coherently, &this should return the memory 
address of the object itself.

It that is correct, could one explain that to me (compiled with DMD 1.0) :

---

import std.stdio;
import std.string;

class Test
{
	this()
	{
		writefln("in this() itself - &this : ", &this);

		foo("called from this(), first shot");
		foo("called from this(), second shot");
	}

	void foo(char[] location)
	{
		writefln(location, " - foo() - &this : ", &this);

		bar(location);
	}

	void bar(char[] location)
	{
		writefln(location, " - bar() - &this : ", &this);
	}
}

int main()
{
	auto test = new Test();

	test.foo("called from main(), first shot");
	test.foo("called from main(), second shot");

	return 0;
}

---

$ ./test
in this() itself - &this : BFAE4394
called from this(), first shot - foo() - &this : BFAE4370
called from this(), first shot - bar() - &this : BFAE4344
called from this(), second shot - foo() - &this : BFAE4370
called from this(), second shot - bar() - &this : BFAE4344
called from main(), first shot - foo() - &this : BFAE438C
called from main(), first shot - bar() - &this : BFAE4360
called from main(), second shot - foo() - &this : BFAE438C
called from main(), second shot - bar() - &this : BFAE4360

---

In accordance with the output, the object location in memory endlessly 
changes ?! I admit to be a little startled...

e-t172
February 10, 2007
Re: &this pointer
By reference it means that 'this' already *is* a pointer.
You don't need to take it's address.  If you do you're just getting the 
address of the pointer not the address of the object.

Try something like
   writefln("%x",cast(void*)this);
if you really want to see the address of the object.

--bb

e-t172 wrote:
> Hi,
> 
> I'm a beginner with D, I've got some basis with C/C++ but I mainly 
> program with PHP. Ad far as I'm concerned, D is a very interesting 
> language, hence I started to develop a personal project in D to get a 
> clear idea of its features ; nonetheless, during my research, I've 
> noticed a behavior I considered very weird about the 'this' reference. 
> According to what I understood from the documentation, it is a reference 
> to the object itself, so coherently, &this should return the memory 
> address of the object itself.
> 
> It that is correct, could one explain that to me (compiled with DMD 1.0) :
> 
> ---
> 
> import std.stdio;
> import std.string;
> 
> class Test
> {
>     this()
>     {
>         writefln("in this() itself - &this : ", &this);
> 
>         foo("called from this(), first shot");
>         foo("called from this(), second shot");
>     }
> 
>     void foo(char[] location)
>     {
>         writefln(location, " - foo() - &this : ", &this);
> 
>         bar(location);
>     }
> 
>     void bar(char[] location)
>     {
>         writefln(location, " - bar() - &this : ", &this);
>     }
> }
> 
> int main()
> {
>     auto test = new Test();
> 
>     test.foo("called from main(), first shot");
>     test.foo("called from main(), second shot");
> 
>     return 0;
> }
> 
> ---
> 
> $ ./test
> in this() itself - &this : BFAE4394
> called from this(), first shot - foo() - &this : BFAE4370
> called from this(), first shot - bar() - &this : BFAE4344
> called from this(), second shot - foo() - &this : BFAE4370
> called from this(), second shot - bar() - &this : BFAE4344
> called from main(), first shot - foo() - &this : BFAE438C
> called from main(), first shot - bar() - &this : BFAE4360
> called from main(), second shot - foo() - &this : BFAE438C
> called from main(), second shot - bar() - &this : BFAE4360
> 
> ---
> 
> In accordance with the output, the object location in memory endlessly 
> changes ?! I admit to be a little startled...
> 
> e-t172
February 10, 2007
Re: &this pointer
e-t172 wrote:
> I've 
> noticed a behavior I considered very weird about the 'this' reference. 
> According to what I understood from the documentation, it is a reference 
> to the object itself,

Correct.

> so coherently, &this should return the memory 
> address of the object itself.

Wrong.
A reference in D is different from a reference in C++. If you take the 
address of a reference in D, you do not get the address of the object 
referenced, but the address of the reference itself.
Think of a D reference as a pointer to the body of the class, with small 
differences. For one, casting between references works differently. For 
another, pointer arithmetic doesn't work on references. But operators 
like (unary) & work the same: in this case, it returns a pointer to the 
memory storing the address being referred to.

To get the address of the object from a reference you can cast it to 
void*. So in your case, replace "&this" with "cast(void*)this" and you 
will see the same address everywhere.
February 10, 2007
Re: &this pointer
Frits van Bommel a écrit :
> A reference in D is different from a reference in C++. If you take the 
> address of a reference in D, you do not get the address of the object 
> referenced, but the address of the reference itself.
> Think of a D reference as a pointer to the body of the class, with small 
> differences. For one, casting between references works differently. For 
> another, pointer arithmetic doesn't work on references. But operators 
> like (unary) & work the same: in this case, it returns a pointer to the 
> memory storing the address being referred to.
> 
> To get the address of the object from a reference you can cast it to 
> void*. So in your case, replace "&this" with "cast(void*)this" and you 
> will see the same address everywhere.

Thank you very much :)

e-t172
February 10, 2007
Re: &this pointer
The phrase "&this" returns the address of the hidden 'this' parameter
passed to the member functions. In other words, you are seeing the address
of a parameter on the stack. 

If you are trying to get the address of the class instance, the phrase 

  "cast(void *)this" 

seems to work, but I'm sure there are other techniques too.

-- 
Derek Parnell
Melbourne, Australia
"Justice for David Hicks!"
skype: derek.j.parnell
February 11, 2007
Re: &this pointer
The following does not work :

import std.stdio;

class Test
{
	this()
	{
		void* ptr = cast(void*) this;

		(*(cast(Test*) ptr)).foo();
	}

	void foo()
	{
		writefln("OK");
	}
}

int main()
{
	auto test = new Test();

	return 0;
}

Result : segmentation fault

Could one explain why ?
February 11, 2007
Re: &this pointer
e-t172 wrote:
> The following does not work :
> 
> import std.stdio;
> 
> class Test
> {
>     this()
>     {
>         void* ptr = cast(void*) this;
> 
>         (*(cast(Test*) ptr)).foo();
>     }
> 
>     void foo()
>     {
>         writefln("OK");
>     }
> }
> 
> int main()
> {
>     auto test = new Test();
> 
>     return 0;
> }
> 
> Result : segmentation fault
> 
> Could one explain why ?

"this" is of type "Test". Therefore to recover it from a void*, do simply:

(cast(Test)ptr).foo();


Brian Byrne
February 11, 2007
Re: &this pointer
On Sun, 11 Feb 2007 21:14:26 +0100, e-t172 wrote:

> The following does not work :
> 
> import std.stdio;
> 
> class Test
> {
> 	this()
> 	{
> 		void* ptr = cast(void*) this;
> 
> 		(*(cast(Test*) ptr)).foo();
> 	}
> 
> 	void foo()
> 	{
> 		writefln("OK");
> 	}
> }
> 
> int main()
> {
> 	auto test = new Test();
> 
> 	return 0;
> }
> 
> Result : segmentation fault
> 
> Could one explain why ?

You need "(cast(Test) ptr).foo();" instead.

This is because 'void* ptr' contains the address of the object, and when
calling object members, you need to use a /reference/ to the object and not
the object itself. Casting 'ptr' to 'Test*' just tells the compiler to
assume that 'ptr' points to a Test object and then using the '*' operator
returns the object and not a reference to the object.

If, however you had cast 'ptr' to just a 'Test', this tells the compiler to
assume that 'ptr' contains a reference to the object and so it can then
find the member to invoke.

-- 

Derek Parnell
Melbourne, Australia
"Justice for David Hicks!"
skype: derek.j.parnell
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