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The meaning of a = b in object oriented languages

 
 
Summercool
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      09-18-2007



The meaning of a = b in object oriented languages.
================================================== ==

I just want to confirm that in OOP, if a is an object, then b = a is
only copying the reference.

(to make it to the most basic form:

a is 4 bytes, let's say, at memory location 0x10000000 to 0x10000003

b is 4 bytes, let's say, at memory location 0x20000000 to 0x20000003

in 0x10000000 to 0x10000003, it is the value 0xF0000000, pointing to
an object

b = a just means
copy the 4 bytes 0xF0 0x00 0x00 0x00 into 0x20000000 to 0x2000003
so that b now points to 0xF0000000 which is the same object.)


so essentially, a is just a pointer to an object.

and b = a just means that put that same pointer into b.

and that's why in Python or Ruby, it is like:

>>> a = {"a" : 1, "b" : 2}
>>> b = a
>>> a

{'a': 1, 'b': 2}
>>> b

{'a': 1, 'b': 2}
>>> a["a"] = 999
>>> a

{'a': 999, 'b': 2}
>>> b

{'a': 999, 'b': 2}

so most or all object oriented language do assignment by reference?
is there any object oriented language actually do assignment by
value? I kind of remember in C++, if you do

Animal a, b;

a = b will actually be assignment by value.
while in Java, Python, and Ruby, there are all assignment by
reference. ("set by reference")

Is that the case: if a is an object, then b = a is only copying the
reference?

 
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Russell Wallace
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      09-18-2007
Summercool wrote:
> so most or all object oriented language do assignment by reference?
> is there any object oriented language actually do assignment by
> value? I kind of remember in C++, if you do
>
> Animal a, b;
>
> a = b will actually be assignment by value.
> while in Java, Python, and Ruby, there are all assignment by
> reference. ("set by reference")
>
> Is that the case: if a is an object, then b = a is only copying the
> reference?


Yes, your understanding is exactly correct; C++ will assign by value
unless you explicitly use pointers, but the other languages will assign
by reference (except for primitive types).

--
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Ben Finney
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      09-18-2007
Summercool <(E-Mail Removed)> writes:

> I just want to confirm that in OOP, if a is an object, then b = a is
> only copying the reference.


Whether the language is OO or not has no bearing on this question. The
semantics of the assignment operator can and do differ between
languages, orthogonal to whether OOP is involved.

--
\ "Our task must be to free ourselves from our prison by widening |
`\ our circle of compassion to embrace all humanity and the whole |
_o__) of nature in its beauty." —Albert Einstein |
Ben Finney
 
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Jim Langston
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      09-18-2007
"Summercool" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed) oups.com...
>
> The meaning of a = b in object oriented languages.
> ================================================== ==
>
> I just want to confirm that in OOP, if a is an object, then b = a is
> only copying the reference.
>
> (to make it to the most basic form:
>
> a is 4 bytes, let's say, at memory location 0x10000000 to 0x10000003
>
> b is 4 bytes, let's say, at memory location 0x20000000 to 0x20000003
>
> in 0x10000000 to 0x10000003, it is the value 0xF0000000, pointing to
> an object
>
> b = a just means
> copy the 4 bytes 0xF0 0x00 0x00 0x00 into 0x20000000 to 0x2000003
> so that b now points to 0xF0000000 which is the same object.)
>
> so essentially, a is just a pointer to an object.
>
> and b = a just means that put that same pointer into b.
>
> and that's why in Python or Ruby, it is like:
>
>>>> a = {"a" : 1, "b" : 2}
>>>> b = a
>>>> a

> {'a': 1, 'b': 2}
>>>> b

> {'a': 1, 'b': 2}
>>>> a["a"] = 999
>>>> a

> {'a': 999, 'b': 2}
>>>> b

> {'a': 999, 'b': 2}
>
> so most or all object oriented language do assignment by reference?
> is there any object oriented language actually do assignment by
> value? I kind of remember in C++, if you do
>
> Animal a, b;
>
> a = b will actually be assignment by value.
> while in Java, Python, and Ruby, there are all assignment by
> reference. ("set by reference")
>
> Is that the case: if a is an object, then b = a is only copying the
> reference?


In C++ the default assignment constructor is virtually the same as the
default copy constructor, which is sometimes called a bitwise copy, although
that is not strictly true. For POD types (Plain Old Data) what you are
showing is true, it's a bitwise copy, very similar to memcpy( destination,
source, sizeof( destination) ). For non POD types, however, that is not
true as objects inside the class or structure will have their assignment
operators called, and they may be overridden. A prime example of this is
std::string. If the std::string member was bitwise copied, then there would
be two instances of a std::string pointing to the same memory locations
(since std::string typically stores the strings data via a pointer).

Assignment operators in C++ should attempt to prevent two pointers poining
to the same memory location. Consier a simple class (untested):

class Foo
{
public:
char* Data;
int DataSize;
Foo( int Size ): DataSize( Size ) { Data = new char[Size]; }
~Foo() { delete Data[]; }
};

Now, if we leave the class at this, we get into problems. The default copy
constructor and assignment operators will do a bitwise copy on the pointer
Data. I.E.

int main()
{
Foo bar1( 10 );
Foo bar2( 20 );
bar2 = bar1; // Lots of problems
}

First off, the default assignment operator will simply copy the pointer from
bar1 (which points to 10 characters) into bar2, overwriting bar2's. Since
we no longer have a pointer to the data from bar2 we can not delete it,
causing a memory leak. Also, at this point bar1 and bar2's Data pointers
point to the same memory location. Changing the contents of one will change
the contents of the other, since they are one in the same. Also, when the
destructors are called, both will attempt to delete[] the same pointer, the
first one will succeed, the second one will cause an error as the pointer
has already been freed. So we need to override the copy constructor and
assignment operators to fix this. So we add to Foo:

Foo& operator=( const Foo& rhs )
{
delete[] Data;
Data = new char[rhs.DataSize];
memcpy( Data, rhs.Data, rhs.DataSize );
DataSize = rhs.DataSize;
}

You can see that we have to manually do some things. We have to delete[]
our pointer, new a new buffer, copy the cotents, copy the DataSize over,
none of which the default assignment operator would of done. The copy
constructor would be similar, we just wouldn't have to delete[] Data;
because nothing has been allocated yet.

Incidently, there may be errors in the code I've shown here if you attempt
to compile it. Be forewarned.


 
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Summercool
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      09-18-2007
On Sep 17, 11:04 pm, Lloyd Linklater <(E-Mail Removed)> wrote:
> SpringFlowers AutumnMoon wrote:
> > Is that the case: if a is an object, then b = a is only copying the
> > reference?

>
> That and it adds a counter.
>
> a = ["foo", "bar"]
> b = a
> b[0] = "bite me"
> p a, b
>
> a = "different"
> p a, b
>
> ***
>
> In the first print, we get
> ["something else", "bar"]
> ["something else", "bar"]
>
> showing that changing b changes a, as expected. However, if we change
> a, b is NOT changed as seen in the second print.
>
> "different"
> ["something else", "bar"]
>
> That means that there is a counter inside that says to separate the two
> or b would have changed with a as a changed with b initially.


i think the line

a = "different"

means a is now set to a pointer to the String object with content
"different".
or that "a is now a reference to the String object."

and b is still a reference to the Array object. so that's why a and b
print out different things. they point to different objects.

i think:

whenever in Ruby, Python, and Java,

a is never an object. a is always a "reference to an object"... this
will solve a lot of puzzles when we don't understand some code
behaviors.

when a writing or a book reads "a is a Hash object; a is an Array
object; or a is an Animal object" it is just a short form to say that
"a is a reference to that object."

b = a means "whatever a is referencing to, now b is referencing it
too".

so that's why a[1] = "foobar" will change what b will display, but
a = "foobar" will not change what b will display. (because a[1] =
"foobar" says "what is a referencing? go there and change its
content that has the index 1" and when b goes there to see it, it is
also changed.)


 
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Laurent Pointal
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      09-18-2007
Summercool a écrit :
>
>
> The meaning of a = b in object oriented languages.
> ================================================== ==

<zip>

Oups, reading the subject I thought it was a Xah Lee post.



 
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Roel Schroeven
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      09-18-2007
Laurent Pointal schreef:
> Summercool a écrit :
>>
>> The meaning of a = b in object oriented languages.
>> ================================================== ==

> <zip>
>
> Oups, reading the subject I thought it was a Xah Lee post.


me too ...

--
The saddest aspect of life right now is that science gathers knowledge
faster than society gathers wisdom.
-- Isaac Asimov

Roel Schroeven
 
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Lew
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      09-18-2007
Summercool wrote:
> when a writing or a book reads "a is a Hash object; a is an Array
> object; or a is an Animal object" it is just a short form to say that
> "a is a reference to that object."
>
> b = a means "whatever a is referencing to, now b is referencing it
> too".
>
> so that's why a[1] = "foobar" will change what b will display, but
> a = "foobar" will not change what b will display.


You can't do both in Java. Is a an array or a String? If a is a String and b
is an array, then neither `a = b' nor `b = a' will compile in Java.

Java is a strongly-typed, compiled language which means it does more static
type checking and thus would reject treating a as both an array and a String.
In that environment the programmer must choose one or the other.

Otherwise what you say is exactly correct.

> (because a[1] = "foobar" says "what is a referencing? go there and change its
> content that has the index 1" and when b goes there to see it, it is
> also changed.)


Speaking just of Java, it's useful to distinguish a variable from an object
(instance). As you point out, the variable represents a reference to the
instance. The variable has a compile-time type in Java, which may be
different from the run-time type of the object, albeit compatible.

C++ is similar in this respect. Python and Ruby are more, shall we say,
flexible in their type systems.

Both jet liners and hang gliders have their uses, both are flight, and neither
is really suitable for the other's purpose.

--
Lew
 
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Lew
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      09-18-2007
Roel Schroeven wrote:
> Laurent Pointal schreef:
>> Summercool a écrit :
>>>
>>> The meaning of a = b in object oriented languages.
>>> ================================================== ==

>> <zip>
>>
>> Oups, reading the subject I thought it was a Xah Lee post.

>
> me too ...


Nah, this dude's all right, so far. As if my opinion mattered.

Stay with it, Summercool. It's what discussion groups are for.

Here's why the reaction: cross-posting of computer-science-type essays,
something Xah Lee does. But he recycles all his decades-old crap and really
doesn't participate in the discussion. This isn't that at all.

--
Lew
 
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Bryan Olson
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      09-19-2007
Jim Langston wrote:
> Assignment operators in C++ should attempt to prevent two pointers poining
> to the same memory location. Consier a simple class (untested):
>
> class Foo
> {
> public:
> char* Data;
> int DataSize;
> Foo( int Size ): DataSize( Size ) { Data = new char[Size]; }
> ~Foo() { delete Data[]; }
> };


[...]
> Foo& operator=( const Foo& rhs )
> {
> delete[] Data;
> Data = new char[rhs.DataSize];
> memcpy( Data, rhs.Data, rhs.DataSize );
> DataSize = rhs.DataSize;
> }
>
> You can see that we have to manually do some things. We have to delete[]
> our pointer, new a new buffer, copy the cotents, copy the DataSize over,
> none of which the default assignment operator would of done.

[...]
> Incidently, there may be errors in the code I've shown here if you attempt
> to compile it. Be forewarned.


There's the "self-assignment" bug. See the popular C++ FAQ.

Follow-ups to comp.lang.c++
--
--Bryan
 
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