Finding the instance reference of an object

[Aaron Brady]

Re: Official definition of call-by-value (Re: Finding the instancereferen ce...)

On Nov 21, 4:33 am, Duncan Booth <duncan.bo...@i nvalid.invalidw rote:

Aaron Brady <castiro...@gma il.comwrote:

a[:] = [1, 2, 3]

>

No, that's not assignment, it's syntactic sugar for a __setslice__
call.  No copies here.

>
Oh dear, perhaps you had better get the Python developers to update the
grammar that Python uses as that seems to think it's an assignment:

Well, the docs don't take my side either.

"object.__setit em__(self, key, value)
Called to implement assignment to self[key]."

But wait, is that true assignment?

"Assignment statements
Assignment statements are used to (re)bind names to values and to
modify attributes or items of mutable objects:"

"If the target is an identifier (name):
.... the name is bound to the object in the current global namespace."

The latter is the case of interest.

[greg]

Re: Official definition of call-by-value (Re: Finding the instancereferen ce...)

Aaron Brady wrote:

But wait, is that true assignment?

It's assignment, but it's not really copying an object. No new
objects are being created -- rather, some of the items within
the lhs object are being rebound to already-existing objects.

It would be possible for the lhs object's __setitem__ method
to be defined so that it created new objects, but then it
would be the __setitem__ method doing that, not the assignment
statement itself.

None of this is really relevant anyway, since the assignment
that call-by-value implies is always to a bare local name, and
there is no way that kind of assignment can ever create a new
object.

--
Greg

[greg]

Re: Official definition of call-by-value (Re: Finding the instancereferen ce...)

Aaron Brady wrote:

Tell me, what happens during a call to the following C++ function?
>
void f( std::vector< int x );

The same thing as would happen if you wrote

std::vector<int x = actual_paramete r_expression;

what happens during a call to the following Python
function?
>
def f( x ): ...

The same thing as would happen if you wrote

x = actual_paramete r_expression

If not, which one is call-by-value?

They're both call-by-value, because they're both equivalent to
assignment according to the rules of the language concerned.

Whether they're "the same" depends on what you mean by "same".
If you define "same" in such a way that they're not, then that
definition of "same" is irrelevant to the matter at hand.

--
Greg

[greg]

Re: Official definition of call-by-value (Re: Finding the instancereferen ce...)

Steven D'Aprano wrote:

You've just *assumed* that assignment in Algol 60 doesn't involving
copying.

I've done no such thing. I've *refrained* from assuming that
the "assignment " in the definition always has to refer to
Algol 60 assignment. You're the one insisting on tying
everything to Algol.

I don't know whether the Algol committee had such a general
meaning in mind when they wrote that document. It's quite
likely they weren't thinking that far ahead. But the fact is
that it *can* be generalized to other languages in an
obvious and useful way, and many language designers after
them have generalized it in exactly that way.

>>(I wait now with bated breath for somebody to point out some Python
>>implementatio n or feature where assignment doesn't make a copy...)

>
Ah crap, I meant *Pascal*.

Not in plain Pascal, but Apple's Object Pascal (and probably
other OO Pascal dialects) have object types that are implicitly
referred to by pointers, like Python objects -- and when you
pass one by value, only the pointer is copied, not the whole
object.

--
Greg

[Aaron Brady]

Re: Official definition of call-by-value (Re: Finding the instancereferen ce...)

On Nov 21, 7:06 pm, greg <g...@cosc.cant erbury.ac.nzwro te:

Aaron Brady wrote:

Tell me, what happens during a call to the following C++ function?

>

void f( std::vector< int x );

>
The same thing as would happen if you wrote
>
   std::vector<int x = actual_paramete r_expression;
>

what happens during a call to the following Python
function?

>

def f( x ): ...

>
The same thing as would happen if you wrote
>
   x = actual_paramete r_expression
>

If not, which one is call-by-value?

>
They're both call-by-value, because they're both equivalent to
assignment according to the rules of the language concerned.

No, you have described call-by-equals-sign, or call-by-assignment.
While call-by-assignment and call-by-value are equivalent in C++, that
does not make your rule hold for all languages. Python is call-by-
assignment too, as well as call-by-sharing. Just because a language
is call-by-assignment, is not sufficient (or necessary) to be call-by-
value. Call-by-value has other characteristics that Python does not
meet.

[greg]

Re: Official definition of call-by-value (Re: Finding the instancereferen ce...)

Aaron Brady wrote:

Call-by-value has other characteristics that Python does not
meet.

The designers of most other dynamic languages don't
seem to share that opinion, since they use the term
call-by-value just as though it *does* mean call-
by-assignment and nothing more.

--
Greg