any way to customize the is operator?

[Lonnie Princehouse]

There doesn't seem to be any way to customize the behavior of "is" as
can be done for other operators... why not?

 

[Fredrik Lundh]

There doesn't seem to be any way to customize the behavior of "is" as
can be done for other operators... why not?

because it does "id(a) == id(b)", and there's no way to customize
the behaviour of id().

(objects are not allowed to lie about who they are, or what they
are).

</F>

 

[Lonnie Princehouse]

(objects are not allowed to lie about who they are, or what they are).

Dangit! I need to find a less honest programming language. Anyone
have a Perl cookbook handy? ...

 

[bruno at modulix]

Dangit! I need to find a less honest programming language. Anyone
have a Perl cookbook handy? ...

+1 QOTW (approved by a fellow Perl programmer FWIW !-)

 

[Sybren Stuvel]

Lonnie Princehouse enlightened us with:

There doesn't seem to be any way to customize the behavior of "is" as
can be done for other operators... why not?

Pure logic: A == A or A != A. An object is another object or not.
Why would you want to change that?

Sybren

 

[Steven D'Aprano]

Dangit! I need to find a less honest programming language. Anyone
have a Perl cookbook handy? ...

No, you need a better algorithm.


Why did you want to customize "is"?

 

[Lonnie Princehouse]

Why did you want to customize "is"?

Well, mostly out of principle ;-)

But also because I'm wrapping a C library which passes around C structs
which are wrapped in shim C++ classes for a Boost.Python layer. Boost
Python does a marvelous job of translating between Python and C++ data
types; when a C structure is returned, it magically translates that
structure into a Python wrapper. The problem is that a _new_ Python
wrapper is created every time a pointer is returned, even if a wrapper
already exists for that particular pointer.

To illustrate: Suppose you have a function in C++ which is a simple
identity function, e.g.

template <typename T>
T *identity (T *x) {
return x;
}

Calling the wrapped version of this function from Python will produce a
Python wrapper which represents the same underlying C++ object, but is
not actually the same Python object:
False

I wanted to override the behavior of "is" so that (a is b) would be
True --- which shouldn't have caused a problem, since the wrapper
class's attributes are read-only from Python. As it is, I've overriden
__cmp__ and __hash__, so at least I get the correct dictionary behavior
True

It's quite possible that there is some way to do this correctly from
the Boost.Python side of things... my understanding of how to use
Boost.Python is minimal.

 

[Steve Holden]

Well, mostly out of principle ;-)

But also because I'm wrapping a C library which passes around C structs
which are wrapped in shim C++ classes for a Boost.Python layer. Boost
Python does a marvelous job of translating between Python and C++ data
types; when a C structure is returned, it magically translates that
structure into a Python wrapper. The problem is that a _new_ Python
wrapper is created every time a pointer is returned, even if a wrapper
already exists for that particular pointer.

To illustrate: Suppose you have a function in C++ which is a simple
identity function, e.g.

template <typename T>
T *identity (T *x) {
return x;
}

Calling the wrapped version of this function from Python will produce a
Python wrapper which represents the same underlying C++ object, but is
not actually the same Python object:



False

I wanted to override the behavior of "is" so that (a is b) would be
True --- which shouldn't have caused a problem, since the wrapper
class's attributes are read-only from Python. As it is, I've overriden
__cmp__ and __hash__, so at least I get the correct dictionary behavior



True

It's quite possible that there is some way to do this correctly from
the Boost.Python side of things... my understanding of how to use
Boost.Python is minimal.

I will adopt my usual stance here of unequivocally stating that there is
no way you can do what you want to. In the case of the "is" operator you
can only expect truth when the left- and right-hand side operands refer
to the same object.

This strategy is usually sound: typically, before the metaphorical ink
has dried on my post some upstart comes along to prove me completely
wrong. So, I have done about all I can to help you. It's up to the rest
of the community now to prove me wrong (as they have so many times in
the past .

regards
Steve

PS: I'm afraid I couldn't resisit shortening yur odds somewhat by doing
what I usually fail to do and examining the source of is_() in
Modules/operator.c. I'm afraid it's not looking good ... the result of
"is" does appear to depend only on the memory addresses of the two
operands being the same:

result = (a1 == a2) ? Py_True : Py_False;
Py_INCREF(result);

sorry ...

 

[Aahz]

Well, mostly out of principle ;-)

But also because I'm wrapping a C library which passes around C structs
which are wrapped in shim C++ classes for a Boost.Python layer. Boost
Python does a marvelous job of translating between Python and C++ data
types; when a C structure is returned, it magically translates that
structure into a Python wrapper. The problem is that a _new_ Python
wrapper is created every time a pointer is returned, even if a wrapper
already exists for that particular pointer.

Then you need to fix Boost.Python. This cannot and will not get fixed at
the Python level -- how can Python possibly know that they are the same
objects?

 

[Terry Reedy]

What I think you need is a custom version of the .__eq__ (self,other)
method that pierces thru the wrappers to compare the underlying C++
objects. Then test a == b.

In Python, 'is' specifically means identity of Python objects.