Instances' __setitem__ methods

S

Spencer Pearson

I was recently trying to implement a dict-like object which would do
some fancy stuff when it was modified, and found that overriding the
__setitem__ method of an instance did not act the way I expected. The
help documentation (from help(dict.__setitem__)) claims that
"d.__setitem__(k,v)" is equivalent to "d[k]=v", but I've produced this
code that, on Python 2.6, acts differently in the two cases.

def print_args( key, value ):
print "print_args called: key = %s, value = %s" %(key,value)

class MyDict( dict ):
def __init__( self ):
dict.__init__( self )
self.__setitem__ = print_args

def __setitem__( self, key, value ):
print "ModelDict.__setitem__ called"
dict.__setitem__( self, key, value )

d = MyDict()

print "d.__setitem__(0,1):",
d.__setitem__(0,1)

print "d[0]=1:",
d[0]=1


I would expect the two setitems to both call print_args, but that's
not what happens. In the first case, it calls print_args, but in the
second case, the __setitem__ declared in MyDict is called instead.

The documentation at http://docs.python.org/reference/datamodel.html#specialnames
says that for new-style classes, "x" is equivalent to
"type(x).__getitem__(x, i)". I assume that "x=y" has similarly been
changed to be equivalent to "type(x).__setitem__(x, i, y)", since that
would produce the results that I'm getting. Is the help documentation
for dict.__setitem__ just outdated, or am I missing some subtlety
here?

Also: when I say "d.f(*args)", am I correct in thinking that d checks
to see if it has an instance attribute called "f", and if it does,
calls f(*args); and if it doesn't, checks whether its parent class
(and then its grandparent, and so on) has a class attribute called
"f", and if it does, calls f(x, *args)?
 
E

Ethan Furman

Spencer said:
I was recently trying to implement a dict-like object which would do
some fancy stuff when it was modified, and found that overriding the
__setitem__ method of an instance did not act the way I expected.

The __magic__ methods are only looked up on the class, never the instance.

~Ethan~
 
C

Chris Rebert

I was recently trying to implement a dict-like object which would do
some fancy stuff when it was modified, and found that overriding the
__setitem__ method of an instance did not act the way I expected. The
help documentation (from help(dict.__setitem__)) claims that
"d.__setitem__(k,v)" is equivalent to "d[k]=v", but I've produced this
code that, on Python 2.6, acts differently in the two cases.

Technically, the strict equivalence is only one-way, as you've shown;
but one generally avoids calling the __magic__ methods directly, so
this subtle distinction is seldom used intentionally.

I would expect the two setitems to both call print_args, but that's
not what happens. In the first case, it calls print_args, but in the
second case, the __setitem__ declared in MyDict is called instead.

The documentation at http://docs.python.org/reference/datamodel.html#specialnames
says that for new-style classes, "x" is equivalent to
"type(x).__getitem__(x, i)". I assume that "x=y" has similarly been
changed to be equivalent to "type(x).__setitem__(x, i, y)", since that
would produce the results that I'm getting. Is the help documentation
for dict.__setitem__ just outdated, or am I missing some subtlety
here?


The sentence of the docs in question begins with "For instance,";
hence, __setitem__ is just the arbitrarily-chosen example used in this
part of the docs. But the point applies equally to all the special
methods. See the very last section of the same webpage:
http://docs.python.org/reference/datamodel.html#special-method-lookup-for-new-style-classes
Also: when I say "d.f(*args)", am I correct in thinking that d checks
to see if it has an instance attribute called "f", and if it does,
calls f(*args); and if it doesn't, checks whether its parent class
(and then its grandparent, and so on) has a class attribute called
"f", and if it does, calls f(x, *args)?

See the first paragraph under the "Classes" entry on
http://docs.python.org/reference/datamodel.html#objects-values-and-types
for perfect accuracy.

Cheers,
Chris
 

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments. After that, you can post your question and our members will help you out.

Ask a Question

Members online

No members online now.

Forum statistics

Threads
473,755
Messages
2,569,534
Members
45,008
Latest member
Rahul737

Latest Threads

Top