append on lists

[Maric Michaud]

Le Tuesday 16 September 2008 14:23:25 Armin, vous avez écrit :
> Alex Marandon wrote:
> > Armin wrote:
> >> Duncan Booth wrote:
> >>
> >> The semantic of [1,2,3,4,7].append(c) and [1,2,3,4,7] + c
> >> (with c = [8,9]) is identical,
> >
> > No it's not, + doesn't alter its operands.
> >
> >  >>> a = 1
> >  >>> b = 2
> >  >>> a + b
> >
> > 3
>
> That's not the point :)

It is, please try to understand it, in python all expressions that mutate an 
object should return None, it's the case for

l.append(x)
l.sort()
l.reverse()

...

all expressions that return something, return a new object, it's the case for

1+2
1.__add__(2) (which is the same)
sorted(l)
l[i:j]
etc...


there are some noticeable exceptions :

For coding facilities, some APIs could return the modified part of the object,
ex : it = c.pop()

Returning the modifyied object itself is mostly considered bad style, because 
it doesn't make clear if this the object or a copy. There is OTHO a use case 
for this, for APIs that allow chaining of operations (don't find by memory 
any example in stdlib, though), alike the ">>" operator in C++, but pyparsing 
is a good example if I remember well. BTW, Pythoneers are not very fond of 
that.

Finally, the very special case of augmented assignment operators :
x.__iadd__(y) which actually return the modified object, but this expression 
isn't intended to be use by callers, refer to the manual to see a good 
example of this, how one should implement __add__ and __iadd__ methods.

a += b
a = a + b

both are statments (doesn't evaluate to any object), but are not equivalent if 
a is mutable.

>
> What's the value of 1.add(b)?  None? Or 3 ??
> (if add works in the same way as append)
> a + b doesn't change a,b ... but a.add(b) -> a=3
> --
> http://mail.python.org/mailman/listinfo/python-list



-- 
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Maric Michaud