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Hi!

I just realized that <dict>.setdefault *always* executes the second
argument - even if it's not necessary, because the requested item in
the first argument exists.

This is not what I expected - why evaluate the second argument if it's
not needed? Also this can lead to side-effects!

Example:

>>> a = {}
>>> b = {}
>>> a.setdefault(1,b.setdefault(1,1))
1

'1' didn't exist in a - so I expect 'b.setdefault(1,1)' to be
evaluated. Great.

>>> a.setdefault(1,b.setdefault(2,1))
1

'1' existed, so it's not necessary to evaluate 'b.setdefault(2,1)'.
But:

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

.... shows that it was executed!

So it's not equivalent to:

if 1not in a:
b.setdefault(2,1)

Is this really by design? If there's a complicated, expensive to
calculate/build 2nd argument (maybe a function call) then it's also
quite ineffective to evaluate it just to throw away...

Thanks!

Tino

Tino Lange a écrit :
> I just realized that <dict>.setdefault *always* executes the second
> argument - even if it's not necessary, because the requested item in
> the first argument exists.

Since setdefault is a method, this is coherent with Python's standard
behaviour: method and function arguments are always evaluated before the
method is called.

--
Alexandre Fayolle
LOGILAB, Paris (France).
http://www.logilab.com http://www.logilab.fr http://www.logilab.org
Développement logiciel avancé - Intelligence Artificielle - Formations

On Thu, 31 Jul 2003 16:54:46 +0000 (UTC), Alexandre Fayolle
<> wrote:

>> I just realized that <dict>.setdefault *always* executes the second
>> argument - even if it's not necessary, because the requested item in
>> the first argument exists.

> method and function arguments are always evaluated before the
>method is called.

Yep. You're right. Of course it's like that.
Anyway - that's not what I expected in the first thought.

It shows that less code is not always the best code...

Thanks!

Tino

On 31 Jul 2003 18:01:51 +0100, (John J. Lee) wrote:

>Because that's the way Python always does function calls?

Of course. From the consistency point of view... perfectly right!
(I thought that maybe because these are very special low-level methods
for built-in types, it might be different....)

>Don't, then. Use key in dict, or dict.has_key(key).

or the C-like style:
>>> a[3] = a.get(1) or b.setdefault(4,1)

Thanks!

Tino

Tino Lange wrote:

> I just realized that <dict>.setdefault *always* executes the second
> argument - even if it's not necessary, because the requested item in
> the first argument exists.
>
> This is not what I expected - why evaluate the second argument if it's
> not needed? Also this can lead to side-effects!

Because Python doesn't have macros or special forms that look like
functions. If you see something that looks like a function, it
evaluates all arguments before it calls the function. Always. Changing
this would cause confusion, not resolve it.

--
Erik Max Francis && && http://www.alcyone.com/max/
__ San Jose, CA, USA && 37 20 N 121 53 W && &tSftDotIotE
/ \ There's a reason why we / Keep chasing morning
\__/ Sandra St. Victor

Quoth Tino Lange:
> On 31 Jul 2003 18:01:51 +0100, (John J. Lee) wrote:
[...]
> >Don't, then. Use key in dict, or dict.has_key(key).
>
> or the C-like style:
> >>> a[3] = a.get(1) or b.setdefault(4,1)

Be careful doing that -- it relies on all possible values for a[1]
being true. Consider:

>>> a = {1: 0}
>>> b = {}
>>> a[3] = a.get(1) or b.setdefault(4, 1)
>>> a
{1: 0, 3: 1}
>>> b
{4: 1}

--
Steven Taschuk o- @
7O )
" (