[Python-Dev] [PEP] += on return of function call result (original) (raw)

Andrew Koenig ark@research.att.com
16 May 2003 08:07:23 -0400

ark> Why can't you do this?

ark> for t in range(5): ark> for r in range(10): ark> foo = log.setdefault(r,'') ark> foo += "test %d\n" % t

Luke> after running this code,

Luke> log = {0: '', 1: '', 2:'', 3: '' ... 9: ''}

Luke> and foo equals "test 5".

Then that is what foo would be if you were able to write

    log.setdefault(r,'') += "test %d\n" % t

as you had wished.

Luke> if, however, you do this:

Luke> for t in range(5): Luke> for r in range(10): Luke> foo = log.setdefault(r,[]) Luke> foo.append("test %d\n" % t)

Luke> then empirically i conclude that you DO end up with the Luke> expected results (but is this true all of the time?)

I presume that is because you are now dealing with vectors instead of strings. In that case, you could also have written

    for t in range(5):
            for r in range(10):
                    foo = log.setdefault(r,[])
                    foo += ["test %d]n" % t]

with the same effect.

Luke> the reason why your example, andrew, does not work, is Luke> because '' is a string - a basic type to which a pointer is Luke> NOT returned i presume that the foo += "test %d"... returns a Luke> DIFFERENT result object such that the string in the dictionary Luke> is DIFFERENT from the string result of foo being updated.

Well, yes. But that is what you would have gotten had you been allowed to write

            log.setdefault(r,"") += <whatever>

in the first place.

Luke> if that makes absolutely no sense whatsoever then think of it Luke> being the difference between integers and pointers-to-integers Luke> in c.

I think this analogy is pointless, as the only people who will understand it are those who didn't need it in the first place :-)

Luke> can anyone tell me if there are any PARTICULAR circumstances where

Luke> foo = log.setdefault(r,[]) Luke> foo.append("test %d\n" % t)

Luke> will FAIL to work as expected?

It will fail if your expectations are incorrect or unrealistic.

Luke> andrew, sorry it took me so long to respond: i initially Luke> thought that under all circumstances for all types of foo, Luke> your example would work.

But it does! At least in the sense of the original query.

The original query was of the form

    Why can't I write an expression like  f(x) += y?

and my answer was, in effect, 

    If you could, it would have the same effect as if you had written

        foo = f(x)
        foo += y

    and then used the value of foo.

Perhaps I'm missing something, but I don't think that anything you've said contradicts this answer.

-- Andrew Koenig, ark@research.att.com, http://www.research.att.com/info/ark