[Python-Dev] [PEP] += on return of function call result

[Andrew Koenig]

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