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

Luke Kenneth Casson Leighton lkcl@samba-tng.org
Thu, 15 May 2003 21:44:18 +0000

On Wed, Apr 02, 2003 at 09:54:35AM -0500, Andrew Koenig wrote:

Luke> example code: Luke> log = {}

Luke> for t in range(5): Luke> for r in range(10): Luke> log.setdefault(r, '') += "test %d\n" % t Luke> pprint(log) Luke> instead, as the above is not possible, the following must be used: Luke> from operator import add Luke> ... Luke> ... Luke> ... Luke> add(log.setdefault(r, ''), "test %d\n" % t) Luke> ... ARGH! just checked - NOPE! add doesn't work. Luke> and there's no function "radd" or "radd" in the Luke> operator module. Why can't you do this? for t in range(5): for r in range(10): foo = log.setdefault(r,'') foo += "test %d\n" % t

after running this code,

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

and foo equals "test 5".

if, however, you do this:

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

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

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

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

can anyone tell me if there are any PARTICULAR circumstances where

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

will FAIL to work as expected?

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

l.

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