[Python-ideas] Literals call local constructors()?

[Andrew Barnert]

On Apr 15, 2014, at 10:24, "Franklin? Lee" <leewangzhong+python at gmail.com> wrote:

> I think C Anthony is saying that he tried to make it happen, and it didn't, but he figured something else out, which is what is relevant to him here.

What's the something else?

It seems to me he was asking why hacking dict literals doesn't work (either from an implementation point of view or from a language design point of view) because he _hadn't_ figured it out. And that's what Chris Angelico answered (from both points of view).

> I am mostly sure he isn't arguing for the ability to hack dict literals

> On Tue, Apr 15, 2014 at 6:40 AM, Chris Angelico <rosuav at gmail.com> wrote:
>> On Tue, Apr 15, 2014 at 8:11 PM, C Anthony Risinger <anthony at xtfx.me> wrote:
>> > This not working is unexpected, if considering the expectation literals:
>> >
>> > {..} and [...]
>> >
>> > ...translate to:
>> >
>> > dict(...) and list(...) [int(..) and str(..)]
>> >
>> > Why isn't/can't this be true?
>> 
>> They don't. They translate into dict-creation bytecodes. In CPython:
>> 
>> Python 3.5.0a0 (default:6a0def54c63d, Mar 26 2014, 01:11:09)
>> [GCC 4.7.2] on linux
>> Type "help", "copyright", "credits" or "license" for more information.
>> >>> import dis
>> >>> def make_dict1():
>> ...     return {'hello':1, 'world':2}
>> ...
>> >>> def make_dict2():
>> ...     return dict(hello=1, world=2)
>> ...
>> >>> dis.dis(make_dict1)
>>   2           0 BUILD_MAP                2
>>               3 LOAD_CONST               1 (1)
>>               6 LOAD_CONST               2 ('hello')
>>               9 STORE_MAP
>>              10 LOAD_CONST               3 (2)
>>              13 LOAD_CONST               4 ('world')
>>              16 STORE_MAP
>>              17 RETURN_VALUE
>> >>> dis.dis(make_dict2)
>>   2           0 LOAD_GLOBAL              0 (dict)
>>               3 LOAD_CONST               1 ('hello')
>>               6 LOAD_CONST               2 (1)
>>               9 LOAD_CONST               3 ('world')
>>              12 LOAD_CONST               4 (2)
>>              15 CALL_FUNCTION          512 (0 positional, 2 keyword pair)
>>              18 RETURN_VALUE
>> 
>> If you want the name dict to be looked up, look up the name dict.
>> Otherwise, code like this would be very VERY confusing:
>> 
>> >>> def make_dict3():
>> ...     dict = {'hello':1, 'world':2}
>> ...     return dict
>> ...
>> >>> dis.dis(make_dict3)
>>   2           0 BUILD_MAP                2
>>               3 LOAD_CONST               1 (1)
>>               6 LOAD_CONST               2 ('hello')
>>               9 STORE_MAP
>>              10 LOAD_CONST               3 (2)
>>              13 LOAD_CONST               4 ('world')
>>              16 STORE_MAP
>>              17 STORE_FAST               0 (dict)
>> 
>>   3          20 LOAD_FAST                0 (dict)
>>              23 RETURN_VALUE
>> 
>> If that had to actually call dict(), it would raise UnboundLocalError
>> for something that doesn't seem to reference locals before assigning
>> to them. As it is, it's exactly the same as make_dict1 except that it
>> does a store/load unnecessarily.
>> 
>> ChrisA
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