The Cost of Dynamism (was Re: Pyhon 2.x or 3.x, which is faster?)

[Steven D'Aprano]

On Sat, 12 Mar 2016 11:10 pm, Chris Angelico wrote:

> On Sat, Mar 12, 2016 at 10:08 PM, BartC <bc at freeuk.com> wrote:
>>>> You're not mistaken. There are no "character constants" in Python.
>>>> (Note that the definition would be Unicode codepoints, rather than
>>>> ASCII values.) I don't often miss them, though.
>>
>>> Yes, a complete non-issue.
>>
>>
>> Really? The issue as I see it is this:
>>
>> Writing: a=65 generates this byte-code for the right-hand-side:
>>
>>     LOAD_CONST      1 (65)       An integer
>>
>> But writing instead: a=ord('A') generates this:
>>
>>     LOAD_GLOBAL     0 (ord)
>>     LOAD_CONST      1 ('A')      A string
>>     CALL_FUNCTION   1
> 
> I think the "non-issue" here is the difference between ASCII and
> Unicode. Either way, there's no way to say "the integer with the
> codepoint of this character" as a literal. But that's actually not
> even all that necessary, because subscripting a text string yields
> one-character strings - you almost never need the ordinals.
> 
> Subscripting a byte string in Py3 yields integers, so you might need
> ordinals for ASCII byte values. But you can get them the same way:
> 
>>>> dis.dis(lambda: b"a"[0])
>   1           0 LOAD_CONST               3 (97)
>               3 RETURN_VALUE
>>>> dis.dis(lambda: u"a"[0])
>   1           0 LOAD_CONST               3 ('a')
>               3 RETURN_VALUE
> 
> Whichever one you need, you can get as a compile-time constant.

Chris, what you're looking at is the result of the CPython keyhole optimizer
doing constant folding. That is **NOT** a language promise. Other
implementations of Python may lack the keyhole optimizer. Future versions
may remove it, or allow the user to disable it.

Python the language has no feature that *guarantees* that you can write 'a'
and get 97 as a compile-time constant.




-- 
Steven