Re: expression in an if statement
Thomas Jollans wrote: What if set has side effects? A compiler could only exclude this possibility if it knew exactly what set will be at run time, And also that 'a' remains bound to the same object, and that object or anything reachable from it is not mutated in any way that could affect the result of set(a). That's quite a lot of information for an optimiser to deduce, particularly in a language as dynamic as Python. -- Greg -- http://mail.python.org/mailman/listinfo/python-list
Re: expression in an if statement
John Nagle na...@animats.com wrote: I was talking to the Facebook guys doing the compiler for PHP, and they said that it was a huge win for them that PHP doesn't allow dynamically replacing a function. I'm not sure if I call all that effort for a 50% speed increase a win. PyPy is seeing speed increases of up to 15 times that of CPython 2.6.2 without reducing the flexibility of the language at all. -- http://mail.python.org/mailman/listinfo/python-list
Re: expression in an if statement
On 8/18/2010 3:12 PM, Thomas Jollans wrote: On Wednesday 18 August 2010, it occurred to John Nagle to exclaim: On 8/18/2010 11:24 AM, ernest wrote: Hi, In this code: if set(a).union(b) == set(a): pass Does Python compute set(a) twice? CPython does. Shed Skin might optimize. Don't know about Iron Python. I doubt any actual Python implementation optimizes this -- how could it? The object set is clearly being called twice, and it happens to be called with the object a as a sole argument twice. What if set has side effects? A compiler could only exclude this possibility if it knew exactly what set will be at run time, which it can't. That just reflects the rather lame state of Python implementations. For some other languages, there are JIT compilers that can optimize such things. If you rebind a function at run time, the compiled code has to be invalidated and recompiled with the new binding. The HotSpot Java JIT compiler did this. See http://books.google.com/books?id=GBISkhhrHh8Cpg=PA786lpg=PA786dq=JIT+compiler+rebindingsource=blots=GhZa3XbrNusig=OVBOnu0vwlVN_B1QC6jc2ltHk_whl=enei=IhluTLCXOIymsQOW6vy1Cwsa=Xoi=book_resultct=resultresnum=9ved=0CDsQ6AEwCA#v=onepageqf=false It's really tough to get this right. There's a huge overhead in either performance or complexity for allowing rebinding of functions during execution. It's also a feature very seldom used after program startup. I was talking to the Facebook guys doing the compiler for PHP, and they said that it was a huge win for them that PHP doesn't allow dynamically replacing a function. John Nagle -- http://mail.python.org/mailman/listinfo/python-list
Re: expression in an if statement
On Thu, 2010-08-19 at 00:12 +0200, Thomas Jollans wrote: On Wednesday 18 August 2010, it occurred to John Nagle to exclaim: On 8/18/2010 11:24 AM, ernest wrote: Hi, In this code: if set(a).union(b) == set(a): pass Does Python compute set(a) twice? CPython does. Shed Skin might optimize. Don't know about Iron Python. I doubt any actual Python implementation optimizes this -- how could it? And why should it if a programmer uses its facilities inefficiently. I would write if set(a).issuperset (b): pass Frederic -- http://mail.python.org/mailman/listinfo/python-list
Re: expression in an if statement
On 19 Ago, 08:40, Frederic Rentsch anthra.nor...@bluewin.ch wrote: On Thu, 2010-08-19 at 00:12 +0200, Thomas Jollans wrote: On Wednesday 18 August 2010, it occurred to John Nagle to exclaim: On 8/18/2010 11:24 AM, ernest wrote: Hi, In this code: if set(a).union(b) == set(a): pass Does Python compute set(a) twice? CPython does. Shed Skin might optimize. Don't know about Iron Python. I doubt any actual Python implementation optimizes this -- how could it? And why should it if a programmer uses its facilities inefficiently. I would write if set(a).issuperset (b): pass Frederic Yes, maybe the example is silly, but situations like this arise frequently and it's good to know. Thanks for the answers. Ernest -- http://mail.python.org/mailman/listinfo/python-list
expression in an if statement
Hi, In this code: if set(a).union(b) == set(a): pass Does Python compute set(a) twice? Thanks in advance. Ernest -- http://mail.python.org/mailman/listinfo/python-list
Re: expression in an if statement
ernest wrote:
In this code:
if set(a).union(b) == set(a): pass
Does Python compute set(a) twice?
a = abc
b = def
_set = set
def set(x):
... print computing set(%r) % x
... return _set(x)
...
if set(a).union(b) == set(a): pass
...
computing set('abc')
computing set('abc')
So yes, set(a) is computed twice.
Peter
--
http://mail.python.org/mailman/listinfo/python-list
Re: expression in an if statement
On 8/18/2010 11:24 AM, ernest wrote: Hi, In this code: if set(a).union(b) == set(a): pass Does Python compute set(a) twice? CPython does. Shed Skin might optimize. Don't know about Iron Python. John Nagle -- http://mail.python.org/mailman/listinfo/python-list
Re: expression in an if statement
On Wednesday 18 August 2010, it occurred to John Nagle to exclaim: On 8/18/2010 11:24 AM, ernest wrote: Hi, In this code: if set(a).union(b) == set(a): pass Does Python compute set(a) twice? CPython does. Shed Skin might optimize. Don't know about Iron Python. I doubt any actual Python implementation optimizes this -- how could it? The object set is clearly being called twice, and it happens to be called with the object a as a sole argument twice. What if set has side effects? A compiler could only exclude this possibility if it knew exactly what set will be at run time, which it can't. I expect that set and a have to be looked up twice, actually: set(a).union(b) might rebind either one of them. This would be considered a very rude and inappropriate thing to do, but Python usually guarantees to allow bad taste and behaviour. I might be wrong on some points here, but this is what I expect the expression (set(a).union(b) == set(a)) has to do, in any conforming implementation of Python. Please correct me if I'm wrong. 1. find out which object set refers to 2. find out which object a refers to 3. call (set) with the single positional argument (a), no keyword arguments 4. get the attribute union of the return value of [3] 5. find out which object b refers to 6. call (.union) with the single positional argument (b). 7. look up __eq__ in the __class__ of the return value of [6] 8. find out which object set refers to 9. find out which object a refers to 10. call (set) with the single positional argument (a), no keyword arguments 11. call [7] with two positional arguments: the return values [6] [10] I'm not 100% sure if there are any guarantees as to when (5) is taken care of -- what would happen if set(a) or even set(a).__getattr__ changed the global b? My list there is obviously referring to Python 3.x, so there is no __cmp__ to worry about. -- http://mail.python.org/mailman/listinfo/python-list
Re: expression in an if statement
On Thu, Aug 19, 2010 at 9:12 AM, Thomas Jollans tho...@jollybox.de wrote: I doubt any actual Python implementation optimizes this -- how could it? The object set is clearly being called twice, and it happens to be called with the object a as a sole argument twice. What if set has side effects? A compiler could only exclude this possibility if it knew exactly what set will be at run time, which it can't. I expect that set and a have to be looked up twice, actually: set(a).union(b) might rebind either one of them. This would be considered a very rude and inappropriate thing to do, but Python usually guarantees to allow bad taste and behaviour. Yep. def test(): ... a = [1] ... b = [1] ... if set(a).union(b) == set(a): pass ... dis.dis(test) 2 0 LOAD_CONST 1 (1) 3 BUILD_LIST 1 6 STORE_FAST 0 (a) 3 9 LOAD_CONST 1 (1) 12 BUILD_LIST 1 15 STORE_FAST 1 (b) 4 18 LOAD_GLOBAL 0 (set) 21 LOAD_FAST0 (a) 24 CALL_FUNCTION1 27 LOAD_ATTR1 (union) 30 LOAD_FAST1 (b) 33 CALL_FUNCTION1 36 LOAD_GLOBAL 0 (set) 39 LOAD_FAST0 (a) 42 CALL_FUNCTION1 45 COMPARE_OP 2 (==) 48 JUMP_IF_FALSE4 (to 55) 51 POP_TOP 52 JUMP_FORWARD 1 (to 56) 55 POP_TOP 56 LOAD_CONST 0 (None) 59 RETURN_VALUE I might be wrong on some points here, but this is what I expect the expression (set(a).union(b) == set(a)) has to do, in any conforming implementation of Python. Please correct me if I'm wrong. You can use dis module to let Python do compiling and explaining for you -- With best regards, Daniel Kluev -- http://mail.python.org/mailman/listinfo/python-list
