Can I rely on...
Bruno Desthuilliers
bruno.42.desthuilliers at websiteburo.invalid
Fri Mar 20 09:00:42 EDT 2009
Emanuele D'Arrigo a écrit :
> Hi everybody,
>
> I just had a bit of a shiver for something I'm doing often in my code
> but that might be based on a wrong assumption on my part.
Do not assume. Either check or use another solution. My 2 cents...
> Take the
> following code:
>
> pattern = "aPattern"
>
> compiledPatterns = [ ]
> compiledPatterns.append(re.compile(pattern))
>
> if(re.compile(pattern) in compiledPatterns):
you don't need the parens around the test.
> print("The compiled pattern is stored.")
>
> As you can see I'm effectively assuming that every time re.compile()
> is called with the same input pattern it will return the exact same
> object rather than a second, identical, object. In interactive tests
> via python shell this seems to be the case but... can I rely on it -
> always- being the case? Or is it one of those implementation-specific
> issues?
I can't tell - I'm not willing to write a serious test for it. IIRC, the
re module maintains a cache of already seen patterns, but I didn't
bother reading the implementation. Anyway: why don't you use a dict
instead, using the "source (ie : string representation) of the pattern
as key ?
ie:
pattern = "aPattern"
compiled_patterns = {}
compiled_patterns[pattern] = re.compile(pattern)
# ...
if pattern in compiled_patterns:
print("The compiled pattern is stored.")
> And what about any other function or class/method? Is there a way to
> discriminate between methods and functions that when invoked twice
> with the same arguments will return the same object and those that in
> the same circumstances will return two identical objects?
Except reading the source code (hey, this is OSS, isn't it), I don't see
any reliable way to know this - unless it is clearly documented of course.
> If the answer is no, am I right to state the in the case portrayed
> above the only way to be safe is to use the following code instead?
>
> for item in compiledPatterns:
> if(item.pattern == pattern):
Once again, using a dict will be *way* more efficient.
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