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${ noResults }
2 Commits (73494c992761a528bf98c8dd48fbdd2f71d4b521)
| Author | SHA1 | Message | Date |
|---|---|---|---|
Dylan Baker
|
a8173630ea |
Don't use len() to test emptiness vs not emptiness
Meson has a common pattern of using 'if len(foo) == 0:' or
'if len(foo) != 0:', however, this is a common anti-pattern in python.
Instead tests for emptiness/non-emptiness should be done with a simple
'if foo:' or 'if not foo:'
Consider the following:
>>> import timeit
>>> timeit.timeit('if len([]) == 0: pass')
0.10730923599840025
>>> timeit.timeit('if not []: pass')
0.030033907998586074
>>> timeit.timeit('if len(['a', 'b', 'c', 'd']) == 0: pass')
0.1154778649979562
>>> timeit.timeit("if not ['a', 'b', 'c', 'd']: pass")
0.08259823200205574
>>> timeit.timeit('if len("") == 0: pass')
0.089759664999292
>>> timeit.timeit('if not "": pass')
0.02340641999762738
>>> timeit.timeit('if len("foo") == 0: pass')
0.08848102600313723
>>> timeit.timeit('if not "foo": pass')
0.04032287199879647
And for the one additional case of 'if len(foo.strip()) == 0', which can
be replaced with 'if not foo.isspace()'
>>> timeit.timeit('if len(" ".strip()) == 0: pass')
0.15294511600222904
>>> timeit.timeit('if " ".isspace(): pass')
0.09413968399894657
>>> timeit.timeit('if len(" abc".strip()) == 0: pass')
0.2023209120015963
>>> timeit.timeit('if " abc".isspace(): pass')
0.09571301700270851
In other words, it's always a win to not use len(), when you don't
actually want to check the length.
|
8 years ago |
Jussi Pakkanen
|
0a14ad63de |
Add simple uninstall target. Closes #753.
|
8 years ago |