On 02/12/2014 12:17 PM, Tobiah wrote:
I do this:
a = 'lasdfjlasdjflaksdjfl;akjsdf;kljasdl;kfjasl'
b = 'lasdfjlasdjflaksdjfl;akjsdf;kljasdl;kfjasl'
print
print id(a)
print id(b)
And get this:
True
140329184721376
140329184721376
This works for longer strings. Does python
compare a new string to every other string
I've made in order to determine whether it
needs to create a new object?
Thanks,
Tobiah
Weird as well, is that in the interpreter,
the introduction of punctuation appears to
defeat the reuse of the object:
>>> b = 'lasdfjlasdjflaksdjflakjsdfkljasdlkfjasl'
>>> a = 'lasdfjlasdjflaksdjflakjsdfkljasdlkfjasl'
>>> a is b
True
>>> a = 'la;sdfjlasdjflaksdjflakjsdfkljasdlkfjasl'
>>> b = 'la;sdfjlasdjflaksdjflakjsdfkljasdlkfjasl'
>>> a is b
False
>>> b = 'la.sdfjlasdjflaksdjflakjsdfkljasdlkfjasl'
>>> a = 'la.sdfjlasdjflaksdjflakjsdfkljasdlkfjasl'
>>> a is b
False
>>> a = 'lasdfjlasdjflaksdjflakjsdfkljasdlkfjasl'
>>> b = 'lasdfjlasdjflaksdjflakjsdfkljasdlkfjasl'
>>> a is b
True
Tobiah
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