Am 31.05.2011 12:08 schrieb Jussi Piitulainen:
The same sharing-an-i thing happens here:
fs = []
for i in range(4):
... fs.append(lambda n : i + n)
...
fs[0](0)
3
And the different private-j thing happens here:
gs = []
for i in range(4):
... gs.append((lambda j : lambda n : j + n)(i))
...
gs[0](0)
0
There is a simpler way: with
>>>> fs = []
>>>> for i in range(4):
> ... fs.append(lambda n, i=i: i + n)
> ...
you give each lambda a different default argument.
>>>> fs[0](0)
> 0
Thomas
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