函数有三个参数：
def func(a, b, c):
____pass
现在我想把b冻结起来，用functools.partial如何实现？
def func(a, b, c):
____pass
现在我想把b冻结起来，用functools.partial如何实现？
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1
hahastudio Sep 19, 2014  2
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2
lianghui Sep 19, 2014 1
写的太罗嗦了：
callback = lambda a,c: func(a,"blablablabla", c) callback("a", "c") |
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