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Re: stacked decorators and consolidating

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Tim Chase wrote:

> I've got some decorators that work fine as such:
> 
>   @dec1(args1)
>   @dec2(args2)
>   @dec3(args3)
>   def myfun(...):
>     pass
> 
> However, I used that sequence quite a bit, so I figured I could do
> something like
> 
>   dec_all = dec1(args1)(dec2(args2)(dec3(args3)))

With these shortcuts

a = dec1(args1)
b = dec2(args2)
c = dec3(args3)
 
to make it look less messy your first attempt is

dec_all = a(b(c))

and the final decorated function will be

a(b(c))(myfun)

when it should be

a(b(c(myfun)))

i. e. instead of decorating myfun three times you are decorating the 
decorator c twice and then use the result of that decoration to decorate 
myfunc.

Does that help? I have my doubts ;)

> to consolidate the whole mess down to
> 
>   @dec_all
>   def myfun(...):
>     pass
> 
> However, this yields different (test-breaking) results.  Messing
> around, I found that if I write it as
> 
>   dec_all = lambda fn: dec1(args1)(dec2(args2)(dec3(args3)(fn)))
> 
> it works and passes all preexisting tests.
> 
> What am I missing that would cause this difference in behavior?
> 
> -tkc

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Re: stacked decorators and consolidating Peter Otten <__peter__@web.de> - 2013-10-29 18:52 +0100

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