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Re: stacked decorators and consolidating

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On 2013-10-29 17:42, MRAB wrote:
> If you apply the stacked decorators you get:
> 
>      myfun = dec1(args1)(dec2(args2)(dec3(args3)(myfun)))
> 
> If you apply dec_all you get:
> 
>      myfun = dec1(args1)(dec2(args2)(dec3(args3)))(myfun)
> 
> See the difference? You need the lambda to fix that.

In this case, they happen to be CherryPy decorators:

  @cherrypy.expose()
  @cherrypy.tools.json_in()
  @cherrypy.tools.json_out()
  def myfunc(...): pass

I'd have figured they would be associative, making the result end up
the same either way, but apparently not.  Thanks for helping shed
some light on the subtle difference.

-tkc

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Re: stacked decorators and consolidating Tim Chase <python.list@tim.thechases.com> - 2013-10-29 12:54 -0500
  Re: stacked decorators and consolidating Gregory Ewing <greg.ewing@canterbury.ac.nz> - 2013-10-30 11:46 +1300

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