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 -- https://mail.python.org/mailman/listinfo/python-list
