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> As for the second decorator, sure test(lambda a: print("You are in block %s" % a)) will work,

not for me...

lanx@nc10-ubuntu:~$ python Python 2.5.2 (r252:60911, Jan 20 2010, 23:16:55) [GCC 4.3.2] on linux2 Type "help", "copyright", "credits" or "license" for more information. >>> def rubyyielder(gen): ... def wrapped_gen(block): ... for elem in gen(): ... block(elem) ... return wrapped_gen ... >>> @rubyyielder ... def test(): ... print("In test") ... yield 1 ... print("back in test") ... yield 2 ... >>> test(lambda a: print("You are in block %s" % a)) File "<stdin>", line 1 test(lambda a: print("You are in block %s" % a)) ^ SyntaxError: invalid syntax

I suppose the lamda syntax has more restrictions...

> If you used a non-keyword such as "send" instead of "yield" you could do Evil and avoid the first decorator. This is left as an exercise.

Simple, I can just port the semantic of my OP and let send execute the callback which is passed to @test.น

Cheers Rolf

( addicted to the Perl Programming Language)


น) at second thought this would require a possibility to access the arguments of the caller. I suppose the caller is an object where arguments are accessible.


deleted undiplomatic irony =)

In reply to Re^10: RFC: Simulating Ruby's "yield" and "blocks" in Perl (Python) by LanX
in thread RFC: Simulating Ruby's "yield" and "blocks" in Perl by LanX

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