http://qs321.pair.com?node_id=1030873


in reply to Re^7: RFC: Simulating Ruby's "yield" and "blocks" in Perl (Python)
in thread RFC: Simulating Ruby's "yield" and "blocks" in Perl

Thanx, it's clear now.

I was just surprised that you need two decorators!

I thought you could simply do something like

 test(def _(x): print("You are in block %s" % x))

(like  test sub { print("You are in block $_[0]") } in Perl)

Just to get closer to the Ruby feeling and to achieve this

>>> def _(x):
...     print("You are in block %s" % x)
... 
>>> test(_)
In test
You are in block 1
back in test
You are in block 2
test lambda a: print("You are in block %s" % a)

[download]

But one needs decorators to manipulate a literal function.

Obviously, there is only this way to do it.

FWIW Python decorators could quite easily be simulated in Perl with attributes.

I will post this soon (if I can't find it already on CPAN) =)

Cheers Rolf

( addicted to the Perl Programming Language)

Replies are listed 'Best First'.
Re^9: RFC: Simulating Ruby's "yield" and "blocks" in Perl (Python)
by MonkOfAnotherSect (Sexton) on Apr 27, 2013 at 14:11 UTC
    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. As for the second decorator, sure 'test(lambda a: print("You are in block %s" % a))' will work, but when you require more than one line you're back to using a decorator unless you want to do something truly horrible.

    Don't want to do something truly horrible.

    -T. "A gentleman is someone who knows how to play the bagpipes but doesn't."

[reply]
      > 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

      [download]

      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.¹

      Cheers Rolf

      ( addicted to the Perl Programming Language)

      Update

      ¹) 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.

      Update

      deleted undiplomatic irony =)

[reply]
[d/l]
[select]
        Ah, you're using an old version where print was still a statement not a function; Python is finicky about how statements are used. v2.6-v2.7 allows "from __future__ import print_function", and in v3.0+ it is a function. If you really must use a lambda in v2.5- then "import sys" and use "sys.stdout.write()"... no you mustn't.

        -T. "We now return you to your regularly scheduled Perl"

[reply]

In Section Meditations