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comp.lang.ruby

Passing a block to a block

Brian Candler

11/4/2008 1:41:00 PM

I notice that if you pass a block to a block, you can't use 'yield' to
yield to that block:

$ cat yield_in_block.rb
foo = lambda { |e| puts e }
blk = lambda { |e| yield e }
blk.call(123, &foo)

$ ruby yield_in_block.rb
yield_in_block.rb:2: no block given (LocalJumpError)
from yield_in_block.rb:3
$ ruby19 yield_in_block.rb
yield_in_block.rb:2:in `block in <main>': no block given (yield)
(LocalJumpError)
from yield_in_block.rb:3:in `call'
from yield_in_block.rb:3:in `<main>'

What's the reason for this? Does 'yield' only invoke the top-level block
passed into a method?

ruby1.9 is happy but only if I explicitly reference the block to yield.

foo = lambda { |e| puts e }
blk = lambda { |e,&y| y[e] }
blk.call(123, &foo)

Thanks,

Brian.
--
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2 Answers

William James

11/4/2008 7:17:00 PM

0



Brian Candler wrote:
> I notice that if you pass a block to a block, you can't use 'yield' to
> yield to that block:
>
> $ cat yield_in_block.rb
> foo = lambda { |e| puts e }
> blk = lambda { |e| yield e }
> blk.call(123, &foo)
>

blk is not a block. It is a Proc object or closure.
(Note that it isn't a Lambda object. Lambda is
for Lisp weenies.)

> $ ruby yield_in_block.rb
> yield_in_block.rb:2: no block given (LocalJumpError)
> from yield_in_block.rb:3
> $ ruby19 yield_in_block.rb
> yield_in_block.rb:2:in `block in <main>': no block given (yield)
> (LocalJumpError)
> from yield_in_block.rb:3:in `call'
> from yield_in_block.rb:3:in `<main>'
>
> What's the reason for this? Does 'yield' only invoke the top-level block
> passed into a method?
>
> ruby1.9 is happy but only if I explicitly reference the block to yield.
>
> foo = lambda { |e| puts e }
> blk = lambda { |e,&y| y[e] }
> blk.call(123, &foo)

foo = proc{|x| puts x }
==>#<Proc:0x02b2ae60@(irb):1>
bar = proc{|x,prc| prc[x] }
==>#<Proc:0x02b24ac4@(irb):2>
bar[ 88, foo ]
88
==>nil

Brian Candler

11/4/2008 8:07:00 PM

0

William James wrote:
> foo = proc{|x| puts x }
> ==>#<Proc:0x02b2ae60@(irb):1>
> bar = proc{|x,prc| prc[x] }
> ==>#<Proc:0x02b24ac4@(irb):2>
> bar[ 88, foo ]

Indeed, there's no problem when you pass them around and invoke them
explicitly.

My question (and I apologise if I didn't word it clearly) was regarding
the 'yield' keyword, to make a callback to a hidden proc which is not
one of the explicit arguments. e.g.

def foo(*a,&b)
yield 123 # similar to b.call(123)
end

pr = Proc.new { |x| puts x }
foo(&pr)

This works fine as shown when passing to a method, but not when passing
to another Proc.
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