From: Eric Mahurin <eric_mahurin@...>
Date: 2005-08-07T11:55:32+09:00
Subject: Re: new block notation

--- Daniel Brockman <daniel@brockman.se> wrote:
> You can convert a lambda to a block by using the `&' syntax
> (as in `&lambda {...}'), and the other way around by calling
> Proc.new or using the `def foo(&block)' syntax.

This gives an idea for an alternative to the ->(..){..} syntax.
 When calling a method, the unary "&" converts a lambda to a
block and in the method/lambda/block definition the "&"
converts a block to a lambda.  Well, how about making the unary
"&" in front of a block converting it to a lambda?

f = &{...}  # equivalent to f = lambda{...}

This is not really a unary "&" operator but rather a &{...}
construct, but it does flow with the current unary "&" syntax,
kind of.

This would distinguish a lambda from hash like the ->(..){..}
syntax does.  You'd still need to resolve the issue of block
arguments taking defaults (or not).  Any of the solutions that
did this within the {..} would work.  I like the { args
<delimiter> code } syntax the best.  Or { <delimiter> args
<delimiter> code } where a delimiter (or pair) that doesn't
create ambiguities is chosen (| is used now and creates
ambiguities with default values).

I just realized an ambiguity with the ->(..){..} syntax. 
Currently it can be used to define a lambda or a block.  If you
have this:

def test(lam=nil,&block)
    ...
end

test ->(..){..}

This looks ambiguous as to whether lam or block should get the
lambda/block.  I assume block wins, but I don't have the latest
1.9 installed on this machine.  I guess the same ambiguity
exists with the older 1.9 that allowed plain {...} for lambdas.
 This says to me that you should have distinguishable different
syntax between block and lambda as it is now:

{ ... }             # hash
method-call { ... } # block
lambda { ... }      # lambda
&{ ... }            # proposed lambda

Simply replacing lambda with "&" keeps this distinction.  The
new ->(..){...} syntax creates ambiguities between block and
hash since this can be either.




		
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