Hi, the following code does not "compile":
···
-----------------------------
def test1
value = return("lalala")
end
-----------------------------
=> SyntaxError: (irb):11: void value expression
Ok, it makes sense. So then, why the following does not fail?
-----------------------------
def test2
value = case 123
when Fixnum
return "returning 'Fixnum' string from method test2"
end
puts "method test2 ends now"
end
--------------------------
test2
=> "returning 'Fixnum' string from method kk"
I'm using Ruby 1.9.2 and in fact I use a method like "test2" it in my
project, but it seems a bit strange for me and I would like to be sure
that it's something correct (even if IMHO it does not make sense) and
will not change in a future Ruby 1.9 version.
Thanks a lot.
--
Iñaki Baz Castillo
<ibc@aliax.net>
Hi, the following code does not "compile":
-----------------------------
def test1
value = return("lalala")
end
-----------------------------
=> SyntaxError: (irb):11: void value expression
Ok, it makes sense. So then, why the following does not fail?
-----------------------------
def test2
value = case 123
when Fixnum
return "returning 'Fixnum' string from method test2"
end
puts "method test2 ends now"
end
--------------------------
> test2
=> "returning 'Fixnum' string from method kk"
Probably because case has branches which actually return a value (even
if it is fall through and nil). Interestingly enough this does not
work with if:
11:07:59 ~$ ruby19 -c x.rb
x.rb:7: void value expression
11:08:04 ~$ cat -n x.rb
1
2 def a
3 x = if rand(10) == 0
4 99
5 else
6 return 55
7 end
8 end
11:08:07 ~$
Could be that you have found a loophole in the parser logic. 
I'm using Ruby 1.9.2 and in fact I use a method like "test2" it in my
project, but it seems a bit strange for me and I would like to be sure
that it's something correct (even if IMHO it does not make sense) and
will not change in a future Ruby 1.9 version.
I'd say stop using this idiom. If you use a "case" expression because
it should yield a value (i.e. "x = case...end") then I would only use
exceptions as other way out. A "return" indicates regular execution
flow which in the case of this usage of "case" would mean "have case
return a value" but not "return from the method".
If you use "case" as a control flow construct (i.e. without assigning
the result) then a "return" inside is OK and Ruby will also accept it.
Same goes for "if".
Kind regards
robert
···
On Thu, Jul 14, 2011 at 10:56 AM, Iñaki Baz Castillo <ibc@aliax.net> wrote:
--
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/
Could be that you have found a loophole in the parser logic.
XD
I'm using Ruby 1.9.2 and in fact I use a method like "test2" it in my
project, but it seems a bit strange for me and I would like to be sure
that it's something correct (even if IMHO it does not make sense) and
will not change in a future Ruby 1.9 version.
I'd say stop using this idiom. If you use a "case" expression because
it should yield a value (i.e. "x = case...end") then I would only use
exceptions as other way out. A "return" indicates regular execution
flow which in the case of this usage of "case" would mean "have case
return a value" but not "return from the method".
If you use "case" as a control flow construct (i.e. without assigning
the result) then a "return" inside is OK and Ruby will also accept it.
Same goes for "if".
Thanks, I'll make my code more "polite" then
···
2011/7/14 Robert Klemme <shortcutter@googlemail.com>:
--
Iñaki Baz Castillo
<ibc@aliax.net>
于 2011年07月14日 17:38, Iñaki Baz Castillo 写道:
If you use a "case" expression because
> it should yield a value (i.e. "x = case...end") then I would only use
> exceptions as other way out. A "return" indicates regular execution
> flow which in the case of this usage of "case" would mean "have case
> return a value" but not "return from the method".
Remember all the block in ruby is actually a snippet ,so return from the block will return from the method which call the block.
For example,type the following code in irb
value =case 123
when Fixnum
return "good"
end
LocalJumpError: unexpected return
The real reason for your question is that, everything in ruby has a value, include statements. When return statements works, it yield a value which is the result of the invocation, the assignment doesn't happen.
value=2
def test2
value=case 123
when Fixnum
return 3
end
end
test2
p value
#output
2