Return in iterator in lambda


#1

I want to create a conditional mathematical function as a lambda
structure. Therefore it may be the case that inside an iterative loop,
the result of the function is known and should be returned. By
implementing this it causes trouble. The test code I created looks like:

func = lambda do
for z in [1,2,3]
return 2
end
“something”
end
puts func.call
puts “after call”

This results in a LocalJumpError (Ruby 1.8.6 and 1.8.7). In ruby 1.9 it
works as expected. Is there a proper way of creating a function like
this in ruby < 1.9?

Regards,

Benjamin


#2

On Tue, Jun 2, 2009 at 10:06 PM, Benjamin Ter kuile
removed_email_address@domain.invalid wrote:

“something”
end
puts func.call
puts “after call”

This results in a LocalJumpError (Ruby 1.8.6 and 1.8.7). In ruby 1.9 it
works as expected. Is there a proper way of creating a function like
this in ruby < 1.9?
Yes there is!
Strange you did not ask how though :wink:
Well I’ll tell you anyway…

s/return/break/

HTH
Robert

Regards,

Benjamin

Posted via http://www.ruby-forum.com/.


Toutes les grandes personnes ont d’abord été des enfants, mais peu
d’entre elles s’en souviennent.

All adults have been children first, but not many remember.

[Antoine de Saint-Exupéry]


#3

On Tue, Jun 2, 2009 at 10:53 PM, Robert D. removed_email_address@domain.invalid
wrote:

end
Well I’ll tell you anyway…

s/return/break/

But why did you put something there, now it does not work anymore :frowning:
Sorry did not see

lambda f do
…each do
break 2
end ||
something
end


Toutes les grandes personnes ont d’abord été des enfants, mais peu
d’entre elles s’en souviennent.

All adults have been children first, but not many remember.

[Antoine de Saint-Exupéry]


Toutes les grandes personnes ont d’abord été des enfants, mais peu
d’entre elles s’en souviennent.

All adults have been children first, but not many remember.

[Antoine de Saint-Exupéry]


#4

Robert D. wrote:

On Tue, Jun 2, 2009 at 10:53 PM, Robert D. removed_email_address@domain.invalid
wrote:

�end
Well I’ll tell you anyway…

s/return/break/

But why did you put something there, now it does not work anymore :frowning:
Sorry did not see

lambda f do
…each do
break 2
end ||
something
end

This will not work for structures like:

data = Array(1…12)
func = lambda do |par|
answer = 0.0
for z in data
answer += Math.log(z)
return 2e22 if answer > 19
end
answer -= data.size*Math.log10(answer)
answer
end
puts func.call 0.2
puts “after call”

I am converting from fortran, which makes it harder to see the easy
elegant solution :slight_smile:
The function presented is not the real one, but is more like the one I
am trying to implement.


#5

Benjamin Ter kuile wrote:

“something”
end
puts func.call
puts “after call”

This results in a LocalJumpError (Ruby 1.8.6 and 1.8.7). In ruby 1.9 it
works as expected. Is there a proper way of creating a function like
this in ruby < 1.9?

Enumerable#find may be more appropriate than #each. This avoids the
issue of how to break out of nested blocks.

func = lambda do |a|
a.find do |z|
z if z >= 1.5
end or “not found”
end

puts func.call([1,2,3])
puts func.call([0,0.2,0.4])


#6

I am not sure what you want to do?

Do you want to return, than I think the above might work with some
tweaking for the non break case, sorry I was lazy.
Or do you just want to break, but than it will not work in Ruby1.9?

Please avoid “for” it confuses me, please use each for my poor brain ;).
Anyway what about this

catch( x ) do
values.each do
throw your_value if some_condition
end
some more code
end

Nahh, I guess if you refactor this into methods it will just be
simpler :wink: or use Joel’s #find if you can, that might give some clean
code too :slight_smile:

Cheers
Robert


Toutes les grandes personnes ont d’abord été des enfants, mais peu
d’entre elles s’en souviennent.

All adults have been children first, but not many remember.

[Antoine de Saint-Exupéry]