Why is local variable created in one branch of if statement available in the other?

ruby-talk
1

I find it hard to understand why this code prints nil instead of
saying "unknown method or local variable foo":

cat hmm.rb

def hmm
  if false
    foo = 'quack'
  else
    p foo
  end
end

hmm

ruby hmm.rb

nil

Thoughts?

···

--
Alex Verkhovsky

2

The decision on whether foo is a method or a variable is made by the parser as it walks through the body of the function, rather than when the function is executed. By default, foo is considered a method, but once it encounters foo = somewhere, it considers foo to be a variable for the remainder of the function.

So, by the time the parser processes the else branch of the if statement, it has already seen an assignment in the earlier if branch, and therefore interprets 'foo' as a variable name rather than a method call.

-mental

···

On Fri, 6 Apr 2007 02:49:27 +0900, "Alexey Verkhovsky" <alexey.verkhovsky@gmail.com> wrote:

I find it hard to understand why this code prints nil instead of
saying "unknown method or local variable foo":

cat hmm.rb

def hmm
  if false
    foo = 'quack'
  else
    p foo
  end
end

3

The variable 'foo' is created during parsing stage so exists as "foo =
'quack'" has been parsed.
Execution then results in the assignment being skipped so foo has the
default nil value.

Cheers
Chris

···

On Apr 5, 1:49 pm, "Alexey Verkhovsky" <alexey.verkhov...@gmail.com> wrote:

I find it hard to understand why this code prints nil instead of
saying "unknown method or local variable foo":

> cat hmm.rb

def hmm
  if false
    foo = 'quack'
  else
    p foo
  end
end

> ruby hmm.rb

nil

Thoughts?

--
Alex Verkhovsky

4

The variable 'foo' is created during parsing stage so exists as "foo =
'quack'" has been parsed.
Execution then results in the assignment being skipped so foo has the
default nil value.

Thanks.