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| Started by | Kevin Mahler <kevin.mahler@yahoo.com> |
|---|---|
| First post | 2011-04-15 01:37 -0500 |
| Last post | 2011-04-15 01:37 -0500 |
| Articles | 1 — 1 participant |
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Re: scope of assignments with lambda Kevin Mahler <kevin.mahler@yahoo.com> - 2011-04-15 01:37 -0500
| From | Kevin Mahler <kevin.mahler@yahoo.com> |
|---|---|
| Date | 2011-04-15 01:37 -0500 |
| Subject | Re: scope of assignments with lambda |
| Message-ID | <90f821efcacb67d7fe0de170d6979e29@ruby-forum.com> |
Fearless Fool wrote in post #988777:
> What I was surprised about was the scoping of variables in assignment
> statements. I guess it makes sense. Assume that p is previously
> undefined, then a reference on the RHS to p like this:
>
> p = p + 1
>
> will raise an error. But wrapped inside a lambda, the reference to p is
> deferred, so it all works:
>
> p = lambda { |n| p.call(n-1) if (n > 0) }
>
> As I said, that's the power of a lexical closure.
"The scoping of variables in assignment" is exactly the same in both
cases.
p = p + 1 does not fail because p is undefined. p is defined. Its
value is nil. The error results from NilClass#+ not existing. This has
nothing to do with "the power of the lexical closure".
Local variables are discovered at compile time. p gets defined at the
beginning of the scope, *before* the line "p = p + 1", and it's
assigned the value of nil. Try
p = p.to_i + 1
and observe the non-error. Even better, try
def foo
puts local_variables #=>bar
bar = 99
end
The reference to p inside the lambda is not deferred. At compile-time,
the p inside the lambda is bound to the local variable p, just as the
p on the right-hand side of "p = p + 1" is bound to the local variable
p. There is nothing special going on as a result of the lambda being
present.
--
Posted via http://www.ruby-forum.com/.
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