# Block and method local variables **URL:** \ **Category:** Ruby\ **Created:** [January 4, 2006, 12:54am UTC](https://www.ruby-forum.com/t/block-and-method-local-variables/51596 "2006-01-04T00:54:48Z")\ **Posts on this page:** 4\ **Page:** 1
**Author:** ![SSSISO-8859-1SQSDan](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/c89c15/32.png) [@SSSISO-8859-1SQSDan](https://www.ruby-forum.com/u/SSSISO-8859-1SQSDan)\ **Post date:** [January 4, 2006, 12:54am UTC](https://www.ruby-forum.com/t/block-and-method-local-variables/51596/1 "2006-01-04T00:54:48Z")
hello all, consider follow code irb(main):528:0\* def abc irb(main):529:1\> ppp=111 irb(main):530:1\> puts ppp irb(main):531:1\> yield irb(main):532:1\> puts ppp irb(main):533:1\> ppp irb(main):534:1\> end =\> nil irb(main):536:0\> abc {} 111 111 =\> 111 irb(main):537:0\> abc { puts ppp } 111 NameError: undefined local variable or method `ppp' for main:Object from (irb):537 from (irb):537:in`abc’ from (irb):537 from :0 irb(main):538:0\> irb(main):539:0\* irb(main):540:0\* ppp = “OOO” =\> “OOO” irb(main):541:0\> abc { puts ppp } 111 OOO 111 =\> 111 irb(main):542:0\> I am right assuming that all “local” variables in the method abc are invisible in the block? I tried a little more irb(main):544:0\* def cba irb(main):545:1\> ppp = 222 irb(main):546:1\> puts ppp irb(main):547:1\> yield ppp irb(main):548:1\> puts ppp irb(main):549:1\> ppp irb(main):550:1\> end =\> nil irb(main):551:0\> cba {} 222 222 =\> 222 irb(main):552:0\> cba {|p|} 222 222 =\> 222 irb(main):553:0\> cba {|p| p = 333} 222 222 =\> 222 irb(main):554:0\> cba {|p| puts p; p = 333} 222 222 222 =\> 222 irb(main):555:0\> if explicitely passed to block ppp is now visible there. but seems to be imposible to change the “method local” variable ppp in this case. the following _does_ work irb(main):557:0\* def xxx irb(main):558:1\> ppp = [1] irb(main):559:1\> p ppp irb(main):560:1\> yield ppp irb(main):561:1\> p ppp irb(main):562:1\> ppp irb(main):563:1\> end =\> nil irb(main):564:0\> xxx {} [1] [1] =\> [1] irb(main):565:0\> xxx {|x| } [1] [1] =\> [1] irb(main):566:0\> xxx {|x| x[0]=2 } [1] [2] =\> [2] irb(main):567:0\> and I kind of understand why it works if I change line 566 to irb(main):566:0\> xxx {|x| x=[2] } than it would not work the reason I came accross this is following I was reading " Parameters to a block may be existing local variables; if so, the new value of the variable will be retained after the block completes. This may lead to unexpected behavior, but there is also a performance gain to be had by using variables that already exist " and in my understanding all variables defined in a method are “local” (C++ background) If not local what are they considered to be then? Regards, Daniel ---
**Author:** ![SSSISO-8859-1SQSDan](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/c89c15/32.png) [@SSSISO-8859-1SQSDan](https://www.ruby-forum.com/u/SSSISO-8859-1SQSDan)\ **Post date:** [January 4, 2006, 2:30am UTC](https://www.ruby-forum.com/t/block-and-method-local-variables/51596/2 "2006-01-04T02:30:52Z")
Okay, I’m too tired… On Wed, 04 Jan 2006 00:58:08 -0000, Ross B. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote: > They are local. I think that test is referring to this: > ^^^^ > That should be ‘text’ > =\> nil > > irb(main):026:0\> d = Demo.new > =\> #Demo:0xb7f49a6c > irb(main):027:0\> [‘one’,‘two’,‘three’].each { |d.last| } > =\> [“one”, “two”, “three”] > irb(main):028:0\> d.last > ^^^^ > And that should be: ``` irb(main):028:0> d.all ``` > =\> [“one”, “two”, “three”] That’ll teach me to refactor in Opera… ---
**Author:** ![SSSISO-8859-1SQSDan](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/c89c15/32.png) [@SSSISO-8859-1SQSDan](https://www.ruby-forum.com/u/SSSISO-8859-1SQSDan)\ **Post date:** [January 4, 2006, 11:15am UTC](https://www.ruby-forum.com/t/block-and-method-local-variables/51596/3 "2006-01-04T11:15:29Z")
Daniel Schüle wrote: > if I change line 566 to > irb(main):566:0\> xxx {|x| x=[2] } > than it would not work Note, that you can use get the return value of the block from yield: def foo p yield( 111 ) end > > foo {|x| x + 10} > > 121 > > =\> nil > are “local” (C++ background) > > If not local what are they considered to be then? They are local. Stress in the setence above must be on “existing”. It means a situation like this: def get\_last(enum) last = nil enum.each {|last|} last end > > get\_last [1,2,43,3,2,4] > > =\> 4 This works because “last” is defined before the block. This does not work: def get\_last(enum) enum.each {|last|} last end > > get\_last [1,2,43,3,2,4] > > NameError: undefined local variable or method `last' for main:Object from (irb):26:in`get\_last’ > > from (irb):28 > > from :0 IOW, if the variable is defined in the surrounding scope that is the one used. If it’s defined only in the block (either as parameter or in the body) it’s visibility is limited to the block. Note that this is how most PL do nested scoping. /\* C _/ int foo() { int x; { /_ x is visible here _/ int y = x; } /_ no y visible here \*/ } Note, that the scope of block parameters may change in the future. There have been lengthy discussions about this but ATM I don’t remember the details. Kind regards ``` robert ``` ---
**Author:** ![SSSISO-8859-1SQSDan](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/s/c89c15/32.png) [@SSSISO-8859-1SQSDan](https://www.ruby-forum.com/u/SSSISO-8859-1SQSDan)\ **Post date:** [January 4, 2006, 2:09am UTC](https://www.ruby-forum.com/t/block-and-method-local-variables/51596/4 "2006-01-04T02:09:39Z")
On Wed, 04 Jan 2006 00:46:16 -0000, Daniel Schüle [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote: > irb(main):534:1\> end > from (irb):537 > irb(main):542:0\> > > I am right assuming that all “local” variables in the method abc > are invisible in the block? Yes. The block captures the scope it is defined in, not the scope it’s called from, i.e. it’s a closure. > irb(main):551:0\> cba {} > =\> 222 Yes. Ignoring implementation details, the block parameter ‘p’ is given a reference to the same object. When you then assign to ‘p’ that reference is replaced by a new reference to the object you assigned. Note that this is only for ‘p’ - ‘ppp’ still has the same reference it always had (to the original object). The reality is apparently slightly more complex but I believe that for most practical purposes (including this)you can ignore that. > irb(main):564:0\> xxx {} > =\> [2] > irb(main):567:0\> > > and I kind of understand why it works > > if I change line 566 to > irb(main):566:0\> xxx {|x| x=[2] } > than it would not work Correct. The original code doesn’t actually assign anything to ‘x’, but instead calls the []= method on it, which modifies the array’s content. Since both ‘ppp’ and ‘x’ reference the same Array instance, your change makes it out of the block. In the second case, you _do_ assign to ‘x’, supplying a new array. ‘ppp’ retains it’s original value, so the new array is (almost) lost. Almost, because of course it’s not _quite_ lost at that point: ``` irb(main):001:0> def xxx irb(main):002:1> ppp = [1] irb(main):003:1> p ppp irb(main):004:1> ppp = yield ppp irb(main):005:1> p ppp irb(main):006:1> end => nil irb(main):007:0> xxx { |x| x = [2] } [1] [2] => nil ``` > are “local” (C++ background) > > If not local what are they considered to be then? They are local. I think that test is referring to this: ``` irb(main):017:0> def test(arg) irb(main):018:1> p arg irb(main):019:1> [1,2,3].select { |arg| arg % 2 == 0 } irb(main):020:1> p arg irb(main):021:1> end => nil irb(main):022:0> test("ten") "ten" 3 => nil ``` An interesting aside to this (IMHO) is this: ``` irb(main):018:0> class Demo irb(main):019:1> def last=(arg) irb(main):020:2> (@last ||= []) << arg irb(main):021:2> end irb(main):022:1> def all irb(main):023:2> @last irb(main):024:2> end irb(main):025:1> end => nil irb(main):026:0> d = Demo.new => # irb(main):027:0> ['one','two','three'].each { |d.last| } => ["one", "two", "three"] irb(main):028:0> d.last => ["one", "two", "three"] ``` Which I guess illustrates that block arguments are handled by assignment to the named variable (or method in this case), and should make more sense of the preceeding example… Cheers,