From: Rick DeNatale <rick.denatale@...>
Date: 2007-10-03T03:23:11+09:00
Subject: Re: got a range from an array, but looking for other ways of doing it

On 9/30/07, SpringFlowers AutumnMoon <summercoolness@gmail.com> wrote:
> SpringFlowers AutumnMoon wrote:
> > Simon Schuster wrote:
> >> a is an array. I'm trying to chop the first and last third off. or,
> >> choose only the middle third. can be a rough third.
>
> that's strange, if i do
>
> a = Array(1..100)
>     ==>[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1
> 6, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 3
> 1, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 4
> 6, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 6
> 1, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 7
> 6, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 9
> 1, 92, 93, 94, 95, 96, 97, 98, 99, 100]
>
> a[0..a.size/2] = a[a.size*2/3..-1] = nil
>     ==>nil
> a
>     ==>[52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64,
> 65, 66]
>
>
> a = Array(1..100)
>     ==>[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1
> 6, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 3
> 1, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 4
> 6, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 6
> 1, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 7
> 6, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 9
> 1, 92, 93, 94, 95, 96, 97, 98, 99, 100]
>
> a[0..a.size/3] = a[a.size*2/3..-1] = nil
>     ==>nil
> a
>     ==>[35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47,
> 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62,
> 63, 64, 65, 66]
>
>
> shouldn't it be  a[0..a.size/2] = a[a.size*2/3..-1] = nil
> really?
>
> because it is equiv to
>
> a[0..a.size/2] = (a[a.size*2/3..-1] = nil)
> a[0..a.size/2] = (nil)

No it's not  because the leftmost = in the first statement is'nt doing
what you think it is:

a = Array(1..100) # => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31,
32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48,
49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65,
66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82,
83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99,
100]

b = a[a.size*2/3..-1] = nil # => nil

a # => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18,
19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35,
36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52,
53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66]

qri Array#[]=
-------------------------------------------------------------- Array#[]=
     array[index]         = obj                     ->  obj
     array[start, length] = obj or an_array or nil  ->  obj or an_array
     or nil
     array[range]         = obj or an_array or nil  ->  obj or an_array
     or nil

Note that Array#[]= returns its second argument, NOT the resulting array.

ALSO

 The value of a ruby assignment expression is the right hand side, NOT
the left hand side.

class Foo
  def a=(b)
    10
  end
end

f = Foo.new

f.send(:a=, 3) # => 10
b = f.a = 3 # => 3


-- 
Rick DeNatale

My blog on Ruby
http://talklikeaduck.denhaven2.com/