let me just give a blip of what i am getting...
a = Array.new(3,)
=> [, , ]
a[2] << "check"
=> ["check"]
a
=> [["check"], ["check"], ["check"]]
this is totally not what i am expecting..
what i am expecting is:
[,,[["check"]]]
as in..
the 2 element should have the "check" string dropped into the next
available position..
how would i go about setting that (or any other) element only?
thanks!
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Sergio Ruiz wrote:
let me just give a blip of what i am getting...
a = Array.new(3,)
=> [, , ]
You've made an array containing three copies of a reference to the same
array instance.
irb(main):001:0> a = Array.new( 3, )
=> [, , ]
irb(main):002:0> a.each_with_index { |x,i| puts "#{i}: #{x.object_id}" }
0: 8484280
1: 8484280
2: 8484280
I believe you'd want:
a = Array.new( 3 ) { |idx| Array.new() }
That will call the block once for each element, and that block will
create a new (different) array instance each time it's called.
irb(main):001:0> a = Array.new( 3 ) { |idx| Array.new }
=> [, , ]
irb(main):002:0> a[2] << "foo"
=> ["foo"]
irb(main):003:0> a
=> [, , ["foo"]]
irb(main):004:0> a.each_with_index { |x,i| puts "#{i}: #{x.object_id}" }
0: 2741440
1: 2741430
2: 2741420
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Alle venerdì 18 maggio 2007, Sergio Ruiz ha scritto:
let me just give a blip of what i am getting...
>> a = Array.new(3,)
=> [, , ]
>> a[2] << "check"
=> ["check"]
>> a
=> [["check"], ["check"], ["check"]]
this is totally not what i am expecting..
what i am expecting is:
[,,[["check"]]]
as in..
the 2 element should have the "check" string dropped into the next
available position..how would i go about setting that (or any other) element only?
thanks!
Look at the documentation for Array.new (ri Array.new). Array.new(n, obj)
creates an array with n entries, all containing the same object, obj. You can
see this comparing the entries' object_id, or using equal?: a[0].equal?(a[1])
=> true. Since all entries contain the same object, when you modify it, the
change shows everywhere in the array.
To fill the array with *different* empty arrays, you need to use the form of
Array.new which takes a block:
Array.new(3){}
In this form, the block is called for each index, with the index as argument
(you can omit it here because you don't need it). The result of the block is
stored in the corresponding entry. The point is that each time the block is
called, a *new* empty array is created and stored in the returned array. This
time, a[0].equal?(a[1]) gives false.
I hope this helps
Stefano
I don't see why. You initialized an array with three copies of the SAME empty
array.
-s
In message <bcda57230145bdb94ab5011f83a1a769@ruby-forum.com>, Sergio Ruiz writes:
a = Array.new(3,)
=> [, , ]
a[2] << "check"
=> ["check"]
a
=> [["check"], ["check"], ["check"]]
this is totally not what i am expecting..
You've made an array containing three copies of a reference to the same
array instance.
got it!
dangit..
this tripped me up all day..
thanks!
--
Posted via http://www.ruby-forum.com/\.
You could just do {} for the block. It DWYM.
-s
In message <f0314c659cd783e3f1a655cd7a5f9451@ruby-forum.com>, Mike Fletcher writes:
a = Array.new( 3 ) { |idx| Array.new() }
That will call the block once for each element, and that block will
create a new (different) array instance each time it's called.
Heh I remember when I made this mistake 
On 18/05/07, Sergio Ruiz <sergio@village-buzz.com> wrote:
> You've made an array containing three copies of a reference to the same
> array instance.got it!
dangit..
this tripped me up all day..
thanks!
--
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