Ok, so I did
a = Array.new(number, [])
It’s really bothering me that if I do a[i] = n, all of a is now n.
Is there a graceful way to get around this?
-Kurt
Kurt M. Dresner wrote:
Ok, so I did
a = Array.new(number, )
It’s really bothering me that if I do a[i] = n, all of a is now n.
Is there a graceful way to get around this?
When I do:
a = Array.new(6, )
a[5] = 3
p a[2]
I get printed out.
- Dan
Kurt M. Dresner wrote:
Ok, so I did
a = Array.new(number, )
It’s really bothering me that if I do a[i] = n, all of a is now n.
Is there a graceful way to get around this?
I guess you want
Hi Kurt,
Ok, so I did
a = Array.new(number, )
It’s really bothering me that if I do a[i] = n, all of a is now n.
Is there a graceful way to get around this?
I’m not sure what you are trying to do here. Maybe you could give a more
detailed description of your problem?
If you are trying to create a new array you have a couple of options.
The default way to construct an array:
a = [1,2,3]
If you want to create an empty array that is a certain size:
a = Array.new(5) #=> [nil, nil, nil, nil, nil]
If you want to create an array that is a certain size and all the elements
have the same value:
a = Array.new(3, “pickles”) #=> [“pickles”, “pickles”, “pickles”]
···
–
John Long
Dan Doel djd15@po.cwru.edu skrev den Sun, 14 Sep 2003 09:11:16 +0900:
Kurt M. Dresner wrote:
Ok, so I did
a = Array.new(number, )
It’s really bothering me that if I do a[i] = n, all of a is now n.
Is there a graceful way to get around this?
When I do:
a = Array.new(6, )
a[5] = 3
p a[2]
I get printed out.
Yeah, but what about
a = Array.new(6, )
a[5] << 3
p a[2]
? 
I think the original poster might wanna try
a = Array.new(6) {Array.new}
but I’m not sure I understand what he wants…
Regards,
Robert
···
- Dan
–
Robert Feldt
Yeah, that’s what I meant. When I do a[i].push(foo) then all of a is
foo. I forgot about the block for the constructor that gets run each
time. Now I am a happy rubyist once again.
-Kurt
···
On Sun, Sep 14, 2003 at 09:25:37AM +0900, Robert Feldt wrote:
Dan Doel djd15@po.cwru.edu skrev den Sun, 14 Sep 2003 09:11:16 +0900:
Kurt M. Dresner wrote:
Ok, so I did
a = Array.new(number, )
It’s really bothering me that if I do a[i] = n, all of a is now n.
Is there a graceful way to get around this?
When I do:
a = Array.new(6, )
a[5] = 3
p a[2]
I get printed out.
Yeah, but what about
a = Array.new(6, )
a[5] << 3
p a[2]?
I think the original poster might wanna try
a = Array.new(6) {Array.new}
but I’m not sure I understand what he wants…
Regards,
Robert
- Dan
–
Robert Feldt======= End of Original Message =======<
Just for completeness, the problem with Array.new(6, ) isn’t Ruby’s
object model - it’s that Array.new(num, object) stores a reference to
the object passed in in each of its cells, not a copy of it. There’s no
ruby-decreed reason it couldn’t have been written to call object.dup
each time it filled a new cell.
martin
···
Kurt M. Dresner kdresner@cs.utexas.edu wrote:
Yeah, that’s what I meant. When I do a[i].push(foo) then all of a is
foo. I forgot about the block for the constructor that gets run each
time. Now I am a happy rubyist once again.