Is it going to have to physically relocate everything after 2 in memory?
Would using a linked list be more efficient if a lot of elements are
going to be removed and inserted?
slice only returns the value within the range or at index but does not
modify the original array. slice![1] would effectively remove value at index
1 from arr.
I don't know how it is implemented.
···
On Sat, Mar 14, 2009 at 3:12 PM, Peter Wu <pwu.380@gmail.com> wrote:
As in how they are organized behind the scenes.
For example, if I say
arr = [1, 2, 3, 4]
arr.slice[1]
Is it going to have to physically relocate everything after 2 in memory?
Would using a linked list be more efficient if a lot of elements are
going to be removed and inserted?
--
Posted via http://www.ruby-forum.com/\.
arr.slice[1] won't work since slice requires arguments, and this is being
interpreted as
(arr.slice)[1]
Is it going to have to physically relocate everything after 2 in memory?
Would using a linked list be more efficient if a lot of elements are
going to be removed and inserted?
So far we haven't done anything which would add or remove anything
arr.slice(1) simply returns a reference to the object in the second element
of arr. arr is untouched
arr.slice(1,3) would return an array containing the second, third and fourth
elements, the original array would be untouched. AND
The MRI ruby implementation of arrays uses a copy on write scheme, so that
both arr and the returned object would actually share memory for the array
of references. If one or the other changed then a copy would be triggered.
Ruby arrays are implemented rather efficiently both in terms of space and
speed. It's usually best to use them simply and only worry about performance
issues when they arise, and when they do, it will probably be something
other than "I should have used a linked list instead of an array"
···
On Sat, Mar 14, 2009 at 10:12 AM, Peter Wu <pwu.380@gmail.com> wrote:
b = a.slice!(0,3)
will end up with a == [4,5] and b == [1,2,3]
b = a.slice(0,3)
will end up with a == [1,2,3,4,5] and b == [1,2,3]
for the last one you can also use b = a[0..2] , you will get the same
result
···
On Mar 14, 2:12 pm, Peter Wu <pwu....@gmail.com> wrote:
As in how they are organized behind the scenes.
For example, if I say
arr = [1, 2, 3, 4]
arr.slice[1]
Is it going to have to physically relocate everything after 2 in memory?
Would using a linked list be more efficient if a lot of elements are
going to be removed and inserted?
--
Posted viahttp://www.ruby-forum.com/.