Hello,
In `ri 'Array#*'`, it says that the result is the concatenation of copies of self if the argument is an int.
So, it is "COPY".
Then why something like '[[0]] * 10', returns an array, that each element is a pointer to the first element?
Does this mean "copy of an Array" is just a duplicated pointer?
Thanks.
I think that here the word "copy" has the same meaning it has in clone and
dup, that is it means a shallow copy: the array itself is copied, but its
contents aren't. In other words, what you get using Array#* with a number is
an array of size n times the original one, whose contents are the contents of
the original array repeated n times.
I hope this helps
Stefano
On Monday 24 March 2008, Magicloud Magiclouds wrote:
Hello,
In `ri 'Array#*'`, it says that the result is the concatenation of
copies of self if the argument is an int.
So, it is "COPY".
Then why something like '[[0]] * 10', returns an array, that each
element is a pointer to the first element?
Does this mean "copy of an Array" is just a duplicated pointer?Thanks.
In source, I got this: MEMCPY(RARRAY(ary2)->ptr+i, RARRAY(ary)->ptr, VALUE, RARRAY(ary)->len);
So, I just got a bunch of pointer to the same object....
Stefano Crocco wrote:
On Monday 24 March 2008, Magicloud Magiclouds wrote:
Hello,
In `ri 'Array#*'`, it says that the result is the concatenation of
copies of self if the argument is an int.
So, it is "COPY".
Then why something like '[[0]] * 10', returns an array, that each
element is a pointer to the first element?
Does this mean "copy of an Array" is just a duplicated pointer?Thanks.
I think that here the word "copy" has the same meaning it has in clone and dup, that is it means a shallow copy: the array itself is copied, but its contents aren't. In other words, what you get using Array#* with a number is an array of size n times the original one, whose contents are the contents of the original array repeated n times.I hope this helps
Stefano