Re: what does rand[] return?

Well if you do ;

print 0 + [10] ;
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you will see that perl treats the reference to the anonymous array [10] as a number . So what you do is like :

rand number
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In my case this number is 27783332 and the array reference is 0x1a7f0a4 ( that you can see by doing print [10]) which is 27783332 in hex ...

If i repeat execution i get the same reference over and over again for this anon array. I don't know why but that's another problem ;)

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Re^2: what does rand[] return? by Happy-the-monk (Canon) on Nov 30, 2005 at 11:53 UTC
If i repeat execution i get the same reference over and over again for this anon array. I don't know why I am as ignorant as you are, but if I was `perl`, I would make life easy and return the next possible free address that was big enough to store your anonymous array reference. If you freed that up - no longer use it - and asked for a new one, I would lazily give you just the one that I had before. Cheers, Sören	[reply]
Re^2: what does rand[] return? by Errto (Vicar) on Nov 30, 2005 at 16:27 UTC
If i repeat execution i get the same reference over and over again for this anon array. I don't know why but that's another problem ;) I'm not an internals guy but I would guess this is an intentional optimization, though Happy-the-monk may also be correct. It would rely on perl noticing the following things: the contents of this array are all constants no other references to this array are taken (i.e. its refcount drops to zero at the end of the loop) the array is not modified in any way Given all that, perl might allocate the array only once and intentionally reuse the same reference on each iteration. One way to find out might be to modify the loop so it creates a second array reference of the same size and then does something with it like store a reference somewhere, and see if that first array still gets the same address every time.	[reply]