Re: Re: Changing each item in an array

diotalevi,
I know that this was probably a contrived example by the Anonymous Monk to illustrate the problem, but in this case I would probably:

for my $item ($one, $two, $three) {
    $item = $item ? 'yes' : 'no';
}
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This gets rid of the array all together and the need to dereference. I realize that the for loop is magically ~~creating the reference and doing the dereference~~ implicitly aliasing for you, but it is transparent. Hmmm - may I should get rid of that ternary operator if my point was for simplicity/clarity to the Anonymous Monk - nah!

Cheers - L~R
updated: Modified verbiage per diotalevi's clarification

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Re: Re: Re: Changing each item in an array by diotalevi (Canon) on Jul 19, 2003 at 00:12 UTC
No, foreach doesn't create a reference and dereference for you. Really. Its an internals thing and it aliases the loop variable. This is really outside the bounds of what I think will confuse the original poster though. I also would normally have used ternary assignment but, again, in the interests of speaking to the original poster's knowledge level I kept the same if/else syntax and just changed the bits that needed to be.	[reply]

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Re: Re: Re: Changing each item in an array
by diotalevi (Canon) on Jul 19, 2003 at 00:12 UTC

No, foreach doesn't create a reference and dereference for you. Really. Its an internals thing and it *aliases* the loop variable. This is really outside the bounds of what I think will confuse the original poster though. I also would normally have used ternary assignment but, again, in the interests of speaking to the original poster's knowledge level I kept the same if/else syntax and just changed the bits that needed to be.

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