Re: Re: Fun With Spaceship

I always wondered what assigning to $x[-1] would do in Perl.

So i started playing around with this...

Works:

$ perl -e 'my @arr=(1,2,3); $arr[-1]=4; print join " ", @arr, "\n";'
1 2 4
$
[download]

Doesn't work:

$ perl -e 'my @arr; $arr[-1]=4; print join " ", @arr, "\n";'
Modification of non-creatable array value attempted, subscript -1 at -
+e line 1.
$
[download]

Does this mean that perl does not create the array yet but merely creates a 'variable of type array' when doing my @array ?(kindof like Vector myVector; in java). It would only create the array and start allocating memory to it when you start inserting elements to that array ..

Jorg

"Do or do not, there is no try" -- Yoda

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(tye)Re: Fun With Spaceship by tye (Sage) on Mar 29, 2001 at 23:27 UTC
`$a[-1]` is "the last element of @a". In the failure case, @a has no elements and so has no last element. I suppose you could argue that it should "create the last element". However, what would you have the following do? `my @a= ( 0, 1 ); $a[-1]= -1; # @a is now ( 0, -1 ) $a[-2]= -2; # @a is now ( -2, -1 ) $a[-3]= -3; # fatal error` [download] Feel free to submit a patch that does unshift for that last line and that for `my @a; $a[-100]= 1` produces `@a= (1,(undef)x99)`. ;) - tye (but my friends call me "Tye")	[reply] [d/l] [select]

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(tye)Re: Fun With Spaceship
by tye (Sage) on Mar 29, 2001 at 23:27 UTC

$a[-1] is "the last element of @a". In the failure case, @a has no elements and so has no last element. I suppose you could argue that it should "create the last element". However, what would you have the following do?

my @a= ( 0, 1 );
$a[-1]= -1;    # @a is now ( 0, -1 )
$a[-2]= -2;    # @a is now ( -2, -1 )
$a[-3]= -3;    # fatal error
[download]

unshift

my @a; $a[-100]= 1

@a= (1,(undef)x99)

tye

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