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Re^4: About List::Util's pure Perl shuffle()by ikegami (Patriarch) |
on Jul 12, 2007 at 13:29 UTC ( [id://626233]=note: print w/replies, xml ) | Need Help?? |
It's similar to print($i, $i++, $i);, at least in effect. Perl must place some form of pointer* to $a[$n]'s value on a stack instead of making a new scalar to save speed and memory. The result is that when $a[$n] is changed by the third expression in the list slice, it also changes the value of the second expression. By explicitly copying the value of $a[$n] into another variable ($x), changes in $a[$n] no longer affect the result of the slice. * — An alias? A pointer to the SV?
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