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?

In reply to Re^4: About List::Util's pure Perl shuffle() by ikegami
in thread About List::Util's pure Perl shuffle() by blazar

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