But wouldn't the original initilization also work?
@array1[0..5] = 1;
No. Think of this in terms of parallel assignment, ignoring for the moment that we are talking of arrays, the above is equivalent to:
($e0, $e1, $e2, $e3, $e4, $e5) = 1;
which is equivalent to:
($e0, $e1, $e2, $e3, $e4, $e5) = (1);
because the list on the left-hand side of the assignment operator ('='), puts the right-hand side in list context. This is then filled in by perl to become:
($e0, $e1, $e2, $e3, $e4, $e5) = (1, undef, undef, undef, undef, undef);
because the list on the right-hand side must have the same number of elements as the list on the left-hand side.
The common idiom to populate an array with a particular value is to use the times operator ('x') which reproduces its left-hand argument the number of times specified by its right-hand argument:
($e0, $e1, $e2, $e3, $e4, $e5) = (1) x 6;
which is equivelent to:
($e0, $e1, $e2, $e3, $e4, $e5) = ((1), (1), (1), (1), (1), (1));
which then gets flatened to:
($e0, $e1, $e2, $e3, $e4, $e5) = (1, 1, 1, 1, 1, 1);
With an array slice, perl basically turns the original:
@array1[0..5] = 1;
into something like:
($array1[0], $array1[1], $array1[2], $array1[3], $array1[4], $array1[5]) = (1, undef, undef, undef, undef, undef);
in the same manner as the first example above.
Regards, Randy.
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