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Can you use the scalar range operator with map and grep? Sure, but you have to be careful.

This does what you'd expect:

grep $_ == 3 .. $_ == 7, 1..10 # ==> (3, 4, 5, 6, 7)
because grep's first argument is evaluated as a scalar.

This is very puzzling:

map $_ == 3 .. $_ == 7, 1..10 # ==> (0, 0, 0, 0, 0, 0, 1, 0, 0, 0)
until you remember that map's first argument is evaluated as a list. What you meant, of course, was:
map scalar($_ == 3 .. $_ == 7), 1..10 # ==> (,,1,2,3,4,5E0,,,)
-- the expected result.

The puzzling case above gets even more puzzling if we show the list that map returns for each input:

map $_ == 3 .. $_ == 7, 1 # ==> (0) map $_ == 3 .. $_ == 7, 2 # ==> (0) map $_ == 3 .. $_ == 7, 3 # ==> () map $_ == 3 .. $_ == 7, 4 # ==> (0) map $_ == 3 .. $_ == 7, 5 # ==> (0) map $_ == 3 .. $_ == 7, 6 # ==> (0) map $_ == 3 .. $_ == 7, 7 # ==> (0, 1) map $_ == 3 .. $_ == 7, 8 # ==> (0) map $_ == 3 .. $_ == 7, 9 # ==> (0) map $_ == 3 .. $_ == 7, 10 # ==> (0)
we get the same ten elements as before, though in an unexpected way. The first line is evaluating:
1 == 3 .. 1 == 7
which is
"" .. ""
Remember that this is being evaluated in list context (because we forgot the scalar()). So, our scalar range operator has accidentally become a list range operator. The final clue is that the list range operator only does string magic (e.g. ("a".."z")) if the strings are non-empty. Otherwise the arguments are converted to integers, in this case zeros:
0 .. 0
which is just a list containing a single zero, as we saw above.

The other cases are now easy to understand.

map $_ == 3 .. $_ == 7, 3
is (1 .. ""), which is converted to (1 .. 0), which gives an empty list. Similarly,
map $_ == 3 .. $_ == 7, 7
is ("" .. 1) which gives (0, 1).


In reply to Re: The Scalar Range Operator by denishowe
in thread The Scalar Range Operator by pbeckingham

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