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)
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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)
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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,,,)
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-- 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)
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we get the same ten elements as before, though in an unexpected way. The first line is evaluating:

1 == 3 .. 1 == 7
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which is

"" .. ""
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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
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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
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is (1 .. ""), which is converted to (1 .. 0), which gives an empty list. Similarly,

map $_ == 3 .. $_ == 7, 7
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is ("" .. 1) which gives (0, 1).