local($/ = chr 4) works fine. "\x04" is assigned to $/, $/ is returned by the assignment, then $/ is localized.

(local $/) = chr 4; would be the proper parenthesisation if you want the localisation to occur first.

the parenthesis following a function are optional, but in some cases, they seem to be downright forbidden.

local($/) = chr 4; would also be valid, but it has different semantics than local $/ = chr 4;

local ($s) is considered a list when found on the LHS of an assignment, causing the list assignment (aassign) to be used. local $s is considered to be a scalar, so a scalar assignment (sassign) would be used then.

$ perl -MO=Concise,-exec -e'local $s = "foo"' 2>&1 | grep assign
5  <2> sassign vKS/2

$ perl -MO=Concise,-exec -e'local ($s) = "foo"' 2>&1 | grep assign
7  <2> aassign[t2] vKS
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In practice:

$ perl -le'my $x = local  $s  = "foo"; print $x'
foo

$ perl -le'my $x = local ($s) = "foo"; print $x'
1
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my, eof and use also vary in semantics based on the presence of parens.

There is another area where I've noticed that parenthesis seem to be forbidden

Red herring. It has nothing to do with parenthesis. map has two calling conventions, both of which work with and without parens:

• map EXPR, LIST
• map BLOCK LIST

Your map has a block *and* a comma, which is wrong.

map( { print $_."\n" } @list );  # Block and no comma works.
map( print($_."\n"),   @list );  # Expression with comma works.

map { print $_."\n" } @list;     # Block and no comma works.
map print($_."\n"),   @list;     # Expression with comma works.
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By the way, why are you using map in void context?

print "$_\n" for @list;
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Ah hah!! ++ Thank you! Now all is clear..

I'm not actually using map in this way, I just remembered having this comma confusion and that was the simplest example I could conjure as a demonstration.