I think the semantics of the escaping operator \ (backslash) are not documented
clearly. Especially I miss this ones:
\$foo ==> $foo
this seems inconsistent to me. I expect
I understand that
raises an error since there are no : (colons)
allowed in a variable-name.
But why raises
also an error? I would expect that \$ disables the function of $
as an operator, so that the following
will not be parsed as
a variable name.
There should be a simple way to generate something like
is undocumented and REALLY ugly.