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Introduce dollar sign escaping #147
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Could you walk me through why only half the back-slashes are required? I'm not sure I understand.
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That's what happens in regular js string escaping too.
\\
gets to be a\
. This is needed to allow somebody to write a\
in front of a$VARIABLE
and still replace the variable.\$VAR
->$VAR
\\$VAR
->\value
\\\$VAR
->\\$VAR
\\\\$VAR
->\\value
and so on.
Hope this makes it clear
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OK. Thank you. What about the other case? If the number of back-slashes is odd, isn't this code going to replace the expression by
$VAR
every time? Shouldn't some back-slashes be retained?Also I'm not sure your examples are quite consistent. In
\$VAR
->$VAR
, the back-slash is escaping the dollar sign. In\\$VAR
->\value
the first back-slash escapes the second and the dollar sign is not escaped. Sounds fine up to there. Then in\\\$VAR
->\\$VAR
, the first black-slash escapes the second, then the third escapes the dollar sign. So the result should be\$VAR
, shouldn't it? Maybe you made a typo? Fourth example looks right.