Specify exactly how RegExp.source should be escaped
allen at wirfs-brock.com
Tue Mar 19 09:23:56 PDT 2013
On Mar 19, 2013, at 8:05 AM, Simon Pieters wrote:
> The spec says about RegExp.source:
> The characters / or backslash \ occurring in the pattern shall be escaped
> in S as necessary to ensure that the String value formed by concatenating
> the Strings "/", S, "/", and F can be parsed (in an appropriate lexical
> context) as a RegularExpressionLiteral that behaves identically to the
> constructed regular expression. For example, if P is "/", then S could be
> "\/" or "\u002F", among other possibilities, but not "/", because ///
> followed by F would be parsed as a SingleLineComment rather than a
> RegularExpressionLiteral. If P is the empty String, this specification can
> be met by letting S be "(?:)".
> The source property of the newly constructed object is set to S.
> Why is the requirement so vague? I would like the spec to state exactly
> how source is to be escaped, maybe with an algorithm like:
Prior to ES5, the escaping wasn't even specified and the spec. simply said that the pattern was implementation defined. We could probably specify it, but somebody would need to develop a proposal that completely specifies the required escaping.
> 1. If S is the empty string, let S be "(?:)".
> 2. Replace all instances of "/" in S with "\/".
> 3. Replace all instances of literal new lines in S with ???
> 4. ???
> Currently, I have no idea what to check for when writing test cases for
> the .source property when testing e.g. empty string or a slash as P.
The key requirement in the specification is that a RegExp created from the .source property string behaves identically to the RegExp created from the original pattern. I would test this by developing a set of interesting patterns that require escaping and then testings that a RegExp created from their .source properties produces identical result to the original patterns.
> Simon Pieters
> Opera Software
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