What string pattern will match an explicit asterisk (*) or at-sign (@)? The obvious example, StringMatchQ["*", "\*"], returns False. I can't find any mention of escaping them in string patterns, only in RegularExpressions, and that was in the advanced tutorial.
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$\begingroup$ Related: mathematica.stackexchange.com/questions/22707/… and mathematica.stackexchange.com/questions/4608/… $\endgroup$ – Sjoerd C. de Vries May 23 '13 at 21:05
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I thought it worth mentioning that the documentation for StringMatchQ gives the following solution to your problem:
Verbatim["p"]specifies the verbatim string"p", with*and@treated literally.
So you'd be able to do something like:
StringMatchQ["*", Verbatim["*"]]
(* True *)
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$\begingroup$ It also mentions
\\*explicitly in the table directly above that. $\endgroup$ – Brett Champion Apr 2 '15 at 18:52 -
$\begingroup$ @BrettChampion It's not in the String Patterns tutorial though, which is where I'd have looked. I didn't put it in my answer because Sjoerd already covered that pretty well. $\endgroup$ – 2012rcampion Apr 2 '15 at 19:13
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Yes, you need to escape a * character in a string pattern because it's a wildcard. To escape it, just prepend a backslash to it. Remember that to insert a backslash in a Mathematica string, you need to type two backslashes:
StringMatchQ["*", "\\*"]
(* => True *)
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$\begingroup$ Come to think about it, this is pretty deep stuff. In a normal string, an asterisk isn't a special character that needs to be escaped, so
"*"is just the character with character code 42, and"\*"is a single character with code 63432."\\*"equals {92, 42}, and only because it is used in a string pattern (where*is special) it is interpreted as an escape sequence and equals the literal"*"(42). Note that"*"=="\\*"yieldsFalse(whereas"*"=="*"yieldsTrue). $\endgroup$ – Sjoerd C. de Vries May 23 '13 at 21:59 -
$\begingroup$ If you use regexes in shell scripts a lot, you'll get pretty familiar with this type of double (or triple) escaping =) $\endgroup$ – 2012rcampion Apr 2 '15 at 16:44
