Java matcher returning extra "." in output

Question

It's my part of java program for project.

String ops="";
String input="1+3-4+(7/8)+cos(sin(50)+2)/2+tan(90)*e^(5+26-4/1*2)";
input=input.replaceAll("cos", "c").replaceAll("sin", "s").replaceAll("tan", "t").replace("e", "2.718");
//System.out.println(input);
Pattern pattern=Pattern.compile("[+-/*()cst^]");
Matcher matcher= pattern.matcher(input);
while (matcher.find()) {ops+=matcher.group();}
System.out.println(ops);

Here I am just reading the input and giving output of +,-,/,*,(,),c,s & t present in it. The output ops should return +-+(/)+c(s()+)/+t()*^(+-/*) while it's returning +-+(/)+c(s()+)/+t()*.^(+-/*).Help me understand the reason please.

From first glance it appears to be a result of subbing "2.718" in for "e", at least that is where the "." is coming from. — FreakyDan, Mar 04 '15 at 16:39

score 4 · Accepted Answer · edited May 23 '17 at 12:06

4

- is used for a range, so +-/ includes all characters between + and / (in the ascii table), i.e. ,-./. To solve the issue you can place the - in first position:

Pattern.compile("[-+/*()cst^]");

Or you could also escape it. More about this: How to match hyphens with Regular Expression?

edited May 23 '17 at 12:06

Community

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answered Mar 04 '15 at 16:40

assylias

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Alternatively `-` can be escaped by `\`. Obviously in java code `\` must be duplicated to pass java and regex escaping. – AlexR Mar 04 '15 at 16:43
So that's the reason.Thanks. – The_ehT Mar 04 '15 at 16:44
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@The_ehT yes it is in the character class section: `[a-zA-Z] a through z or A through Z, inclusive (range)` – assylias Mar 04 '15 at 16:45

Java matcher returning extra "." in output

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