CODEWITHSUNDEEP
rubyruby-1.9.3
DoLoveSky
DoLoveSky

Reputation: 777

How do \ / operator work in ruby Regular expression?

I am using Ruby 1.9.3. Just going thorugh the Ruby tutorials. Now I just got stuck to a statement on which regular expression is working and giving out put also. But confusion with the \/ operators logic.

RegExp-1

Today's date is: 1/15/2013. (String)

(?<month>\d{1,2})\/(?<day>\d{1,2})\/(?<year>\d{4}) (Expression)

RegExp-2

s = 'a' * 25 + 'd' 'a' * 4 + 'c' (String)

/(b|a+)*\/ =~ s #=> ( expression)

Now couldn't understand how \/ and =~ operator works in Ruby.

Could anyome out of here help me to understand the same?

Thanks

Upvotes: 0

Views: 115

Answers (1)

Eric
Eric

Reputation: 636

\ serves as an escape character. In this context, it is used to indicate that the next character is a normal one and should not serve some special function. normally the / would end the regex, as regex's are bookended by the /. but preceding the / with a \ basically says "i'm not telling you to end the regex when I use this /, i want that as part of the regex."

As Lee pointed out, your second regex is invalid, specifically because you never end the regex with a proper /. you escape the last / so that it's just a plaintext character, so the regex is hanging. it's like doing str = "hello.

as another example, normally ^ is used in regex to indicate the beginning of a string, but doing \^ means you just want to use the ^ character in the regex.

=~ says "does the regex match the string?" If there is a match, it returns the index of the start of the match, otherwise returns nil. See this question for details.

EDIT: Note that the ?<month>, ?<day>, ?<year> stuff is grouping. seems like you could use a bit of brush-up on regex, check out this appendix of sorts to see what all the different special characters do.

Upvotes: 2

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