Regex query

Hi,

A Simple question on regular expression in ruby. Some can make me
understand the below code snippets. I was under the impression that
both these statement would give the same answer. But it doesnt. Why ?

irb(main):001:0> ‘banana’ =~ /(an)*/
=> 0
irb(main):002:0> ‘banana’ =~ /(an)+/
=> 1

-Manjo

On 1/10/07, Manoj P M [email protected] wrote:

A Simple question on regular expression in ruby. Some can make me
understand the below code snippets. I was under the impression that
both these statement would give the same answer. But it doesnt. Why ?

irb(main):001:0> ‘banana’ =~ /(an)*/
=> 0

In this case you are looking for ‘’, ‘an’, ‘anan’, ‘ananan’ and so on.
So it matches at the beginning of the string (index 0) with the empty
string.

Luis

On 1/10/07, Manoj P M removed_email_address[email protected] wrote:

Hi,

A Simple question on regular expression in ruby. Some can make me
understand the below code snippets. I was under the impression that
both these statement would give the same answer. But it doesnt. Why ?

irb(main):001:0> ‘banana’ =~ /(an)*/
=> 0

this is happening because * means 0 or more times match the preceding
literal. here always the regex match =~ returns the position in the
string
where it finds the match and starts from 0 (as in C and ruby the array
parameter starts count from 0 like if array=‘sachin’ and then array[0]
will
give ‘s’)
so in the first case it matches ‘b’ and returns zero since it tries to
match
‘an’ matched 0 times so returns the position taken by ‘b’

irb(main):002:0> ‘banana’ =~ /(an)+/

=> 1

in this case the place where it matches ‘an’ is starting at b(0) a(1)
n(2)
a(3) n(4) a(5) so returns 1 since ‘an’ is matched beginning at 1
location.

I hope this clarifies

-Manjo


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