java - How Regex backreference work -

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java - How Regex backreference work -

i aware of fact backreference ovverride values if backtraking occurs , ovverided output new backreference.

but if take regex example:

([abc]+).*\1

then string:"abc me bca"

the output is:"abc me bca"

can explain how possible because per steps:

-[abc] matches input. -because quantifier + repeate 1 or more time , 1 time again matches b , c.then stops @ whitespace it's not either 'a', 'b' or 'c'. -.* eats input string after abc , farther goes \1(the backreference). - .* backtracking \1 fails , because .* i.e 0 or more through charecters , 1 time again + of [abc]+ backtracking. -in backtracking of [abc]+ until 'a' after removing 'b' , 'c' still no match \1 bca.

so how output came "abc me bca"..?

first character a , lastly character a.so match.

what happens abc captured in grouping compared bca.it fails.

so engine backtracks 1.now ab compared ca.it fails.so engine

will backtrack again.a compared lastly a , passes.so engine

stores a in grouping satisfies match criteria .note \1 gets

stored in first group.it not fixed value.see demo.

https://regex101.com/r/sj9gm7/72

java regex

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