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For expression (?<!foo),Negative Lookbehind ,asserts that what immediately precedes the current position in the string is not foo.
Now to match string not ending with characters abc.
The formal expression is \w+(?<!abc)\b or \b\w+(?<!abc)\b. 

echo "xxabc jkl" | grep -oP  '\w+(?<!abc)\b'
jkl

Let's try another form: \w+\b(?<!abc). 

echo "xxabc jkl" | grep -oP  '\w+\b(?<!abc)'
jkl

It get correct answer too.
In my opinion, \w+\b(?<!abc) is not a Negative Lookbehind,nothing behind (?<!abc),it can't called Negative Lookbehind.
1.Is expression \w+\b(?<!abc) a Negative Lookbehind?
2.Why \w+\b(?<!abc) has the same effct as \w+(?<!abc)\b?

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    Keep in mind that `\b` is a zero-width match - it doesn't matter whether it comes before or after, the lookbehind will be checked for a match at exactly the same point in the string. – jasonharper Aug 12 '17 at 05:38

1 Answers1

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1)It is a negative lookback window. Negative lookback is used if you want to match something not proceeded by something else. ?<!abc is a negative look back window. So it will only be true if the proceeding tokens are not "abc" which is true with the string provided: "xxabc jkl" technically, a space is proceeding "jkl" Therefore it matches.

2) \b looks for a word boundary. Word boundaries apply for spaces and end of strings which are present in both of your regular expressions. If you really want to see a difference, try taking the space out of the input. So use "xxabcjkl"

shockawave123
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