regex - How negative lookbehind works in below example? -

why on applying regular expression(rx) on data(d) gives output(o) ?
regular expression (rx):

s/(?<!\#include)[\s]*\<[\s]*([^\s\>]*)[\s]*\>/\<$1\>/g

data (d):

#include  <a.h>  // 2 spaces after e

output (o):

#include <a.h>  // 1 space still there

expected output is:

#include<a.h>  // no space after include

the condition (?<!\#include) true you've passed first of 2 spaces, therefore match starts there.

#include  <a.h>          ^^^^^^- matched regex.

that means space not removed replace operation.

if use positive lookbehind assertion instead, desired result:

s/(?<=#include)\s*<\s*([^\s>]*)\s*>/<$1>/g;

which can rewritten use more efficient \k:

s/#include\k\s*<\s*([^\s>]*)\s*>/<$1>/g;

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