You did successfully modify the value of what pos() is locally bounded to, but it is just your own local copy, which only exists and is meaningful within your {} scope. When you reenter that {} scope, the value for pos() is taken from what is stored by regexp engine itself.

This does make sense, and it makes the regexp itself more robust. 
$\ = "\n";
$, = ",";
print 'a string' =~ m/(.)(?{print "before mod, pos = ".pos();pos()+=4;
+print "after mod, pos = ".pos();})/g;
print pos(); #cause error, it is undef, as now we are inside any regex
+p

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In reply to Re: immutable pos() inside regex by pg
in thread immutable pos() inside regex by diotalevi

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