grep: return code with -v and empty input

[Edward Avis]

According to the documentation, the -v flag is supposed to invert the
exit status:

>Normally, exit status is 0 if matches were found, and 1 if no matches
>were found (the `-v' option inverts the sense of the exit status).

However it doesn't seem to work like that for empty input.  Here's some
commands I ran, the shell I used was bash2:

% if echo a | grep -q a; then echo found; else echo not found; fi
found
% if echo b | grep -q a; then echo found; else echo not found; fi
not found
% if : | grep -q a; then echo found; else echo not found; fi
not found

So far so good - the pattern was 'not found' in empty input, as you'd
expect.  But what about with the -v flag and the if-test inverted?

% if echo a | grep -qv a; then echo not found; else echo found; fi
found
% if echo b | grep -qv a; then echo not found; else echo found; fi
not found
% if : | grep -qv a; then echo not found; else echo found; fi
found

I suspect this may be a documentation bug; perhaps it should say
something like:

'If the -v flag is given, then a "matching line" is one which does _not_
match the pattern.  The exit status is still 0 if there were no
"matching lines" and 1 if there were some.'

-- 
Ed Avis
address@hidden