When use sed sub-command 'n', the GNU sed doesn't print extra character
because n flushed the pattern space.
In busybox, sed would print extra character when n hits the EOF.
Because in "case 'n'", when the next_line is null, it wouldn't execute the
break, it would go the "case 'q'" and goto the discard_commands, in
discard_commands it output the old pattern space that contain the last
character, so we see it print extra character.
------------------------------------------------------------------------
case 'n':
if (!G.be_quiet)
sed_puts(pattern_space, last_gets_char);
if (next_line) {
free(pattern_space);
pattern_space = next_line;
last_gets_char = next_gets_char;
next_line = get_next_line(&next_gets_char, &last_puts_char,
last_gets_char);
substituted = 0;
linenum++;
break;
} else # add by myself
goto again; # add by myself
/* fall through */
/* Quit. End of script, end of input. */
case 'q':
/* Exit the outer while loop */
free(next_line);
next_line = NULL;
goto discard_commands;
------------------------------------------------------------------------
So in order to comply with GNU sed, in case 'n', when the next_line is null,
we add "else" at the end of the second "if":
goto again;
to avoid print the old pattern space which have been printed.
Signed-off-by: Dengke Du <[email protected]>
---
editors/sed.c | 3 ++-
testsuite/sed.tests | 6 +-----
2 files changed, 3 insertions(+), 6 deletions(-)
diff --git a/editors/sed.c b/editors/sed.c
index 7bbf820..2e4a805 100644
--- a/editors/sed.c
+++ b/editors/sed.c
@@ -1282,7 +1282,8 @@ static void process_files(void)
substituted = 0;
linenum++;
break;
- }
+ } else
+ goto again;
/* fall through */
/* Quit. End of script, end of input. */
diff --git a/testsuite/sed.tests b/testsuite/sed.tests
index 34479e5..96ff7a5 100755
--- a/testsuite/sed.tests
+++ b/testsuite/sed.tests
@@ -73,13 +73,9 @@ testing "sed t (test/branch clears test bit)" "sed -e
's/a/b/;:loop;t loop'" \
testing "sed T (!test/branch)" "sed -e 's/a/1/;T notone;p;: notone;p'" \
"1\n1\n1\nb\nb\nc\nc\n" "" "a\nb\nc\n"
-test x"$SKIP_KNOWN_BUGS" = x"" && {
-# Normal sed end-of-script doesn't print "c" because n flushed the pattern
-# space. If n hits EOF, pattern space is empty when script ends.
-# Query: how does this interact with no newline at EOF?
testing "sed n (flushes pattern space, terminates early)" "sed -e 'n;p'" \
"a\nb\nb\nc\n" "" "a\nb\nc\n"
-}
+
# non-GNU sed: N does _not_ flush pattern space, therefore c is eaten @ script
end
# GNU sed: N flushes pattern space, therefore c is printed too @ script end
testing "sed N (flushes pattern space (GNU behavior))" "sed -e 'N;p'" \
--
2.8.1
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