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);
- substituted = 0;
- linenum++;
- break;
+ if (next_line == NULL) {
+ /* If no next line, jump to end of script and exit. */
+ goto discard_line;
}
- /* fall through */
+ free(pattern_space);
+ pattern_space = next_line;
+ last_gets_char = next_gets_char;
+ next_line = get_next_line(&next_gets_char, &last_puts_char);
+ substituted = 0;
+ linenum++;
+ break;
/* Quit. End of script, end of input. */
case 'q':
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'" \