printf: Ignore any options

This was misguidedly "fixed" in
9e08609f85, which made printf error out
with any "-"-prefixed words as the first argument.

Note: This means currently `printf --help` doesn't print the help.
This also matches `echo`, and we currently don't have anything to make
a literal `--help` execute a builtin help except for keywords. Oh well.

Fixes #9132
This commit is contained in:
Fabian Boehm 2022-08-10 16:55:56 +02:00
parent c288443b4d
commit 37f7818bbb
2 changed files with 17 additions and 11 deletions

View File

@ -649,19 +649,10 @@ int builtin_printf_state_t::print_formatted(const wchar_t *format, int argc, con
maybe_t<int> builtin_printf(parser_t &parser, io_streams_t &streams, const wchar_t **argv) {
const wchar_t *cmd = argv[0];
int argc = builtin_count_args(argv);
help_only_cmd_opts_t opts;
int optind;
int retval = parse_help_only_cmd_opts(opts, &optind, argc, argv, parser, streams);
if (retval != STATUS_CMD_OK) return retval;
argv++;
argc--;
if (opts.print_help) {
builtin_print_help(parser, streams, cmd);
return STATUS_CMD_OK;
}
argc -= optind;
argv += optind;
if (argc < 1) {
return STATUS_INVALID_ARGS;
}

View File

@ -119,3 +119,18 @@ printf '%d\n' 0g
# CHECKERR: 0g: value not completely converted (can't convert 'g')
echo $status
# CHECK: 1
# Test that we ignore options
printf -a
printf --foo
# CHECK: -a--foo
echo
set -l helpvar --help
printf $helpvar
echo
# CHECK: --help
printf --help
echo
# CHECK: --help