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fix getopt[_long] clobbering of optopt on success

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getopt is only specified to modify optopt on error, and some software
apparently infers an error from optopt!=0.

getopt_long is changed analogously. the resulting behavior differs
slightly from the behavior of the GNU implementation of getopt_long,
which keeps an internal shadow copy of optopt and copies it to the
public one on return, but since the GNU implementation also exhibits
this shadow-copy behavior for plain getopt where is is non-conforming,
I think this can reasonably be considered a bug rather than an
intentional behavior that merits mimicing.
This commit is contained in:
Rich Felker
2017-01-04 19:43:59 -05:00
parent 150747b41e
commit 786fda875a
2 changed files with 5 additions and 2 deletions
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src/misc/getopt_long.c
+3 -1
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@@ -75,9 +75,9 @@ static int __getopt_long_core(int argc, char *const *argv, const char *optstring
if (cnt==1) {
i = match;
optind++;
optopt = longopts[i].val;
if (*opt == '=') {
if (!longopts[i].has_arg) {
optopt = longopts[i].val;
if (colon || !opterr)
return '?';
__getopt_msg(argv[0],
@@ -89,6 +89,7 @@ static int __getopt_long_core(int argc, char *const *argv, const char *optstring
optarg = opt+1;
} else if (longopts[i].has_arg == required_argument) {
if (!(optarg = argv[optind])) {
optopt = longopts[i].val;
if (colon) return ':';
if (!opterr) return '?';
__getopt_msg(argv[0],
@@ -107,6 +108,7 @@ static int __getopt_long_core(int argc, char *const *argv, const char *optstring
return longopts[i].val;
}
if (argv[optind][1] == '-') {
optopt = 0;
if (!colon && opterr)
__getopt_msg(argv[0], cnt ?
": option is ambiguous: " :