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sh: When setting option via long name, don't go via letter.

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Looking up the letter makes no sense and prevents adding options that only
have a long name, no letter.
This commit is contained in:
jilles 2015-08-14 21:44:15 +00:00
parent 5d6a758aba
commit 675e2bd4e4
1 changed files with 20 additions and 13 deletions
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33
bin/sh/options.c
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@ -73,6 +73,7 @@ char *minusc; /* argument to -c option */
static void options(int);
static void minus_o(char *, int);
static void setoption(int, int);
static void setoptionbyindex(int, int);
static int getopts(char *, char *, char **, char ***, char **);
@ -269,7 +270,7 @@ minus_o(char *name, int val)
} else {
for (i = 0; i < NOPTS; i++)
if (equal(name, optlist[i].name)) {
setoption(optlist[i].letter, val);
setoptionbyindex(i, val);
return;
}
error("Illegal option -o %s", name);
@ -278,26 +279,32 @@ minus_o(char *name, int val)
static void
setoption(int flag, int val)
setoptionbyindex(int idx, int val)
{
int i;
if (flag == 'p' && !val && privileged) {
if (optlist[idx].letter == 'p' && !val && privileged) {
if (setgid(getgid()) == -1)
error("setgid");
if (setuid(getuid()) == -1)
error("setuid");
}
optlist[idx].val = val;
if (val) {
/* #%$ hack for ksh semantics */
if (optlist[idx].letter == 'V')
Eflag = 0;
else if (optlist[idx].letter == 'E')
Vflag = 0;
}
}
static void
setoption(int flag, int val)
{
int i;
for (i = 0; i < NOPTS; i++)
if (optlist[i].letter == flag) {
optlist[i].val = val;
if (val) {
/* #%$ hack for ksh semantics */
if (flag == 'V')
Eflag = 0;
else if (flag == 'E')
Vflag = 0;
}
setoptionbyindex(i, val);
return;
}
error("Illegal option -%c", flag);