command line processing library - getopt

Question

Can someone help me with the getopt function?

When I do the following in main:

char *argv1[] = {"testexec","-?"};
char *argv2[] = {"testexec","-m","arg1"};
int  cOption;
/* test for -? */

setvbuf(stdout,(char*)NULL,_IONBF,0);
printf("\n argv1 ");
while (( cOption = getopt (2, argv1, "m:t:n:fs?")) != -1) {
    switch(cOption){
        case 'm':
            printf("\n -m Arg : %s \n",optarg);
            break;
        case '?':
            printf("\n -? Arg ");
            break;
        case 'n':
            printf("\n -n Arg : %s \n",optarg);
            break;
    }
}

printf("\n argv2 ");

while (( cOption = getopt (3, argv2, "m:t:n:fs?")) != -1) {
    switch(cOption){
        case 'm':
            printf("\n -m Arg : %s \n",optarg);
            break;
        case '?':
            printf("\n -? Arg : %s \n",optarg);
            break;
        case 'n':
            printf("\n -n Arg : %s \n",optarg);
            break;
    }
}

I'm running this code on rhel3 which uses old libc version. I don't know which one to be exact.

Now the problem is getopt doesn't work the second time with argv2. But if I comment out the first getopt call with argv1 , it works.

Can someone tell me what am I doing wrong here?

Answer 1

argv1 and 2 must end in 0:

char* argv1[] = {"par1", "par2", 0};

Edit: OK, I read the getopt man page and I found this:

The variable optind is the index of the next element to be processed in argv. The system initializes this value to 1. The caller can reset it to 1 to restart scanning of the same argv, or when scanning a new argument vector.

So, making optind=1 between the two calls at getopt makes it work as expected.