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.