Using getopt_long() second pass does not work - c

Question

I wanted to use getopt_long() twice in a routine so that a verbosity flag can be set before parsing everything else but somehow the second call of the function does not process the arguments as expected. Here is a simplified code that demonstrates the issue. Anyone has any clues?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <getopt.h>
static void processInput(int argc, const char **argv) {
int k;
int verbose = 0;
struct option long_options[] = {
{"help" , no_argument, NULL, 'h'},
{"verbose", no_argument, NULL, 'v'},
{0, 0, 0, 0}
};
// Construct short_options from long_options
char str[1024] = "";
for (k = 0; k < sizeof(long_options) / sizeof(struct option); k++) {
struct option *o = &long_options[k];
snprintf(str + strlen(str), 1023, "%c%s", o->val, o->has_arg == required_argument ? ":" : (o->has_arg == optional_argument ? "::" : ""));
}
// First pass: just check for verbosity level
int opt, long_index = 0;
while ((opt = getopt_long(argc, (char * const *)argv, str, long_options, &long_index)) != -1) {
printf("Processing %c\n", opt);
switch (opt) {
case 'v':
verbose++;
break;
default:
break;
}
}
//
// Add verbose flag here to show something while setting up
//
// Second pass: now we go through all of them.
long_index = 0;
while ((opt = getopt_long(argc, (char * const *)argv, str, long_options, &long_index)) != -1) {
printf("Processing %c\n", opt);
switch (opt) {
case 'h':
printf("Help text here\n");
break;
case 'v':
// Add a dummy line here so that '-v' does not trigger the default task
break;
default:
fprintf(stderr, "I don't understand: -%c optarg = %s\n", opt, optarg);
exit(EXIT_FAILURE);
break;
}
}
}
int main(int argc, const char **argv) {
processInput(argc, argv);
return EXIT_SUCCESS;
}

Answer 1

The global variable optind is increased to argc by the end of getopt_long() routine so the second pass simply would not go further. In order to have the function reprocess everything from the beginning again, the global variable optind needed to be reset. So, add
// Second pass: now we go through all of them.
optind = 1;
long_index = 0;

Answer 2

As stated is the man page of getopt_long :
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 you have to set optind to 1 before the second pass.
// Second pass: now we go through all of them.
optind = 1;
long_index = 0;
while ((opt = getopt_long(argc, (char * const *)argv, str, long_options, &long_index)) != -1) {
printf("Processing %c\n", opt);
switch (opt) {
case 'h':
printf("Help text here\n");
break;
case 'v':
// Add a dummy line here so that '-v' does not trigger the default task
break;
default:
fprintf(stderr, "I don't understand: -%c optarg = %s\n", opt, optarg);
exit(EXIT_FAILURE);
break;
}
}
Little warning : Further on the man, there is this note :
A program that scans multiple argument vectors, or rescans the same vector more than once, and wants to make use of GNU extensions such as '+' and '-' at the start of optstring, or changes the value of POSIXLY_CORRECT between scans, must reinitialize getopt() by resetting optind to 0, rather than the traditional value of 1. (Resetting to 0 forces the invocation of an internal initialization routine that rechecks POSIXLY_CORRECT and checks for GNU extensions in optstring.)
So be carefull and choose what it most suit your situation.