Value printed by gdb does not consist with the right value

Applications Programming
1

Hello, I find the value printed by gdb does not consist with the right value.The following is the output.

(gdb) 
7	    while ( ( optc = getopt(argc, argv, ":b:B:h" ) ) != -1 ) {
(gdb) 
8	        printf( "%c    %d    %s\n", optc, optind, optarg);
(gdb) 
B    5    1-2
7	    while ( ( optc = getopt(argc, argv, ":b:B:h" ) ) != -1 ) {
(gdb) p optind
$1 = 1
(gdb)

The optind printed on stdout is 5, however, it is 1 printed by gdb. What,s wrong?
my source file is:

#include <stdio.h>
#include <getopt.h>
void main( int argc, char** argv) {
    int i = 0;
    char optc = 0;
    i = 1;
    while ( ( optc = getopt(argc, argv, ":b:B:h" ) ) != -1 ) {
        printf( "%c    %d    %s\n", optc, optind, optarg);
    }

   if (optind < argc) {
       printf("non-option ARGV-elements: ");
       while (optind < argc)
           printf("%s ", argv[optind++]);
       printf("\n");
   }

}

Thanks!

2

By the time you're printing it in gdb getopt() has been called again, which will change the value.