gdb does not print right value!

915086731 · 10-17-2011, 03:16 AM

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

Code:

(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:

Code:

#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!

meenakshi.khurana · 10-17-2011, 06:23 AM

Here, gdb is resolving wrong optind variable.

Change your loop to

Code:

while ( ( optc = getopt(argc, argv, ":b:B:h" ) ) != -1 ) {
        printf( "%c    %d    %s %x %x\n", optc, optind, optarg,&optc,&optind);
    }

And then run in gdb.

Now print the address of optind & optc in gdb (print &optind) as well as via this loop. You will get to know that address of optc is same but of optind, its something else.So, your program and gdb resolved to different addresses when they want to access the optind variable

Here, your optind is actually linked to optind@@GLIBC_2.0. In gdb write this:

(gdb) p optind

and hit the tab key (twice), it'll autocomplete the symbols. Here you can see, you have optind and optind@@GLIBC_2.0.

Now print address and value of optind@@GLIBC_2.0

(gdb) print 'optind@@GLIBC_2.0'

(gdb) print &'optind@@GLIBC_2.0'

915086731 · 10-17-2011, 07:22 AM

Thank for your help, but I don't know why it's address is not the same! Can you explain in detail?

Code:

(gdb) p &optind
$2 = (int *) 0x4c8a90fc
(gdb) n
B    4    1-2 bffff1db 8049848

meenakshi.khurana · 10-17-2011, 11:14 AM

Write following command at gdb prompt:

<CODE>
(gdb) info var optind
All variables matching regular expression "optind":

File /usr/include/getopt.h:
int optind;

Non-debugging symbols:
0x08049860 optind@@GLIBC_2.0
(gdb)
</CODE>

From the above output, you can view gdb thinks there are two copies. One is in getopt.c in libc.so. That's the one that GDB is printing. The other is in your binary.

This is basically a bug and and it stands for any variable defined in a shared library and used in the executable.

For further details, please refer following link:

http://sourceware.org/ml/gdb/2003-12/msg00165.html