Examining strings using gdb

I would like to know how to view command line arguments passed to a function using gdb.

To test this I compiled the following code

.global main
main:
    pushl %ebp
    movl %esp, %ebp
    movl $1, %eax
    movl $0, %ebx
    int $0x80

with

gcc -Wall -Wextra -g -m32 args.s

to invoke the assembler and linker.

My System is x86_64 ubuntu (but for learning I Started with creating 32 bit applications).

I launched gdb and passed 2 command line arguments with

start test 12345

At this point I expected the stack to look like this:

Address argument 2 <-- $ebp+20
Address argument 1 <-- $ebp+16
Address name of programm <-- $ebp+12
argc <-- $ebp+8
old eip <-- $ebp+4
old ebp <-- $ebp

Reading the argument counter

(gdb) x /d $ebp+8

gives the expected Output:

0xffffd110: 3

Now I tried to check the command line arguments which have been passed. I expected

x /x $ebp+16

to give the address of the first character of the first passed argument in memory. The output was:

0xffffd118: 0xb4

But When I tried to access that region

(gdb) x /x 0xb4

I got:

0xb4:   Cannot access memory at address 0xb4

Which leads me to the conclusion that I tried to access a location which is different from the location where my commandline arguments are stored.

Can anyone give me some pointers to the true location of the command line arguments?

Solution

The main function of C gets the arguments in a different manner than the first called function of the Linux loader. Recapitulate the declaration of main in C: int main(int argc, char *argv[]) (as mentioned in the standard). The main function will get a value and a pointer to an array of strings, i.e. a pointer to several pointers to a string.

Tl;dr:

argc: x /d $ebp+8
argv[0]: x /s **(int)($ebp+12)
argv[1]: x /s *(*(int)($ebp+12)+4)
argv[2]: x /s *(*(int)($ebp+12)+8)