args[0] always contains full path when running in GDB

Question

When I run execute normally (without GDB), the args[0] is what I entered on the command prompt. But when I run using GDB, it includes full path. Is there any way to GDB to set args[0] same as command prompt?

My code is as follows:

#include <stdio.h>

// defining main with arguments
int main(int argc, char* argv[])
{
        printf("You have entered %d arguments:\n", argc);

        for (int i = 0; i < argc; i++) {
                printf("%s\n", argv[i]);
        }
        return 0;
}

Stand alone:

./a.out
You have entered 1 arguments:
./a.out

With GDB:

gdb ./a.out
GNU gdb (GDB) Red Hat Enterprise Linux 7.6.1-120.el7
(gdb) run
Starting program: /home/prasads/./a.out
You have entered 1 arguments:
/home/prasads/./a.out

Is there a way without having to change the code that I can get GDB to same output as command prompt?

Answer 1

Is there a way without having to change the code that I can get GDB to same output as command prompt?

Yes: you can use exec-wrapper to set argv[0] to anything you want.

Update:

exec-wrapper seems to be used for setting env variables; Not sure how I can use exec-wrapper to change argv[0]

Here is an example:

#include <stdio.h>

int main(int argc, char *argv[])
{
  for (int j = 0; j < argc; ++j) printf("%d: %s\n", j, argv[j]);
  return 0;
}

gcc echo-argv.c -o echo-argv
gdb --batch -ex run --args ./echo-argv foo bar baz

0: /tmp/echo-argv
1: foo
2: bar
3: baz
[Inferior 1 (process 3580209) exited normally]

cat wrapper.sh
#!/bin/bash

exec -a "xxx" "$@"

gdb --batch -ex 'set exec-wrapper ./wrapper.sh' -ex run --args ./echo-argv foo bar baz

0: xxx
1: foo
2: bar
3: baz
[Inferior 1 (process 3580521) exited normally]

Replace xxx with ./a.out or whatever else you want.

P.S. This GDB bug may also be relevant here.