Question

Is there a way to define a new data type (C structure or union) in gdb. The idea is to define a structure and then make gdb print data from an address interpreted as the newly defined structure.

For example, lets say we have a sample structure.

struct sample {
  int i;
  struct sample *less;
  struct sample *more;
}

And if 0x804b320 is the address of an array of struct sample. The binary doesn’t have debugging information so that gdb understands struct sample. Is there any way to define struct sample in a gdb session? So that we can print p *(struct sample *)0x804b320

Answer 1

Yes, here is how to make this work:

// sample.h
struct sample {
  int i;
  struct sample *less;
  struct sample *more;
};

// main.c
#include <stdio.h>
#include <assert.h>
#include "sample.h"
int main()
{
  struct sample sm;
  sm.i = 42;
  sm.less = sm.more = &sm;

  printf("&sm = %p\n", &sm);
  assert(sm.i == 0);  // will fail
}

gcc main.c   # Note: no '-g' flag

gdb -q ./a.out
(gdb) run
&sm = 0x7fffffffd6b0
a.out: main.c:11: main: Assertion `sm.i == 0' failed.

Program received signal SIGABRT, Aborted.
0x00007ffff7a8da75 in raise ()
(gdb) fr 3
#3  0x00000000004005cc in main ()

No local variables, no type struct sample:

(gdb) p sm
No symbol "sm" in current context.
(gdb) p (struct sample *)0x7fffffffd6b0
No struct type named sample.

So we get to work:

// sample.c
#include "sample.h"
struct sample foo;

gcc -g -c sample.c

(gdb) add-symbol-file sample.o 0
add symbol table from file "sample.o" at
    .text_addr = 0x0

(gdb) p (struct sample *)0x7fffffffd6b0
$1 = (struct sample *) 0x7fffffffd6b0
(gdb) p *$1
$2 = {i = 42, less = 0x7fffffffd6b0, more = 0x7fffffffd6b0}

Voilà!