Examining mmaped addresses using GDB

Question

I\'m using the driver I posted at Direct Memory Access in Linux to mmap some physical ram into a userspace address. However, I can\'t use GDB to look at any of the address;

Answer 1

I believe Linux does not make I/O memory accessible via ptrace(). You could write a function that simply reads the mmap'ed address and have gdb invoke it. Here's a slightly modified version of your foo-user.c program along with the output from a gdb session.

#include 
#include 
#include 
#include 
#include 

char *mptr;

char peek(int offset)
{
    return mptr[offset];
}

int main(void)
{
    int fd;
    fd = open("/dev/foo", O_RDWR | O_SYNC);
    if (fd == -1) {
        printf("open error...\n");
        return 1;
    }
    mptr = mmap(0, 1 * 1024 * 1024, PROT_READ | PROT_WRITE,
             MAP_FILE | MAP_SHARED, fd, 4096);
    printf("On start, mptr points to 0x%lX.\n", (unsigned long) mptr);
    printf("mptr points to 0x%lX. *mptr = 0x%X\n", (unsigned long) mptr,
           *mptr);
    mptr[0] = 'a';
    mptr[1] = 'b';
    printf("mptr points to 0x%lX. *mptr = 0x%X\n", (unsigned long) mptr,
           *mptr);
    close(fd);
    return 0;
}



$ make foo-user CFLAGS=-g
$ gdb -q foo-user
(gdb) break 27
Breakpoint 1 at 0x804855f: file foo-user.c, line 27.
(gdb) run
Starting program: /home/me/foo/foo-user 
On start, mptr points to 0xB7E1E000.
mptr points to 0xB7E1E000. *mptr = 0x61

Breakpoint 1, main () at foo-user.c:27
27          mptr[0] = 'a';
(gdb) n
28          mptr[1] = 'b';
(gdb) print peek(0)
$1 = 97 'a'
(gdb) print peek(1)
$2 = 98 'b'