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; i.e., x 0x12345678 (where 0x12345678 is the return value of mmap) fails with an error "Cannot access memory at address 0x12345678".
Is there any way to tell GDB that this memory can be viewed? Alternatively, is there something different I can do in the mmap (either the call or the implementation of foo_mmap there) that will allow it to access this memory?
Note that I'm not asking about /dev/mem (as in the first snippet there) but about a mmap to memory acquired via ioremap(), virt_to_phys() and remap_pfn_range()
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 <sys/stat.h>
#include <fcntl.h>
#include <unistd.h>
#include <stdio.h>
#include <sys/mman.h>
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'