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...
");
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.
", (unsigned long) mptr);
printf("mptr points to 0x%lX. *mptr = 0x%X
", (unsigned long) mptr,
*mptr);
mptr[0] = 'a';
mptr[1] = 'b';
printf("mptr points to 0x%lX. *mptr = 0x%X
", (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'
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…