Am 23.08.2011 19:07, schrieb Al Viro:
On Tue, Aug 23, 2011 at 06:58:18PM +0200, Richard Weinberger wrote:
What about this hack/solution?
While booting UML can check whether the host's vDSO contains
a SYSCALL instruction.
If so, UML will not make the host's vDSO available to it's
processes...
Note that this is *only* for 32bit side of things. 64bit one works fine...
I wouldn't search for SYSCALL in vdso, BTW - not when we have a good way
to trigger that crap and recognize it.
At boot time, fork a child. Have it traced with PTRACE_SYSCALL. Let it
put recognizable values in registers and call __kernel_vsyscall(). Then
let the parent do one more PTRACE_SYSCALL, then PTRACE_POKEUSER and set ebp
to 0x69696969. PTRACE_CONT the sucker and let it report what it sees in ecx.
If it's what we'd put there - fine, it looks safe. If it's 0x69696969 -
we have a problem, no vdso for us.
BTW: IMHO we can completely disable the vDSO for 32bit guests.
I did some benchmarks, there is no performance gain at all
within UML.
The attached program runs some syscalls for 10 seconds and prints
the number of iterations.
Some results (5 runs on my Intel Core2):
vdso: 360099 362057 365982 367132 368907
none: 344195 355759 358974 366630 420027
Thanks,
//richard
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <signal.h>
#include <sys/types.h>
#include <sys/stat.h>
static unsigned long long iter;
static void done(int sig)
{
printf("%llu\n", iter);
exit(0);
}
int main()
{
signal(SIGALRM, done);
alarm(10);
for(;;){
close(dup(0));
getpid();
getuid();
umask(022);
iter++;
}
return 0;
}
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