Found by Bodo Stroesser. Description from Bodo:
>>
On i386, if a signal handler is started, the kernel saves the fpu-state
of the interrupted routine in the sigcontext on the stack. Calling
unlazy_fpu() and setting current->used_math=0, the kernel supplies the
signal-handler with a cleared virtual fpu.
On sigreturn(), the old fpu-state of the interrupted routine is
restored.
If a process never used the fpu, it virtually has a cleared fpu.
If such a process is interrupted by a signal handler, no fpu-context is
saved and sigcontext->fpstate is set to NULL.
Assume, that the signal handler uses the fpu. Then, AFAICS, on sigreturn
current->used_math will be 1. Since sigcontext->fpstate still is NULL,
restore_sigcontext() doesn't call restore_i387(). Thus, no
clear_fpu() is done, current->used_math is not reset.
Now, the interrupted processes fpu no longer is cleared!
<<
Fix by AK. Just clear the FPU again when this happens.
patch for i386 and x86-64.
Signed-off-by: Andi Kleen <ak@suse.de>
Signed-off-by: Andrew Morton <akpm@osdl.org>
Signed-off-by: Linus Torvalds <torvalds@osdl.org>
if (verify_area(VERIFY_READ, buf, sizeof(*buf)))
goto badframe;
err |= restore_i387(buf);
+ } else {
+ struct task_struct *me = current;
+ if (me->used_math) {
+ clear_fpu(me);
+ me->used_math = 0;
+ }
}
}
if (verify_area(VERIFY_READ, buf, sizeof(*buf)))
goto badframe;
err |= restore_i387_ia32(current, buf, 0);
+ } else {
+ struct task_struct *me = current;
+ if (me->used_math) {
+ clear_fpu(me);
+ me->used_math = 0;
+ }
}
}
if (verify_area(VERIFY_READ, buf, sizeof(*buf)))
goto badframe;
err |= restore_i387(buf);
+ } else {
+ struct task_struct *me = current;
+ if (me->used_math) {
+ clear_fpu(me);
+ me->used_math = 0;
+ }
}
}