In do_ade(), is_fpu_owner() isn't preempt-safe. For example, when an
unaligned ldc1 is executed, do_cpu() is called and then FPU will be
enabled (and TIF_USEDFPU will be set for the current process). Then,
do_ade() is called because the access is unaligned. If the current
process is preempted at this time, TIF_USEDFPU will be cleard. So when
the process is scheduled again, BUG_ON(!is_fpu_owner()) is triggered.
This small program can trigger this BUG in a preemptible kernel:
int main (int argc, char *argv[])
{
double u64[2];
while (1) {
asm volatile (
".set push \n\t"
".set noreorder \n\t"
"ldc1 $f3, 4(%0) \n\t"
".set pop \n\t"
::"r"(u64):
);
}
return 0;
}
V2: Remove the BUG_ON() unconditionally due to Paul's suggestion.
Signed-off-by: Huacai Chen <chenhc@lemote.com>
Signed-off-by: Jie Chen <chenj@lemote.com>
Signed-off-by: Rui Wang <wangr@lemote.com>
Cc: <stable@vger.kernel.org>
Cc: John Crispin <john@phrozen.org>
Cc: Steven J. Hill <Steven.Hill@imgtec.com>
Cc: linux-mips@linux-mips.org
Cc: Fuxin Zhang <zhangfx@lemote.com>
Cc: Zhangjin Wu <wuzhangjin@gmail.com>
Cc: stable@vger.kernel.org
Signed-off-by: Ralf Baechle <ralf@linux-mips.org>
case sdc1_op:
die_if_kernel("Unaligned FP access in kernel code", regs);
BUG_ON(!used_math());
- BUG_ON(!is_fpu_owner());
lose_fpu(1); /* Save FPU state for the emulator. */
res = fpu_emulator_cop1Handler(regs, ¤t->thread.fpu, 1,