[IA64] Remove getting break_num by decoding instruction
break.b always sets cr.iim to 0 and the current code tries to get the break_num by decoding instruction. However, their seems to be a race condition while reading the regs->cr_iip, as on other cpu the break.b at regs->cr_iip might have been replaced with the original instruction as a result of unregister_kprobe() and hence decoding instruction to obtain break_num will result in wrong value in this case. Also includes changes to kprobes.c which now has to handle break number zero. Signed-off-by: Anil S Keshavamurthy <anil.s.keshavamurthy@intel.com> Signed-off-by: Tony Luck <tony.luck@intel.com>
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@ -740,7 +740,7 @@ int __kprobes kprobe_exceptions_notify(struct notifier_block *self,
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switch(val) {
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case DIE_BREAK:
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/* err is break number from ia64_bad_break() */
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if (args->err == 0x80200 || args->err == 0x80300)
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if (args->err == 0x80200 || args->err == 0x80300 || args->err == 0)
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if (pre_kprobes_handler(args))
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ret = NOTIFY_STOP;
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break;
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@ -132,24 +132,6 @@ __kprobes ia64_bad_break (unsigned long break_num, struct pt_regs *regs)
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siginfo_t siginfo;
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int sig, code;
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/* break.b always sets cr.iim to 0, which causes problems for
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* debuggers. Get the real break number from the original instruction,
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* but only for kernel code. User space break.b is left alone, to
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* preserve the existing behaviour. All break codings have the same
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* format, so there is no need to check the slot type.
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*/
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if (break_num == 0 && !user_mode(regs)) {
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struct ia64_psr *ipsr = ia64_psr(regs);
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unsigned long *bundle = (unsigned long *)regs->cr_iip;
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unsigned long slot;
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switch (ipsr->ri) {
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case 0: slot = (bundle[0] >> 5); break;
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case 1: slot = (bundle[0] >> 46) | (bundle[1] << 18); break;
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default: slot = (bundle[1] >> 23); break;
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}
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break_num = ((slot >> 36 & 1) << 20) | (slot >> 6 & 0xfffff);
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}
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/* SIGILL, SIGFPE, SIGSEGV, and SIGBUS want these field initialized: */
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siginfo.si_addr = (void __user *) (regs->cr_iip + ia64_psr(regs)->ri);
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siginfo.si_imm = break_num;
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