时间戳

首页 > 解决方案 > Disabling/Enabling interrupts on x86 architectures

x86 - Disabling/Enabling interrupts on x86 architectures

问题描述

I am using NetBSD 5.1 for x86 systems. While studying some driver related code, I see that we use splraise and spllower to block or allow interrupts. I searched some of the mechanisms on internet to understand how these mechanisms work in reality. Did not get any real info on that.

When I disassembled I got the mechanism but still do not understand how all these assembly instruction yield me the result. I know x86 instruction individually, but not how the whole stuff works in its entirety.

Need your help in understanding its principles for x86 system. I understand that we need to disable Interrupt Enable (IE) bit, but this assembly seems to be doing more than just this work. Need help.

  (gdb) x/50i splraise
   0xc0100d40:  mov    0x4(%esp),%edx
   0xc0100d44:  mov    %fs:0x214,%eax
   0xc0100d4a:  cmp    %edx,%eax
   0xc0100d4c:  ja     0xc0100d55
   0xc0100d4e:  mov    %edx,%fs:0x214
   0xc0100d55:  ret
   0xc0100d56:  lea    0x0(%esi),%esi
   0xc0100d59:  lea    0x0(%edi,%eiz,1),%edi
   (gdb) p spllower
   $38 = {<text variable, no debug info>} 0xc0100d60
   0xc0100d60:  mov    0x4(%esp),%ecx
   0xc0100d64:  mov    %fs:0x214,%edx
   0xc0100d6b:  cmp    %edx,%ecx
   0xc0100d6d:  push   %ebx
   0xc0100d6e:  jae,pn 0xc0100d8f
   0xc0100d71:  mov    %fs:0x210,%eax
   0xc0100d77:  test   %eax,%fs:0x244(,%ecx,4)
   0xc0100d7f:  mov    %eax,%ebx
   0xc0100d81:  jne,pn 0xc0100d91
   0xc0100d84:  cmpxchg8b %fs:0x210
   0xc0100d8c:  jne,pn 0xc0100d71
   0xc0100d8f:  pop    %ebx
   0xc0100d90:  ret
   0xc0100d91:  pop    %ebx
   0xc0100d92:  jmp    0xc0100df0
   0xc0100d97:  mov    %esi,%esi
   0xc0100d99:  lea    0x0(%edi,%eiz,1),%edi
   0xc0100da0:  mov    0x4(%esp),%ecx
   0xc0100da4:  mov    %fs:0x214,%edx
   0xc0100dab:  cmp    %edx,%ecx
   0xc0100dad:  push   %ebx
   0xc0100dae:  jae,pn 0xc0100dcf
   0xc0100db1:  mov    %fs:0x210,%eax
   0xc0100db7:  test   %eax,%fs:0x244(,%ecx,4)
   0xc0100dbf:  mov    %eax,%ebx
   0xc0100dc1:  jne,pn 0xc0100dd1
   0xc0100dc4:  cmpxchg8b %fs:0x210
   0xc0100dcc:  jne,pn 0xc0100db1
   0xc0100dcf:  pop    %ebx
   0xc0100dd0:  ret
   0xc0100dd1:  pop    %ebx
   0xc0100dd2:  jmp    0xc0100df0
   0xc0100dd7:  mov    %esi,%esi
   0xc0100dd9:  lea    0x0(%edi,%eiz,1),%edi
   0xc0100de0:  nop
   0xc0100de1:  jmp    0xc0100df0

The code seems to be using a helper function cx8_spllower starting at address 0xc0100da0.

标签： x86kerneldriverinterruptnetbsd

解决方案

[E|R]FLAGS.IE使用 eg清除CLI会禁用 CPU 上的所有（可屏蔽）中断。出于多种原因，它可能是不可取的（例如，您希望允许一些，或者您可能不希望CLI在 VM 中进行虚拟化开销）。

实现目标的另一种方法是告诉中断控制器（旧的 8259 PIC 或 Pentium APIC/IOAPIC）您不想为优先级低于某个级别的中断提供服务。为此，您需要与控制器进行通信，这本身可能会产生额外的开销（与真实和虚拟硬件的通信都很慢）。

后者的一种变体是将当前中断级别/优先级保留在一个变量中并让中断进入，但仅真正服务于那些级别/优先级不低于该变量中当前中断的那些。那些未服务的将被标记为待处理，并且当当前级别/优先级下降到足够低时将得到完全服务。这就是为什么splraise()比spllower().

此级别/优先级变量似乎在不同位置和代码版本中的命名不同：（CPL不要与 CPU 的当前特权级别混淆）、SPL(?) ILEVEL、。

这是我目前对实现的有限理解。还有更多细节。

以下是我找到并用于答案的一些线索：

推荐阅读