为什么在X86上执行单步指令？

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因此，有一个 int 3，它是用于调试器中的断点的中断指令。

So there is "int 3" which is an interrupt instruction used for breakpoints in debuggers.

但是还有一个 int 1，它用于单个调试器。踩。但是为什么需要这个呢？我已经读到，在EFLAGS寄存器中设置陷阱标志（TF）将启用单步执行，并且会为每条指令捕获到OS中。那么为什么需要一个单独的中断类型呢？

But then there is also "int 1" which is used for single stepping. But why is this needed? I've read that setting the Trap Flag (TF) in EFLAGS register will enable single stepping and will trap into the OS for each instruction. So why is a separate interrupt type needed?

谢谢！

推荐答案

int 3 是一个特殊的1字节中断。如果存在，调用它将进入调试器，否则应用程序通常会崩溃。

int 3 is a special 1-byte interrupt. Invoking it will break into the debugger if one is present, otherwise the application will typically crash.

当调试器设置陷阱标志时，这将导致处理器自动执行每条指令后 int 1 中断。这使调试器可以单步执行指令，而不必插入 int 3 指令。您不必显式调用此中断。

When the debugger sets the trap flag, this causes the processor to automatically execute an int 1 interrupt after every instruction. This allows the debugger to single-step by instructions, without having to insert an int 3 instruction. You do not have to invoke this interrupt explicitly.