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I seem to recall hearing in the past that an x86 program can detect whether it is being debugged. This was related to malware analysis.

Yet researching registers, I read up on the DRs and found the statement:

The debug registers are privileged resources; the MOV instructions that access them can only be executed at privilege level zero. An attempt to read or write the debug registers when executing at any other privilege level causes a general protection fault.

So, could someone explain perhaps if there is some alternative feature in x86 that allows a program to become aware that it is being debugged? I think I recall reading something about debuggers actually patching things into a program's instruction memory, but I'm not yet sure (I haven't had time to study the inner workings of a debugger yet).

Could someone clarify whether there is a fundamental feature of x86 that leaks information suggesting the attachment of a debugger to a program? Or perhaps it's just due to something more invasive debuggers might be doing?

Peter Cordes
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J.Todd
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    TF (single-step Trap Flag) is part of EFLAGS, and setting software breakpoints will leave `0xcc int3` bytes in the machine code. Having a look at any basic debug-detection example would show this, or mention it in comments. Hardware breakpoints set with DR aren't directly observable to user-space code being debugged, AFAIK, but I think there are still some indirect things code can do to detect debuggers. (Mainstream debuggers like GDB default to using software breakpoints.) – Peter Cordes Dec 24 '21 at 21:43
  • @PeterCordes Hmm, so due to the 15 byte limit (which I learned thanks to your 2016 comment "The ISA for 386 and later does impose a 15-byte limit."), could a program prevent its own debugging? As an answer put it: "For example, you can take the innocuous looking instruction: `89 E5` `mov %sp,%bp` And turn it into a really long instruction: `66 66 66 66 … 66 66 89 E5` `mov %sp,%bp`" So couldn't we do this for every instruction in order to make it impossible for `0xcc` bytes to be appended? – J.Todd Dec 24 '21 at 21:57
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    Appended? No, overwritten (and restored after the breakpoint is hit). Memory is an array, not a linked list; you can't insert a byte and push all later bytes over by one, not without `memmove` of an unknown number of later bytes, and that would change their addresses, breaking PC-relative stuff and jump targets from elsewhere. – Peter Cordes Dec 24 '21 at 22:10
  • Ah, right. I was imaging 0xcc as some byte that could be added to an instruction without changing its behavior aside from adding an interrupt afterwards. I suppose this misunderstanding came from studying a bit about the [modular](https://www.researchgate.net/profile/Kent-Griffin/publication/262272054/figure/fig2/AS:665149623001101@1535595436482/x86-instruction-format.png) nature of the x86 instruction. – J.Todd Dec 24 '21 at 22:13
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    That's true unless it's a jump, or it faults, in which case the `int3` would never be reached. When you set a breakpoint on an instruction, you want to trap *before* it executes; that's how breakpoints work. That's why debuggers replace the first byte of this instruction, not the first byte of the following instruction. But regardless, `int3` is a separate instruction anyway, so instruction length is actually a non-issue, regardless of trying to append or prepend instead of just overwrite. Seems like some practical experience using a debugger would help you think things through... – Peter Cordes Dec 24 '21 at 22:18

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No, there is no mechanism built-in into the x86 architecture allowing a program to detect whether it’s being debugged, but there are numerous heuristics allowing you to determine whether it’s probably being debugged, e.g.

  • check TracerPid in /proc/self/status is nonzero,
  • examine rdtsc difference (obviously generally unsuitable), or
  • inspect the trap flag

but these methods aren’t infallible. Ultimately it could be the case your program isn’t even run on a bare-metal x86 processor.

Kai Burghardt
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