How does the tct command work under the hood? - debugging

The windbg command tct executes a program until it reaches a call instruction or a ret instruction. I am wondering how the debugger implements this functionality under the hood.
I could imagine that the debugger scans the instructions from the current instructions for the next call or ret and sets according breakpoints on the found instructions. However, I think this is unlikely because it would also have to take into account jmp instructions so that there are an arbitrary number of possible call or ret instructions where such a breakpoint would have to be set.
On the other hand, I wonder if the x86/x64 CPU provides a functionality that raises an exception to be caught by the debugger whenever the CPU is about to process a call or ret instruction. Yet, I have not heard of such a functionality.

I'd guess that it single-steps repeatedly, until the next instruction is a call or ret, instead of trying to figure out where to set a breakpoint. (Which in the general case could be as hard as solving the Halting Problem.)
It's possible it could optimize that by scanning forward over "straight line" code and setting a breakpoint on the next jmp/jcc/loop or other control-transfer instruction (e.g. xabort), and also catching signals/exceptions that could transfer control to an SEH handler.
I'm also not aware of any HW support for breaking on a certain type of instruction or opcode: the x86 debug registers DR0..7 allow hardware breakpoints at code addresses without rewriting machine code to int3, and also hardware watchpoints (to trap data load/store to a specific address or range of addresses). But not filtering by opcode.

Related

SPARC single-stepping mode

Is there a SPARC equivalent to x86's single step mode? What I want is to stop execution after every instruction and move control flow to a trap handler or something similar.
I thought of using the ta instruction in the delayed execution slot but this would not work when the previous instruction is a branching instruction with the annul bit set.
Sparc lacks a single step bit in PSR, so it's harder to single step. But I've used a trick to help get closer. Set TPC to the address of the instruction you want to single step, and set TNPC to an address someplace else where you've placed a trap instruction. When you execute the retry instruction to get back to the process context, it will single step the one instruction you want, then it will next execute the trap instruction which will bring you right back to the kernel, where you can do whatever you want. (n.b this is for sparc64, not sure about sparc32). This is a nice trick because you don't have modify existing instructions in the user's address space. This was important to me since I was single stepping instructions in the kernel.
Another idea I had, but never tried, was to simply set TNPC to an illegal address. Then after the instruction at TPC was executed, you'd get an automatic trap back into the kernel. And since the trap handling code knows that the process is being single stepped, there would be no confusion over a "real" illegal address trap.

Is it valid to write below ESP?

For a 32-bit windows application is it valid to use stack memory below ESP for temporary swap space without explicitly decrementing ESP?
Consider a function that returns a floating point value in ST(0). If our value is currently in EAX we would, for example,
PUSH EAX
FLD [ESP]
ADD ESP,4 // or POP EAX, etc
// return...
Or without modifying the ESP register, we could just :
MOV [ESP-4], EAX
FLD [ESP-4]
// return...
In both cases the same thing happens except that in the first case we take care to decrement the stack pointer before using the memory, and then to increment it afterwards. In the latter case we do not.
Notwithstanding any real need to persist this value on the stack (reentrancy issues, function calls between PUSHing and reading the value back, etc) is there any fundamental reason why writing to the stack below ESP like this would be invalid?
TL:DR: no, there are some SEH corner cases that can make it unsafe in practice, as well as being documented as unsafe. #Raymond Chen recently wrote a blog post that you should probably read instead of this answer.
His example of a code-fetch page-fault I/O error that can be "fixed" by prompting the user to insert a CD-ROM and retry is also my conclusion for the only practically-recoverable fault if there aren't any other possibly-faulting instructions between store and reload below ESP/RSP.
Or if you ask a debugger to call a function in the program being debugged, it will also use the target process's stack.
This answer has a list of some things you'd think would potentially step on memory below ESP, but actually don't, which might be interesting. It seems to be only SEH and debuggers that can be a problem in practice.
First of all, if you care about efficiency, can't you avoid x87 in your calling convention? movd xmm0, eax is a more efficient way to return a float that was in an integer register. (And you can often avoid moving FP values to integer registers in the first place, using SSE2 integer instructions to pick apart exponent / mantissa for a log(x), or integer add 1 for nextafter(x).) But if you need to support very old hardware, then you need a 32-bit x87 version of your program as well as an efficient 64-bit version.
But there are other use-cases for small amounts of scratch space on the stack where it would be nice to save a couple instructions that offset ESP/RSP.
Trying to collect up the combined wisdom of other answers and discussion in comments under them (and on this answer):
It is explicitly documented as being not safe by Microsoft: (for 64-bit code, I didn't find an equivalent statement for 32-bit code but I'm sure there is one)
Stack Usage (for x64)
All memory beyond the current address of RSP is considered volatile: The OS, or a debugger, may overwrite this memory during a user debug session, or an interrupt handler.
So that's the documentation, but the interrupt reason stated doesn't make sense for the user-space stack, only the kernel stack. The important part is that they document it as not guaranteed safe, not the reasons given.
Hardware interrupts can't use the user stack; that would let user-space crash the kernel with mov esp, 0, or worse take over the kernel by having another thread in the user-space process modify return addresses while an interrupt handler was running. This is why kernels always configure things so interrupt context is pushed onto the kernel stack.
Modern debuggers run in a separate process, and are not "intrusive". Back in 16-bit DOS days, without a multi-tasking protected-memory OS to give each task its own address space, debuggers would use the same stack as the program being debugged, between any two instructions while single-stepping.
#RossRidge points out that a debugger might want to let you call a function in the context of the current thread, e.g. with SetThreadContext. This would run with ESP/RSP just below the current value. This could obviously have side-effects for the process being debugged (intentional on the part of the user running the debugger), but clobbering local variables of the current function below ESP/RSP would be an undesirable and unexpected side-effect. (So compilers can't put them there.)
(In a calling convention with a red-zone below ESP/RSP, a debugger could respect that red-zone by decrementing ESP/RSP before making the function call.)
There are existing program that intentionally break when being debugged at all, and consider this a feature (to defend against efforts to reverse-engineer them).
Related: the x86-64 System V ABI (Linux, OS X, all other non-Windows systems) does define a red-zone for user-space code (64-bit only): 128 bytes below RSP that is guaranteed not to be asynchronously clobbered. Unix signal handlers can run asynchronously between any two user-space instructions, but the kernel respects the red-zone by leaving a 128 byte gap below the old user-space RSP, in case it was in use. With no signal handlers installed, you have an effectively unlimited red-zone even in 32-bit mode (where the ABI does not guarantee a red-zone). Compiler-generated code, or library code, of course can't assume that nothing else in the whole program (or in a library the program called) has installed a signal handler.
So the question becomes: is there anything on Windows that can asynchronously run code using the user-space stack between two arbitrary instructions? (i.e. any equivalent to a Unix signal handler.)
As far as we can tell, SEH (Structured Exception Handling) is the only real obstacle to what you propose for user-space code on current 32 and 64-bit Windows. (But future Windows could include a new feature.)
And I guess debugging if you happen ask your debugger to call a function in the target process/thread as mentioned above.
In this specific case, not touching any other memory other than the stack, or doing anything else that could fault, it's probably safe even from SEH.
SEH (Structured Exception Handling) lets user-space software have hardware exceptions like divide by zero delivered somewhat similarly to C++ exceptions. These are not truly asynchronous: they're for exceptions triggered by instructions you ran, not for events that happened to come after some random instruction.
But unlike normal exceptions, one thing a SEH handler can do is resume from where the exception occurred. (#RossRidge commented: SEH handlers are are initially called in the context of the unwound stack and can choose to ignore the exception and continue executing at the point where the exception occurred.)
So that's a problem even if there's no catch() clause in the current function.
Normally HW exceptions can only be triggered synchronously. e.g. by a div instruction, or by a memory access which could fault with STATUS_ACCESS_VIOLATION (the Windows equivalent of a Linux SIGSEGV segmentation fault). You control what instructions you use, so you can avoid instructions that might fault.
If you limit your code to only accessing stack memory between the store and reload, and you respect the stack-growth guard page, your program won't fault from accessing [esp-4]. (Unless you reached the max stack size (Stack Overflow), in which case push eax would fault, too, and you can't really recover from this situation because there's no stack space for SEH to use.)
So we can rule out STATUS_ACCESS_VIOLATION as a problem, because if we get that on accessing stack memory we're hosed anyway.
An SEH handler for STATUS_IN_PAGE_ERROR could run before any load instruction. Windows can page out any page it wants to, and transparently page it back in if it's needed again (virtual memory paging). But if there's an I/O error, your Windows attempts to let your process handle the failure by delivering a STATUS_IN_PAGE_ERROR
Again, if that happens to the current stack, we're hosed.
But code-fetch could cause STATUS_IN_PAGE_ERROR, and you could plausibly recover from that. But not by resuming execution at the place where the exception occurred (unless we can somehow remap that page to another copy in a highly fault-tolerant system??), so we might still be ok here.
An I/O error paging in the code that wants to read what we stored below ESP rules out any chance of reading it. If you weren't planning to do that anyway, you're fine. A generic SEH handler that doesn't know about this specific piece of code wouldn't be trying to do that anyway. I think usually a STATUS_IN_PAGE_ERROR would at most try to print an error message or maybe log something, not try to carry on whatever computation was happening.
Accessing other memory in between the store and reload to memory below ESP could trigger a STATUS_IN_PAGE_ERROR for that memory. In library code, you probably can't assume that some other pointer you passed isn't going to be weird and the caller is expecting to handle STATUS_ACCESS_VIOLATION or PAGE_ERROR for it.
Current compilers don't take advantage of space below ESP/RSP on Windows, even though they do take advantage of the red-zone in x86-64 System V (in leaf functions that need to spill / reload something, exactly like what you're doing for int -> x87.) That's because MS says it isn't safe, and they don't know whether SEH handlers exist that could try to resume after an SEH.
Things that you'd think might be a problem in current Windows, and why they're not:
The guard page stuff below ESP: as long as you don't go too far below the current ESP, you'll be touching the guard page and trigger allocation of more stack space instead of faulting. This is fine as long as the kernel doesn't check user-space ESP and find out that you're touching stack space without having "reserved" it first.
kernel reclaim of pages below ESP/RSP: apparently Windows doesn't currently do this. So using a lot of stack space once ever will keep those pages allocated for the rest of your process lifetime, unless you manually VirtualAlloc(MEM_RESET) them. (The kernel would be allowed to do this, though, because the docs say memory below RSP is volatile. The kernel could effectively zero it asynchronously if it wants to, copy-on-write mapping it to a zero page instead of writing it to the pagefile under memory pressure.)
APC (Asynchronous Procedure Calls): They can only be delivered when the process is in an "alertable state", which means only when inside a call to a function like SleepEx(0,1). calling a function already uses an unknown amount of space below E/RSP, so you already have to assume that every call clobbers everything below the stack pointer. Thus these "async" callbacks are not truly asynchronous with respect to normal execution the way Unix signal handlers are. (fun fact: POSIX async io does use signal handlers to run callbacks).
Console-application callbacks for ctrl-C and other events (SetConsoleCtrlHandler). This looks exactly like registering a Unix signal handler, but in Windows the handler runs in a separate thread with its own stack. (See RbMm's comment)
SetThreadContext: another thread could change our EIP/RIP asynchronously while this thread is suspended, but the whole program has to be written specially for that to make any sense. Unless it's a debugger using it. Correctness is normally not required when some other thread is messing around with your EIP unless the circumstances are very controlled.
And apparently there are no other ways that another process (or something this thread registered) can trigger execution of anything asynchronously with respect to the execution of user-space code on Windows.
If there are no SEH handlers that could try to resume, Windows more or less has a 4096 byte red-zone below ESP (or maybe more if you touch it incrementally?), but RbMm says nobody takes advantage of it in practice. This is unsurprising because MS says not to, and you can't always know if your callers might have done something with SEH.
Obviously anything that would synchronously clobber it (like a call) must also be avoided, again same as when using the red-zone in the x86-64 System V calling convention. (See https://stackoverflow.com/tags/red-zone/info for more about it.)
in general case (x86/x64 platform) - interrupt can be executed at any time, which overwrite memory bellow stack pointer (if it executed on current stack). because this, even temporary save something bellow stack pointer, not valid in kernel mode - interrupt will be use current kernel stack. but in user mode situation another - windows build interrupt table (IDT) suchwise that when interrupt raised - it will be always executed in kernel mode and in kernel stack. as result user mode stack (below stack pointer) will be not affected. and possible temporary use some stack space bellow it pointer, until you not do any functions calls. if exception will be (say by access invalid address) - also space bellow stack pointer will be overwritten - cpu exception of course begin executed in kernel mode and kernel stack, but than kernel execute callback in user space via ntdll.KiDispatchExecption already on current stack space. so in general this is valid in windows user mode (in current implementation), but you need good understand what you doing. however this is very rarely i think used
of course, how correct noted in comments that we can, in windows user mode, write below stack pointer - is just the current implementation behavior. this not documented or guaranteed.
but this is very fundamental - unlikely will be changed: interrupts always will be executed in privileged kernel mode only. and kernel mode will be use only kernel mode stack. the user mode context not trusted at all. what will be if user mode program set incorrect stack pointer ? say by
mov rsp,1 or mov esp,1 ? and just after this instruction interrupt will be raised. what will be if it begin executed on such invalid esp/rsp ? all operation system just crashed. exactly because this interrupt will be executed only on kernel stack. and not overwrite user stack space.
also need note that stack is limited space (even in user mode), access it bellow 1 page (4Kb)already error (need do stack probing page by page, for move guard page down).
and finally really there is no need usually access [ESP-4], EAX - in what problem decrement ESP first ? even if we need access stack space in loop huge count of time - decrement stack pointer need only once - 1 additional instruction (not in loop) nothing change in performance or code size.
so despite formal this is will be correct work in windows user mode, better (and not need) use this
of course formal documentation say:
Stack Usage
All memory beyond the current address of RSP is considered volatile
but this is for common case, including kernel mode too. i wrote about user mode and based on current implementation
possible in future windows and add "direct" apc or some "direct" signals - some code will be executed via callback just after thread enter to kernel (during usual hardware interrupt). after this all below esp will be undefined. but until this not exist. until this code will be work always(in current builds) correct.
In general (not specifically related to any OS); it's not safe to write below ESP if:
It's possible for the code to be interrupted and the interrupt handler will run at the same privilege level. Note: This is typically very unlikely for "user-space" code, but extremely likely for kernel code.
You call any other code (where either the call or the stack used by the called routine can trash the data you stored below ESP)
Something else depends on "normal" stack use. This can include signal handling, (language based) exception unwinding, debuggers, "stack smashing protector"
It's safe to write below ESP if it's not "not safe".
Note that for 64-bit code, writing below RSP is built into the x86-64 ABI ("red zone"); and is made safe by support for it in tool chains/compilers and everything else.
When a thread gets created, Windows reserves a contiguous region of virtual memory of a configurable size (the default is 1 MB) for the thread's stack. Initially, the stack looks like this (the stack grows downwards):
--------------
| committed |
--------------
| guard page |
--------------
| . |
| reserved |
| . |
| . |
| |
--------------
ESP will be pointing somewhere inside the committed page. The guard page is used to support automatic stack growth. The reserved pages region ensures that the requested stack size is available in virtual memory.
Consider the two instructions from the question:
MOV [ESP-4], EAX
FLD [ESP-4]
There are three possibilities:
The first instruction executes successfully. There is nothing that uses the user-mode stack that can execute between the two instructions. So the second instruction will use the correct value (#RbMm stated this in the comments under his answer and I agree).
The first instruction raises an exception and an exception handler does not return EXCEPTION_CONTINUE_EXECUTION. As long as the second instruction is immediately after the first one (it is not in the exception handler or placed after it), then the second instruction will not execute. So you're still safe. Execution continues from stack frame where the exception handler exists.
The first instruction raises an exception and an exception handler returns EXCEPTION_CONTINUE_EXECUTION. Execution continues from the same instruction that raised the exception (potentially with a context modified by the handler). In this particular example, the first will be re-executed to write a value below ESP. No problem. If the second instruction raised an exception or there are more than two instructions, then the exception might occur a place after a value is written below ESP. When the exception handler gets called, it may overwrite the value and then return EXCEPTION_CONTINUE_EXECUTION. But when execution resumes, the value written is assumed to still be there, but it's not anymore. This is a situation where it's not safe to write below ESP. This applies even if all of the instructions are placed consecutively. Thanks to #RaymondChen for pointing this out.
In general, if the two instructions are not placed back-to-back, if you are writing to locations beyond ESP, there is no guarantee that the written values won't get corrupted or overwritten. One case that I can think of where this might happen is structured exception handling (SEH). If a hardware-defined exception (such as divide by zero) occurs, the kernel exception handler will be invoked (KiUserExceptionDispatcher) in kernel-mode, which will invoke the user-mode side of the handler (RtlDispatchException). When switching from user-mode to kernel-mode and then back to user-mode, whatever value was in ESP will be saved and restored. However, the user-mode handler itself uses the user-mode stack and will iterate over a registered list of exception handlers, each of which uses the user-mode stack. These functions will modify ESP as required. This may lead to losing the values you've written beyond ESP. A similar situation occurs when using software-define exceptions (throw in VC++).
I think you can deal with this by registering your own exception handler before any other exception handlers (so that it is called first). When your handler gets called, you can save your data beyond ESP elsewhere. Later, during unwinding, you get the cleanup opportunity to restore your data to the same location (or any other location) on the stack.
You need also to similarly watch out for asynchronous procedure calls (APCs) and callbacks.
Several answers here mention APCs (Asynchronous Procedure Calls), saying that they can only be delivered when the process is in an "alertable state", and are not truly asynchronous with respect to normal execution the way Unix signal handlers are
Windows 10 version 1809 introduces Special User APCs, which can fire at any moment just like Unix signals. See this article for low level details.
The Special User APC is a mechanism that was added in RS5 (and exposed through NtQueueApcThreadEx), but lately (in an insider build) was exposed through a new syscall - NtQueueApcThreadEx2. If this type of APC is used, the thread is signaled in the middle of the execution to execute the special APC.

How does a debugger set breakpoints if the image is in read-only memory?

How does a debugger set breakpoints if the image is in read-only memory? I know there are hardware breakpoints, but in the debugger I use (OllyDbg) those have to be set specially using a different dialog than normal breakpoints.
Explanation:
Here is a routine in a debugger that is comparing itself to a copy of itself. EDX points to the running image, EBX points to the known good copy of the image. The breakpoint on 4010CE only is reached if there is a mismatch. The character being compared is in the AL register. As you can see the debugger shows EB F6 at 10CE, but this is false. 10CE actually has CC in it, as you can see by looking at the AL register. This is because the debugger has secretely inserted the CC to perform the breakpoint.
The debugger first has to change the memory protection of the page it wants to write to. This can be done with VirtualProtectEx. After that it is able to write with WriteProcessMemory and then set the protection back to the original value.
Let me preface this with a disclaimer that I'm not familiar with your particular toolset.
If you haven't enabled hardware breakpoints, the only remaining breakpoint type is a software breakpoint. These are only hit (on x86 because that's what I'm most familiar with) when you replace the first byte of an instruction with a trap instruction, and will only be routed through the breakpoint mechanism of your OS to your debugger if the correct trap instruction for your OS is used and the debugger has already registered itself with the OS as a debugger for this process. In order to cause the software breakpoint to happen at the correct moment, the trap instruction must be written into your code segment over the first byte of your correct instruction.
The two answers that got here first explain the two scenarios which could get you here (at least, the only two I can think of):
The kernel always has write access everywhere, except for hardware-protected pages (ie on some sort of ROM), which your process' memory is almost certainly not. It has the ability to write the breakpoint instruction regardless of the permissions exposed to the user process being debugged.
The debugger must use some syscall to change the access rights on the memory of the target process before inserting the breakpoint.
Personally, I'm guessing the first thing is happening. The segment permissions are only in place to protect your target process from itself, not from a debugger process or from the kernel. Debugging mechanisms in operating systems pretty regularly violate "normal" permissions to allow the debugger to do whatever it wants to the target process. This, of course, is why some operating systems require you to enter a password before you're allowed to use the debugger in certain scenarios.
However, you can test if it's the second one by attempting to write to the code segment from inside the target process after a breakpoint has been set. If the write succeeds, you know the permissions have been lowered by the OS (to allow the process to be debugged). It would be pretty awkward for the OS to require a debugger to jump through this hoop since it can already insert arbitrary code into the writeable parts of memory and then force a jump to it by generating a stack frame overflow.
The debugger takes advantage of the WriteProcessMemory() function to alter the instruction in place. It'll keep a copy of the instruction. When the bp is hit it will reset the old byte value and set EIP back to the previous instruction so the real instruction can execute.

how does debugger resume from breakpoint?

Assume a debugger(common x86 ring3 debugger such as olly, IDA, gdb...) sets a
software breakpoint to virtual address 0x1234.
this is accomplished by replacing the whatever opcode at 0x1234 to '0xCC'
now let's assume that debugee process runs this 0xCC instruction and raises
software exception and debugger catches this.
debugger inspects memory contents, registers and do some stuff.. and
now it wants to resume the debugee process.
this is as far as I know. from now, its my assumption.
debugger recovers the original opcode(which was replaced to 0xCC) of
debugee in order to resume the execution.
debugger manipulates the EIP of debugee's CONTEXT to point the
recovered instruction.
debugger handles the exception and now, debugee resumes from breakpoint.
but debugger wants the breakpoint to remain.
how can debugger manage this?
To answer the original question directly, from the GDB internals manual:
When the user says to continue, GDB will restore the original
instruction, single-step, re-insert the trap, and continue on.
in short and common people words:
Since getting into debug state is atomic operation in X86 and in ARM the processor gets into it and exit of debug state as same as any other instruction in the architecture.
see gdb documentation explains how it works and can be used.
Here are some highlights from ARM and X86 specifications:
in ARM:
SW (Software) breakpoints are implemented by temporarily replacing the
instruction opcode at the breakpoint location with a special
"breakpoint" instruction immediately prior to stepping or executing
your code. When the core executes the breakpoint instruction, it will
be forced into debug state. SW breakpoints can only be placed in RAM
because they rely on modifying target memory.
A HW (Hardware) breakpoint is set by programming a watchpoint unit to monitor the core
busses for an instruction fetch from a specific memory location. HW
breakpoints can be set on any location in RAM or ROM. When debugging
code where instructions are copied (Scatterloading), modified or the
processor MMU remaps areas of memory, HW breakpoints should be used.
In these scenarios SW breakpoints are unreliable as they may be either
lost or overwritten.
In X86:
The way software breakpoints work is fairly simple. Speaking about x86
specifically, to set a software breakpoint, the debugger simply writes
an int 3 instruction (opcode 0xCC) over the first byte of the target
instruction. This causes an interrupt 3 to be fired whenever execution
is transferred to the address you set a breakpoint on. When this
happens, the debugger “breaks in” and swaps the 0xCC opcode byte with
the original first byte of the instruction when you set the
breakpoint, so that you can continue execution without hitting the
same breakpoint immediately. There is actually a bit more magic
involved that allows you to continue execution from a breakpoint and
not hit it immediately, but keep the breakpoint active for future use;
I’ll discuss this in a future posting.
Hardware breakpoints are, as you might imagine given the name, set
with special hardware support. In particular, for x86, this involves a
special set of perhaps little-known registers know as the “Dr”
registers (for debug register). These registers allow you to set up to
four (for x86, this is highly platform specific) addresses that, when
either read, read/written, or executed, will cause the processor to
throw a special exception that causes execution to stop and control to
be transferred to the debugger

What happens if TF(trap flag) is set to 0 in 8086 microprocessors?

Here I searched that:
Trap Flag (T) – This flag is used for on-chip debugging. Setting trap
flag puts the microprocessor into single step mode for debugging. In
single stepping, the microprocessor executes a instruction and enters
into single step ISR.
If trap flag is set (1), the CPU automatically generates an internal
interrupt after each instruction, allowing a program to be inspected
as it executes instruction by instruction.
If trap flag is reset (0), no function is performed.
https://en.wikipedia.org/wiki/Trap_flag
Now I am coding on emu-8086. As explained, TF must be set in order to debugger work.
Should I set a TF always myself or it is set automatically?
If I somehow set a TF to 0, will the whole computer systems debuggers work or just emu-8086 wont debug?
I've never used emu8086 but by looking at some screenshot of it and judging by its name it's probably an emulator - this means it is not running the code natively.
Each instruction is changing the state of a virtual 8086 CPU (represented as a data structure in memory) and not the state of your real CPU.
With this emulation, emu8086 doesn't need to rely on the TF flag to single-step your program, it just needs to stop after one step of emulation and wait for you to hit another button.
This is also why you can find a thing such as "Step back".
If you were wondering what would happen if a debugged program (and not an emulated one) sets the TF flag then the answer is that it depends on the debugger.
The correct behaviour is the one where the debuggee receives the exceptions but this is hard to handle correctly (since the debugger itself uses the TF flag).
Some debugger just don't care and swallow the exception (i.e. they don't forward it to the program under debug) assuming that a well written program doesn't need to use the TF flag.
Unfortunately malwares routinely use a set of anti-debug technique including setting the TF and checking it back/waiting for exceptions to detect the presence of a debugger.
A truly transparent debugger has to handle the RFLAGS register carefully.
When debugging with breakpoints the TF is not set while the program is executing, so there is nothing to worry about.
However when single stepping the TF is set during the next instruction, this is problematic during a pushfd/q and the debugger must explicitly handle that case to avoid detection.
If the debuggee sets the TF the debugger must pass the debug exception to the program - under current OS the TF won't last more than an instruction because the OS will catch the exception,
trasnform it in a signal and dispatch it to the program while clearing the TF. So the debugger can simply do a check before stepping into a popfd/q instruction.
Where the TF doesn't get cleared by the OS the debugger must effectively emulate RFLAGS with a copy.
The debugger sets TF according to what it needs to do. The code being debugged should not modify TF.

Resources