How do segmented stacks work

Question

How do segmented stacks work? This question also applies to Boost.Coroutine so I am using the C++ tag as well here. The main doubt comes from this article It looks like what they do is keep some space at the bottom of the stack and check if it has gotten corrupted by registering some sort of signal handler with the memory allocated there (perhaps via mmap and mprotect?) And then when they detect that they have run out of space they continue by allocating more memory and then continuing from there. 3 questions about this

1. Isn't this construct a user space thing? How do they control where the new stack is allocated and how do the instructions the program is compiled down to get aware of that?

   A push instruction is basically just adding a value to the stack pointer and then storing the value in a register on the stack, then how can the push instruction be aware of where the new stack starts and correspondingly how can the pop know when it has to move the stack pointer back to the old stack?

2. They also say

   After we've got a new stack segment, we restart the goroutine by retrying the function that caused us to run out of stack

   what does this mean? Do they restart the entire goroutine? Won't this possibly cause non deterministic behavior?

3. How do they detect that the program has overrun the stack? If they keep a canary-ish memory area at the bottom then what happens when the user program creates an array big enough that overflows that? Will that not cause a stack overflow and is a potential security vulnerability?

If the implementations are different for Go and Boost I would be happy to know how either of them deal with this situation

Answer 1

I'll give you a quick sketch of one possible implementation.

First, assume most stack frames are smaller than some size. For ones that are larger, we can use a longer instruction sequence at entry to make sure there is enough stack space. Let's assume we're on an architecture that that has 4k pages and we're choosing 4k - 1 as the maximum size stack frame handled by the fast path.

The stack is allocated with a single guard page at the bottom. That is, a page that is not mapped for write. At function entry, the stack pointer is decremented by the stack frame size, which is less than the size of a page, and then the program arranges to write a value at the lowest address in the newly allocated stack frame. If the end of the stack has been reached, this write will cause a processor exception and ultimately be turned into some sort of upcall from the OS to the user program -- e.g. a signal in UNIX family OSes.

The signal handler (or equivalent) has to be able to determine this is a stack extension fault from the address of the instruction that faulted and the address it was writing to. This is determinable as the instruction is in the prolog of a function and the address being written to is in the guard page of the stack for the current thread. The instruction being in the prolog can be recognized by requiring a very specific pattern of instructions at the start of functions, or possibly by maintaining metadata about functions. (Possibly using traceback tables.)

At this point the handler can allocate a new stack block, set the stack pointer to the top of the block, do something to handle unchaining the stack block, and then call the function that faulted again. This second call is safe because the fault is in the function prolog the compiler generated and no side effects are allowed before validating there is enough stack space. (The code may also need to fixup the return address for architectures that push it onto the stack automatically. If the return address is in a register, it just needs to be in the same register when the second call is made.)

Likely the easiest way to handle unchaining is to push a small stack frame onto the new extension block for a routine that when returned to unchains the new stack block and frees the allocated memory. It then returns the processor registers to the state they were in when the call was made that caused the stack to need to be extended.

The advantage of this design is that the function entry sequence is very few instructions and is very fast in the non-extending case. The disadvantage is that in the case where the stack does need to be extended, the processor incurs an exception, which may cost much much more than a function call.

Go doesn't actually use a guard page if I understand correctly. Rather the function prolog explicitly checks the stack limit and if the new stack frame won't fit it calls a function to extend the stack.

Go 1.3 changed its design to not use a linked list of stack blocks. This is to avoid the trap cost if the extension boundary is crossed in both directions many times in a certain calling pattern. They start with a small stack, and use a similar mechanism to detect the need for extension. But when a stack extension fault does occur, the entire stack is moved to a larger block. This removes the need for unchaining entirely.

There are quite a few details glossed over here. (E.g. one may not be able to do the stack extension in the signal handler itself. Rather the handler can arrange to have the thread suspended and hand it off to a manager thread. One likely has to use a dedicated signal stack to handle the signal as well.)

Another common pattern with this sort of thing is the runtime requiring there to be a certain amount of valid stack space below the current stack frame for either something like a signal handler or for calling special routines in the runtime. Go works this way and the stack limit test guarantees a certain fixed amount of stack space is available below the current frame. One can e.g. call plain C functions on the stack so long as one guarantees they do not consume more than the fixed stack reserve amount. (One can use this to call C library routines in theory, though most of these have no formal specification of how much stack they might use.)

Dynamic allocation in the stack frame, such as alloca or stack allocated variable length arrays, add some complexity to the implementation. If the routine can compute the entire final size of the frame in the prolog then it is fairly straightforward. Any increase in the frame size while the routine is running likely has to be modeled as a new call, though with Go's new architecture that allows moving the stack, it is possible the alloca point in the routine can be made such that all the state allows a stack move to happen there.