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spacechild1 | 4 days ago
The obvious issue is that you can't know how much space is left on the stack, so you basically have to guess and pick an arbitrary "safe" size limit. This gets even more tricky when functions may be called recursively.
The more subtle issue is that the stack memory returned by alloca() has function scope and therefore you must never call it directly in a loop.
I use alloca() on a regular basis, but I have to say there are safer and better alternatives, depending on the particular use case: arena/frame allocators, threadlocal pseudo-stacks, static vectors, small vector optimizations, etc.
12_throw_away|4 days ago
Oh, huh. I've never actually tried it, but I always assumed it would be possible to calculate this, at least for a given OS / arch. You just need 3 quantities, right? `remaining_stack_space = $stack_address - $rsp - $system_stack_size`.
But I guess there's no API for a program to get its own stack address unless it has access to `/proc/$pid/maps` or similar?
fluntcaps|4 days ago
wat10000|3 days ago
Stack memory is weird in general. It's usually a fixed amount determined when the thread starts, with the size typically determined by vibes or "seems to work OK." Most programmers don't have much of a notion of how much stack space their code needs, or how much their program needs overall. We know that unbounded non-tail recursion can overflow the stack, but how about bounded-but-large? At what point do you need to start considering such things? A hundred recursive calls? A thousand? A million?
It's all kind of sketchy, but it works well enough in practice, I suppose.
chuckadams|4 days ago
Joker_vD|4 days ago
Does such thing even exist? And non-64 bit platforms the address space is small enough that with several threads of execution you may just be unable to grow your stack even up to $system_stack_size because it'd bump into something else.
unknown|4 days ago
[deleted]
norir|4 days ago
chuckadams|4 days ago
cyberax|3 days ago
This is not a problem for Go, because it has resizable stacks.