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3 points by almkglor 6132 days ago | link | parent

re: call/cc - I think a bit of the lambda the ultimate series of papers eventually boils down to the realization that a machine language jump-to-subroutine is equivalent to a call/cc, and the target of the call/cc just has to access the return address on the stack as a function address.


3 points by kens1 6132 days ago | link

I don't get it. There's the whole stack copying for call/cc, so call/cc is much more expensive.

(I read the "Lambda the ultimate GOTO" paper you referenced earlier; it's about goto vs structured programming, not call/cc. As an aside, it's very interesting to reflect on just how controversial structured programming was.)

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4 points by soegaard 6131 days ago | link

Implementing call/cc efficiently has been well-reasearched in the Scheme community. For a very well-written account of a non-stack-copying implementation see

R. Kent Dybvig. "Three Implementation Models for Scheme". PhD. Thesis. http://www.cs.indiana.edu/~dyb/papers/3imp.pdf

Then continue at ReadScheme at "Compiler Technology/Implementation Techniques and Optimization" to see further developments (Look especially for Clinger's papers).

http://library.readscheme.org/page8.html

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2 points by almkglor 6131 days ago | link

Who said anything about copying stack? For that matter - who said local variables should be kept in the stack anyway?

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1 point by kens1 6131 days ago | link

In MIT Scheme, the stack gets copied; at least that's what I was told last week. Whether or not you use a stack, the state will need to be copied.

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1 point by almkglor 6131 days ago | link

In a function call, the state (the current computation being done) is saved anyway, and therefore "copied" if that is your preferred term. So ideally, call/cc should have the same overhead as an ordinary function call; the only difference is that in call/cc the continuation state is the value given to the function, while in a function call it's just one of the values given to the function.

Note however that much of the theoretical analyses of call/cc make a basic assumption of a "spaghetti stack", which would mean that partially unwound stacks would be saved implicitly as long as any continuation exists which refers to that stack, and all stacks themselves are subject to garbage collection. Most machines don't actually have a spaghetti stack and can't make a spaghetti stack anyway ^^. That said a spaghetti stack could be implemented as a simple list, with push == cons and pop = cdr.

Alternatively store the local variables on a garbage-collected heap, and include a reference to the local variables with the continuation/return address (you'll probably need to save the pointer-to-local-variables anyway, since the target function is likely to use that pointer for its own locals). Again, no additional overhead over plain function calls, except perhaps to restructure the return address and the pointer-to-local-variables.

Don't know about MIT Scheme, but if I were to implement call/cc on stock hardware and compiling down to machine language that's what I'd do ^^

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3 points by kens1 6131 days ago | link

I'm still totally not understanding your claim that call/cc should have the same overhead as an ordinary function call.

I read the Clinger "Implementation Strategies for Continuations" paper and they found call/cc about 10 times slower than function calls on the tak/ctak tests. I tried those tests on PLT Scheme and the overhead I saw is even worse: .7 seconds with function calls vs 51.8 seconds with continuations on (tak 24 16 8).

Clinger says about the stack strategy: "When a continuation is captured, however, a copy of the entire stack is made and stored in the heap. ... Variations on the stack strategy are used by most implementations of Scheme and Smalltalk-80."

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3 points by almkglor 6131 days ago | link

Components of function call: (1) put arguments somewhere (2) put return address somewhere (3) jump to function.

Components of call/cc: (1) put return address as argument somewhere (2) put return address somewhere (3) jump to function.

That said, continuations generally assume that "put X somewhere" means somewhere != stack. Yes, even return addresses are not intended to be on the stack when using continuations. The main problem is when compiling down to C, which always assumes that somewhere == stack. If you're compiling down to machine language where you have access to everything, then you can just ignore the stack, but not in C, which inherently expects the stack.

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