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No. It is well known that the C preprocessor is not turing complete. Looking in https://github.com/orangeduck/CPP_COMPLETE/blob/master/RECR.h https://github.co
by CJefferson 14y ago
No.
It is well known that the C preprocessor is not turing complete. Looking in https://github.com/orangeduck/CPP_COMPLETE/blob/master/RECR.h https://github.com/orangeduck/CPP_COMPLETE/blob/master/RECR.... shows us this technique is not turing complete, as these functions define a maximum recursion depth.
- krenoten 14y agoIt is impossible to implement a system without a de-facto recursion depth limit.
- asynchrony 14y agoRecursion depth limit is really a memory limit, so you just need someone to come and install more RAM when you run out in order to obtain de facto infinite memory, hence infinite recursion depth.
- mistercow 14y agoEven that is not infinite because there is a limit to how much memory you can address. You don't have to bring physical realities into it.
- asynchrony 14y agoThe system will just spontaneously update to the next power-of-two bits when address space gets short. Problem solved.
- mistercow 14y agoThat would break binary compatibility with all existing programs, though. So I guess the claim should be that it's impossible to implement a practically useful system without a de-facto recursion depth limit.
- tjgq 14y agoHow is a "limit to how much memory [one] can address" not a physical, rather than logical, constraint? Really, the crux of the matter is that you can have a function call itself an unbounded number of times in C (i.e. unbounded recursion). That you eventually run into a stack overflow is a limitation of the machine, not of the language; the definition of C does not prescribe a maximum number of recursive calls. This, together with conditionals, makes the C language Turing-complete.
- scott_s 14y agoYes, but as I point out elsewhere, Turing Completeness separates the expression from the machine. Machines will always have limits. But the expression of computation does not have to have limits.
- vidarh 14y agoBut this is not a de facto recursion limit of the system - the limit is encoded in the program itself, in the macros used to implement the limited recursion. void f() { f(); } is only limited by the system, and if compiled on a system that automatically expands the stack to fill all available memory, you could in theory keep adding memory, and the system could handle more recursions. The recursion limit is external to the program. But the linked c-preprocessor programs would have to be explicitly modified by adding more lines of code to achieve the same thing.
- finitocom 14y agoHow does tail-recursion fit into this discussion of infinite recursion?
- mistercow 14y agoI'm not sure how that disqualifies this from Turing completeness any more than finite pointer sizes disqualify every other language from Turing completeness.
- evincarofautumn 14y agoThe C preprocessor is a pushdown automaton, which if you add fixpoints (unbounded recursion) becomes a bounded storage machine. Our actual computers are such machines, which if you give them an infinite amount of memory are Turing machines. A language can be Turing-complete even if the implementation of that language is not.
- mistercow 14y agoCan C be implemented such that pointers do not have a fixed size? That is to say, suppose you have a hypothetical machine that uses bignum memory addresses. Can C be implemented to run on that machine without cheating and using a fixed pointer size?
- scott_s 14y agoYou don't need to reason about C's pointers to consider its Turing Complete. That it allows arbitrary storage, it has conditionals and allows full looping (either unbounded loops or full recursion) is enough. That is, a language that had C's semantics sans pointers would still be Turing Complete.