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You 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 (eith
by scott_s 14y ago
You 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.
- mistercow 14y agoAre you saying that C with neither pointers nor arrays would still be Turing complete? I am skeptical.
- knz42 14y agoIt's not, and it's not obvious. https://tdotc.wordpress.com/2012/11/18/on-the-turing-completeness-of-c/ https://tdotc.wordpress.com/2012/11/18/on-the-turing-complet...
- scott_s 14y agoI actually pondered on this for a long time, and I haven't come to a conclusion. I cannot, however, counter any of your arguments, as I thought some of the same things myself. This is a surprising train of thought.