Question

If a language has control structures and variables, but no support for arrays, lists, memory access and allocation, etc, can it be Turing-complete?

Maybe if there was no limit to the amount of variables you can create, you can simulate arrays by creating variables like array_1, array_2, … array_6000 and manually loop through them, and somehow create complex data structures and recursion?

Edit: Even if you cannot access variables by name manipulation (array_10+i is not allowed)?

Answer 1

Certainly. Have a look at Lambda Calculus, which is one of the most minimal Turing Complete languages I’ve ever seen. Basically, all you have are lambdas (function literals); no assignment, no declaration, no data structures. It’s all very very slimmed-down.

You can, however, simulate a linear data structure like a List by chaining functions together. It gets pretty verbose, but it’s certainly possible and it’s much nicer than having a large series of sequentially named variables.

Generally speaking, whether or not a language is Turing Complete has nothing to do with whether it has Arrays. Functional languages like SML and Haskell lack arrays, just like Lambda Calculus, and these are actually useful languages! Saying a language is “Turing Complete” is merely another way of saying that there is no Turing Computable function which cannot be expressed in said language. This is a surprisingly loose qualification, allowing many languages which would be completely impractical (like Lambda Calculus).