A Basic Just-In-Time Compiler (2015)

(nullprogram.com)

31 points | by ibobev 3 hours ago ago

5 comments

  • kscarlet an hour ago ago

    It seems that JIT is overloaded with at least 2 meaning.

    The canonical definition of JIT is "compilation during execution of a program". Usually, a program is being interpreted first, then switches to compiled code in the middle of execution. This is not what this article does.

    What this article does is sometimes called on-the-fly AOT, or just on-the-fly compilation. I'd prefer not overloading the term "JIT".

    • MobiusHorizons 28 minutes ago ago

      I always took the distinction to be the executable memory bit as opposed to writing an executable and then launching it in the usual way. Of course high performance runtimes that contain JITs do typically interpret first in order to get type information and reduce startup latency, but that’s more a function of the language not being efficiently AOT compileable rather than fundamental to the concept of a JIT

      • kscarlet 20 minutes ago ago

        In that sense almost every compiled Lisp/Scheme implementation, GHC, etc. or any other interactive programming system, count as JIT. But virtually nobody in those circles refer to such implementations as JIT. Instead, people says "I wish our implementation was JIT to benefit from all those optimizations it enables"!

        • MobiusHorizons 10 minutes ago ago

          Do they generate machine code in ram and jump to it? Or do they interpret byte code?

          EDIT: at least GHC seems to be a traditional AOT compiler.

  • possiblywrong an hour ago ago

    Following is the closed form solution linked in the article (from a since-deleted Reddit comment):

        from functools import reduce
    
        def recurrence(ops, a0, n):
            def transform(x, op):
                return eval(repr(x) + op)
            ops = ops.split()
            m = reduce(transform, [op for op in ops if op[0] in ('*', '/')], 1)
            b = reduce(transform, ops, 0)
            for k in range(n + 1):
                print('Term 0: ', a0 * m ** k + b * (m ** k - 1) / (m - 1))
    
    > This is really only interesting if a particular (potentially really large) term of the sequence is desired, as opposed to all terms up to a point. The key observation is that any sequence of the given set of operations reduces to a linear recurrence which has the given solution.