Haskell uses lazy-evaluation to implement recursion, so treats anything as a promise to provide a value when needed (this is called a thunk). Thunks get reduced only as much as necessary to proceed, no more. This resembles the way you simplify an expression mathematically, so it’s helpful to think of it that way. The fact … Read more
JIT compilation is a tricky balancing act between not spending too much time doing the compilation phase (thus slowing down short lived applications considerably) vs. not doing enough analysis to keep the application competitive in the long term with a standard ahead-of-time compilation. Interestingly the NGen compilation steps are not targeted to being more aggressive … Read more
The compiler is simply able to transform this int fac_times (int n, int acc) { if (n == 0) return acc; else return fac_times(n – 1, acc * n); } into something like this: int fac_times (int n, int acc) { label: if (n == 0) return acc; acc *= n–; goto label; }
All current mainstream compilers perform tail call optimisation fairly well (and have done for more than a decade), even for mutually recursive calls such as: int bar(int, int); int foo(int n, int acc) { return (n == 0) ? acc : bar(n – 1, acc + 2); } int bar(int n, int acc) { return … Read more
How do you empty a vase containing five flowers? Answer: if the vase is not empty, you take out one flower and then you empty a vase containing four flowers. How do you empty a vase containing four flowers? Answer: if the vase is not empty, you take out one flower and then you empty … Read more
No, and it never will since Guido van Rossum prefers to be able to have proper tracebacks: Tail Recursion Elimination (2009-04-22) Final Words on Tail Calls (2009-04-27) You can manually eliminate the recursion with a transformation like this: >>> def trisum(n, csum): … while True: # Change recursion to a while loop … if n … Read more
You have three (or so) options to break out of loops. Suppose you want to sum numbers until the total is greater than 1000. You try var sum = 0 for (i <- 0 to 1000) sum += i except you want to stop when (sum > 1000). What to do? There are several options. … Read more
Tail-call optimization is where you are able to avoid allocating a new stack frame for a function because the calling function will simply return the value that it gets from the called function. The most common use is tail-recursion, where a recursive function written to take advantage of tail-call optimization can use constant stack space. … Read more
Consider a simple function that adds the first N natural numbers. (e.g. sum(5) = 0 + 1 + 2 + 3 + 4 + 5 = 15). Here is a simple JavaScript implementation that uses recursion: function recsum(x) { if (x === 0) { return 0; } else { return x + recsum(x – 1); … Read more