There are a couple of problems there. One is that you’re falling prey to The Horror of Implicit Globals by failing to declare your build variable in the function. But the logic has issues as well, here’s a minimal reworking: var keys_short = [“ch”,”d”,”u”,”tz”]; var keys_long = [“children”,”data”,”user_id”,”time_zone”]; function refit_keys(o){ var build, key, destKey, ix, … Read more
Whenever you call a function, the arguments are pushed on the stack, which means that data on the stack segment is “allocated”. When the function is called, the return adress is also pushed on the stack, by the CPU, so it knows where to return to. In your example case this means, that no arguments … Read more
setTimeout does not have a minimal delay of 0ms. The minimal delay is anywhere in the range of 5ms-20ms dependent on browsers. My own personal testing shows that setTimeout doesn’t place your back on the event stack immediately Live Example It has an arbitary minimal time delay before it gets called again var s = … Read more
The recursion limit is not the limit on recursion but the maximum depth of the python interpreter stack.There is something on the stack before your function gets executed. Spyder executes some python stuff before it calls your script, as do other interpreters like ipython. You can inspect the stack via methods in the inspect module. … Read more
When you use a function recursively, the compiler takes care of stack management for you, which is what makes recursion possible. Anything you can do recursively, you can do by managing a stack yourself (for indirect recursion, you just have to make sure your different functions share that stack). So, no, there is nothing that … Read more
You can use Directory.GetFiles to replace your method. Directory.GetFiles(dirPath, “*”, SearchOption.AllDirectories)
You have to return all the way up the stack: func2.call(this, sWord); should be: return func2.call(this, sWord);
It is a guard against a stack overflow, yes. Python (or rather, the CPython implementation) doesn’t optimize tail recursion, and unbridled recursion causes stack overflows. You can check the recursion limit with sys.getrecursionlimit: import sys print(sys.getrecursionlimit()) and change the recursion limit with sys.setrecursionlimit: sys.setrecursionlimit(1500) but doing so is dangerous — the standard limit is a … Read more
9 min, including test import math def nthperm(li, n): n -= 1 s = len(li) res = [] if math.factorial(s) <= n: return None for x in range(s-1,-1,-1): f = math.factorial(x) d = n / f n -= d * f res.append(li[d]) del(li[d]) return res #now that’s fast… nthperm(range(40), 123456789012345678901234567890)
You’ll need your recursion to also build something that can be sorted by at the end: declare @t TABLE ( [CHILD] [int] NOT NULL, [PARENT] [int] NOT NULL ) insert @t values ( 0, -1), — I added this ( 1, 2 ), ( 2, 0 ), ( 3, 1 ), ( 4, 2 ), … Read more