Using Named Function Expressions: You can give a function expression a name that is actually private and is only visible from inside of the function ifself: var factorial = function myself (n) { if (n <= 1) { return 1; } return n * myself(n-1); } typeof myself === ‘undefined’ Here myself is visible only … Read more
I played around with domi27’s idea and came up with this. I made a nested array as my tree, [‘link’][‘sublinks’] is null or another array of more of the same. Templates The sub-template file to recurse with: <!–includes/menu-links.html–> {% for link in links %} <li> <a href=”https://stackoverflow.com/questions/8326482/{{ link.href }}”>{{ link.name }}</a> {% if link.sublinks %} … Read more
You can increment the stack depth allowed – with this, deeper recursive calls will be possible, like this: import sys sys.setrecursionlimit(10000) # 10000 is an example, try with different values … But I’d advise you to first try to optimize your code, for instance, using iteration instead of recursion.
Yes, that’s para. Compare with catamorphism, or foldr: para :: (a -> [a] -> b -> b) -> b -> [a] -> b foldr :: (a -> b -> b) -> b -> [a] -> b para c n (x : xs) = c x xs (para c n xs) foldr c n (x : … Read more
Update: As of January 1, 2020 Safari is the only browser that supports tail call optimization. The chromium team explicitly states that Tail Call Optimization is not under active development and can be tracked here. The implementation for Firefox can be tracked here Original Post Yes, ES2015 offers tail call optimization in strict mode. Dr. … Read more
You should wrap your recursive function call into a setTimeout, setImmediate or process.nextTick function to give node.js the chance to clear the stack. If you don’t do that and there are many loops without any real async function call or if you do not wait for the callback, your RangeError: Maximum call stack size exceeded … Read more
def getPermutations(string, prefix=””): if len(string) == 1: yield prefix + string else: for i in xrange(len(string)): for perm in getPermutations(string[:i] + string[i+1:], prefix+string[i]): yield perm Or without an accumulator: def getPermutations(string): if len(string) == 1: yield string else: for i in xrange(len(string)): for perm in getPermutations(string[:i] + string[i+1:]): yield string[i] + perm
Update: Given that this post is quite old, and I’ve modified this utility a lot for my own use during that time, I thought I should post a new version. My newest code can be found on The MathWorks File Exchange: dirPlus.m. You can also get the source from GitHub. I made a number of … Read more
Yes, you can do this: def infinity(start): yield start for x in infinity(start + 1): yield x This will error out once the maximum recursion depth is reached, though. Starting from Python 3.3, you’ll be able to use def infinity(start): yield start yield from infinity(start + 1) If you just call your generator function recursively … Read more
It is returning None because when you recursively call it: if my_var != “a” and my_var != “b”: print(‘You didn\’t type “a” or “b”. Try again.’) get_input() ..you don’t return the value. So while the recursion does happen, the return value gets discarded, and then you fall off the end of the function. Falling off … Read more