tail-recursion

问题 This question already has answers here : Variadic curried sum function (12 answers) Closed 4 years ago . I want to make a function which adds arguments. Invoking this function should be functionAdd(2)(3)(4)...(n); And the result 2+3+4...+n I'm trying this function myfunction(num){ var summ =+ num; if(num !== undefined){ return myfunction(summ); } }; But it doesn't works, an error of ovwerflow. And I don't understand where I should out from this function; 回答1: You can use the .valueOf to do

问题 I have been attempting to perform multiple (different) string replacement with recursion and I have hit a roadblock. I have sucessfully gotten the first replacement to work, but the subsequent replacements never fire. I know this has to do with the recursion and how the with-param string is passed back into the call-template. I see my error and why the next xsl:when never fires, but I just cant seem to figure out exactly how to pass the complete modified string from the first xsl:when to the

问题 I have been attempting to perform multiple (different) string replacement with recursion and I have hit a roadblock. I have sucessfully gotten the first replacement to work, but the subsequent replacements never fire. I know this has to do with the recursion and how the with-param string is passed back into the call-template. I see my error and why the next xsl:when never fires, but I just cant seem to figure out exactly how to pass the complete modified string from the first xsl:when to the

问题 Let's say I have a simple function like this: int all_true(int* bools, int len) { if (len < 1) return TRUE; return *bools && all_true(bools+1, len-1); } This function can be rewritten in a more obviously tail-recursive style as follows: int all_true(int* bools, int len) { if (len < 1) return TRUE; if (!*bools) return FALSE; return all_true(bools+1, len-1); } Logically, there is zero difference between the two; assuming bools contains only TRUE or FALSE (sensibly defined), they do exactly the

问题 I watched the talk Three Beautiful Quicksorts and was messing around with quicksort. My implementation in python was very similar to c (select pivot, partition around it and recursing over smaller and larger partitions). Which I thought wasn't pythonic . So this is the implementation using list comprehension in python. def qsort(list): if list == []: return [] pivot = list[0] l = qsort([x for x in list[1:] if x < pivot]) u = qsort([x for x in list[1:] if x >= pivot]) return l + [pivot] + u

问题 I am trying to understand tail-recursion in Haskell. I think I understand what it is and how it works but I'd like to make sure I am not messing things up. Here is the "standard" factorial definition: factorial 1 = 1 factorial k = k * factorial (k-1) When running, for example, factorial 3 , my function will call itself 3 times(give it or take). This might pose a problem if I wanted to calculate factorial 99999999 as I could have a stack overflow. After I get to factorial 1 = 1 I will have to

问题 In Haskell, if I write fac n = facRec n 1 where facRec 0 acc = acc facRec n acc = facRec (n-1) (acc*n) and compile it with GHC, will the result be any different than if I used fac 0 = 1 fac n = n * fac (n-1) I could easily do fac n = product [1..n] and avoid the whole thing, but I'm interested in how an attempt at tail recursion works out in a lazy language. I get that I can still get a stack overflow because thunks are building up, but does anything actually happen differently (in terms of

问题 I'm very new to Scala, so forgive my ignorance! I'm trying to iterate of pairs of integers that are bounded by a maximum. For example, if the maximum is 5, then the iteration should return: (0, 0), (0, 1), ..., (0, 5), (1, 0), ..., (5, 5) I've chosen to try and tail-recursively return this as a Stream: @tailrec def _pairs(i: Int, j: Int, maximum: Int): Stream[(Int, Int)] = { if (i == maximum && j == maximum) Stream.empty else if (j == maximum) (i, j) #:: _pairs(i + 1, 0, maximum) else (i, j)

问题 I can't think of a true RAII language that also has tail call optimization in the specs, but I know many C++ implementations can do it as an implementation-specific optimization. This poses a question for those implementations that do: given that destructors are invoked at the end of a automatic variable's scope and not by a separate garbage collection routine, doesn't it violate TCO's constraint that a recursive call must be the last instruction at the end of a function? For example:-

问题 I almost understand how tail recursion works and the difference between it and a normal recursion. I only don't understand why it doesn't require stack to remember its return address. // tail recursion int fac_times (int n, int acc) { if (n == 0) return acc; else return fac_times(n - 1, acc * n); } int factorial (int n) { return fac_times (n, 1); } // normal recursion int factorial (int n) { if (n == 0) return 1; else return n * factorial(n - 1); } There is nothing to do after calling a