nested-loops

How to break out of nested loops?

依然范特西╮ 提交于 2019-11-26 19:41:17
If I use a break statement, it will only break inner loop and I need to use some flag to break the outer loop. But if there are many nested loops, the code will not look good. Is there any other way to break all of the loops? (Please don't use goto stmt .) for(int i = 0; i < 1000; i++) { for(int j = 0; j < 1000; j++) { if(condition) { // both of the loops need to break and control will go to stmt2 } } } stmt2 What about: if(condition) { i = j = 1000;break; } No, don't spoil the fun with a break . This is the last remaining valid use of goto ;) If not this then you could use flags to break out

Why does Python have a limit on the number of static blocks that can be nested?

不羁岁月 提交于 2019-11-26 19:39:57
问题 The number of statically nested blocks in Python is limited to 20. That is, nesting 19 for loops will be fine (although excessively time consuming; O(n^19) is insane), but nesting 20 will fail with: SyntaxError: too many statically nested blocks What is the underlying reason for having such a limit? Is there a way to increase the limit? 回答1: This limit not only applies to for loops, but to all other control flow blocks as well. The limit for the number of nested control flow blocks is defined

How to break outer loops from inner structures that respond break (loops/switch)

与世无争的帅哥 提交于 2019-11-26 19:32:54
问题 How to I break an outer loop from within an nested structure that responds to the break statement in Swift? For example: while someCondition { if someOtherCondition { switch (someValue) { case 0: // do something case 1: // exit loop case 2...5: // do something else default: break } } else { someCondition = false } } The break will only get me out of the switch , and in Swift, it has to be used as empty cases are not allowed. How can I entirely exit the loop from within the switch ? 回答1: Swift

Variable amount of nested for loops

安稳与你 提交于 2019-11-26 19:10:11
问题 Edit: I'm sorry, but I forgot to mention that I'll need the values of the counter variables. So making one loop isn't a solution I'm afraid. I'm not sure if this is possible at all, but I would like to do the following. To a function, an array of numbers is passed. Each number is the upper limit of a for loop, for example, if the array is [2, 3, 5] , the following code should be executed: for(var a = 0; a < 2; a++) { for(var b = 0; b < 3; b++) { for(var c = 0; c < 5; c++) { doSomething([a, b,

Iterate Multi-Dimensional Array with Nested Foreach Statement

爱⌒轻易说出口 提交于 2019-11-26 18:33:33
I think this might be a pretty simple question, but I haven't been able to figure it out yet. If I've got a 2-dimensional array like so: int[,] array = new int[2,3] { {1, 2, 3}, {4, 5, 6} }; What's the best way to iterate through each dimension of the array with a nested foreach statement? If you want to iterate over every item in the array as if it were a flattened array, you can just do: foreach (int i in array) { Console.Write(i); } which would print 123456 If you want to be able to know the x and y indexes as well, you'll need to do: for (int x = 0; x < array.GetLength(0); x += 1) { for

How to nest multiple parfor loops

不羁的心 提交于 2019-11-26 16:18:50
问题 parfor is a convenient way to distribute independent iterations of intensive computations among several "workers". One meaningful restriction is that parfor -loops cannot be nested, and invariably, that is the answer to similar questions like there and there. Why parallelization across loop boundaries is so desirable Consider the following piece of code where iterations take a highly variable amount of time on a machine that allows 4 workers. Both loops iterate over 6 values, clearly hard to

Is there a way to loop through a multidimensional array without knowing it&#39;s depth?

只愿长相守 提交于 2019-11-26 14:23:25
问题 So far, if I have to loop through a multidimensional array, I use a foreach loop for each dimension. e.g for two dimensions foreach($array as $key=>$value) { foreach($value as $k2=>$v2) { echo } } What do I do when I don't know the depth of the array? ie the depth is variable. The only thing I can think of is to code a whole stack of loops and to break the loop if the next value is not an array.This seems a little silly. Is there a better way? 回答1: Yes, you can use recursion. Here's an

Breaking/exit nested for in vb.net

天大地大妈咪最大 提交于 2019-11-26 12:06:02
问题 How do I get out of nested for or loop in vb.net? I tried using exit for but it jumped or breaked only one for loop only. How can I make it for the following: for each item in itemList for each item1 in itemList1 if item1.text = \"bla bla bla\" then exit for end if end for end for 回答1: Unfortunately, there's no exit two levels of for statement, but there are a few workarounds to do what you want: Goto . In general, using goto is considered to be bad practice (and rightfully so), but using

How to break nested loops in JavaScript?

对着背影说爱祢 提交于 2019-11-26 11:35:24
I tried this: for(i = 0; i < 5; i++){ for(j = i + 1; j < 5; j++){ break(2); } alert(1); } only to get: SyntaxError : missing ; before statement So, how would I break a nested loop in JavaScript? Noon Silk You should be able to break to a label, like so: function foo () { dance: for(var k = 0; k < 4; k++){ for(var m = 0; m < 4; m++){ if(m == 2){ break dance; } } } } You need to name your outer loop and break that loop, rather than your inner loop - like this. outer_loop: for(i=0;i<5;i++) { for(j=i+1;j<5;j++) { break outer_loop; } alert(1); } See Aaron's. Otherwise: j=5;i=5 instead of break .

A puzzle related to nested loops

落花浮王杯 提交于 2019-11-26 11:18:58
问题 For a given input N, how many times does the enclosed statement executes? for i in 1 … N loop for j in 1 … i loop for k in 1 … j loop sum = sum + i ; end loop; end loop; end loop; Can anyone figure out an easy way or a formula to do this in general. Please explain. 回答1: First, I written a C code to generate sum: int main(){ int i =0, k =0, j =0, n =0; int N =0; int sum =0; N =10; for (n=1; n <= N; n++){ // unindented code here sum =0; for (i=1; i<=n; i++) for (j=1; j<=i; j++) for (k=1; k<=j;