break

Early-breaking from Rust's match

好久不见. 提交于 2020-07-29 12:20:14
问题 I want to switch through many possible cases for x and there's one case (here x == 0 ) where I want to check the result of some additional code to determine what to do next. One possibility is to return early from the match. I'd use break to do this early-returning in C, but this isn't allowed in Rust. return returns from the parent function (in this case main() ) and not from the match only (i.e. the println! at the end isn't run!). I could just negate the sub-condition (here y == 0 ) and

Early-breaking from Rust's match

一世执手 提交于 2020-07-29 12:19:06
问题 I want to switch through many possible cases for x and there's one case (here x == 0 ) where I want to check the result of some additional code to determine what to do next. One possibility is to return early from the match. I'd use break to do this early-returning in C, but this isn't allowed in Rust. return returns from the parent function (in this case main() ) and not from the match only (i.e. the println! at the end isn't run!). I could just negate the sub-condition (here y == 0 ) and

Python - Determine Tic-Tac-Toe Winner

我的未来我决定 提交于 2020-07-06 20:42:37
问题 I am trying to write a code that determines the winner of a tic-tac-toe game. (This is for a college assignment) I have written the following function to do so: This code only checks for horizontal lines, I haven't added the rest. I feel that this is something that needs a bit of hardcoding. def iswinner(board, decorator): win = True for row in range(len(board)): for col in range(len(board)): if board[row][col] == decorator: win = True else: win = False break Where "board" is a 2D array of

break; causing segment fault

北城余情 提交于 2020-06-29 06:28:08
问题 I was doing parenthesis checker using the stack. One of the if else statement containing a break; statement is causing segment fault. I tried removing the break program runs fine but prints the wrong answer as that break is needed to print correct output. What is the cause of such a segment fault? the break doesn't access any memory unit.right? Questions link #include <iostream> #include<stack> using namespace std; int main() { //code int n; char c,comp; cin>>n; while(n--) { stack<char>s;

Using label and break in if-else statement

為{幸葍}努か 提交于 2020-06-25 07:08:39
问题 I have this code which generates random number from an array, at a particular condition (when x and y both are equal to zero). I want the control to jump to the label. But the control never jumps to the label in any condition. I wanted to know that whether I am doing it right or not? int[] arr = {0, 1, 2}; Random rn = new Random(); label: { //some code if (x != 0 && y !=0) { //some code } else { break label; } } 回答1: Try to Avoid using labels. What you could do there, is: while(true) { if(x =

Using label and break in if-else statement

别来无恙 提交于 2020-06-25 07:07:25
问题 I have this code which generates random number from an array, at a particular condition (when x and y both are equal to zero). I want the control to jump to the label. But the control never jumps to the label in any condition. I wanted to know that whether I am doing it right or not? int[] arr = {0, 1, 2}; Random rn = new Random(); label: { //some code if (x != 0 && y !=0) { //some code } else { break label; } } 回答1: Try to Avoid using labels. What you could do there, is: while(true) { if(x =

list comprehensions with break

风流意气都作罢 提交于 2020-05-16 02:35:08
问题 I have the following code that I would like to write in one line with a list comprehension. list1 = [4, 5, 6, 9, 10, 16, 21, 23, 25, 27] list2 = [1, 3, 5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56] list3 = [] for i in list1: for j in list2: if j>i: # print(i,j) list3.append(j) break print(list1) print(list3) The output is: [4, 5, 6, 9, 10, 16, 21, 23, 25, 27] [5, 7, 7, 11, 11, 17, 24, 24, 26, 56] It's the break statement that throws me off, I don't know where to put it. Thank you 回答1: You

list comprehensions with break

假装没事ソ 提交于 2020-05-16 02:35:03
问题 I have the following code that I would like to write in one line with a list comprehension. list1 = [4, 5, 6, 9, 10, 16, 21, 23, 25, 27] list2 = [1, 3, 5, 7, 8, 11, 12, 13, 14, 15, 17, 20, 24, 26, 56] list3 = [] for i in list1: for j in list2: if j>i: # print(i,j) list3.append(j) break print(list1) print(list3) The output is: [4, 5, 6, 9, 10, 16, 21, 23, 25, 27] [5, 7, 7, 11, 11, 17, 24, 24, 26, 56] It's the break statement that throws me off, I don't know where to put it. Thank you 回答1: You

break statement in “if else” - java

给你一囗甜甜゛ 提交于 2020-05-10 08:41:09
问题 I keep getting an error, if without else . I tried else if as well for (;;){ System.out.println("---> Your choice: "); choice = input.nextInt(); if (choice==1) playGame(); if (choice==2) loadGame(); if (choice==3) options(); if (choice==4) credits(); if (choice==5) System.out.println("End of Game\n Thank you for playing with us!"); break; else System.out.println("Not a valid choice!\n Please try again...\n");=[;'mm } also if you have a better idea on how to present this code please do not

break与continue

走远了吗. 提交于 2020-04-02 09:38:35
continue直接跳出本次循环,进入下次循环。 break直接跳出循环。 for( i=1;i<=10;i++) { if(i==6) break; printf(“%d”,i); } //输出结果:12345 for(var i=1;i<=10;i++) { if(i==6) continue; printf(“%d”,i); } //输出结果:1234578910 来源: https://www.cnblogs.com/two-fire/p/10930719.html