continue

Should developers avoid using continue in C# or its equivalent in other languages to force the next iteration of a loop? Would arguments for or against overlap with arguments about Goto ? I think there should be more use of continue! Too often I come across code like: for (...) { if (!cond1) { if (!cond2) { ... highly indented lines ... } } } instead of for (...) { if (cond1 || cond2) { continue; } ... } Use it to make the code more readable! Is continue any more harmful than, say, break ? If anything, in the majority of cases where I encounter/use it, I find it makes code clearer and less

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: How would I do a "hit any key" (or grab a menu option) in Python? raw_input requires you hit return. Windows msvcrt has getch() and getche(). Is there a portable way to do this using the standard libs? 回答1: try: # Win32 from msvcrt import getch except ImportError: # UNIX def getch(): import sys, tty, termios fd = sys.stdin.fileno() old = termios.tcgetattr(fd) try: tty.setraw(fd) return sys.stdin.read(1) finally: termios.tcsetattr(fd, termios.TCSADRAIN, old) 回答2: try: os.system('pause') #windows, doesn't require enter except whatever_it_is:

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I have been trying to find all connected components using 8 neighbors in a binary image, without using the function "bwlabel". For example, my input matrix is: a = 1 1 0 0 0 0 0 1 1 0 0 1 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 I would to have something like this: a = 1 1 0 0 0 0 0 1 1 0 0 2 2 0 1 1 0 0 0 2 0 1 1 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 There are 3 connected objects in this image. 回答1: This is a common problem in image processing. There are many variations, such as flood filling a region in an

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: When I copy text to the clipboard onPrimaryClipChanged method is called twice. Any ideas why? @Override public boolean onCreateOptionsMenu ( Menu menu ) { getMenuInflater (). inflate ( R . menu . main , menu ); final ClipboardManager cliboardManager = ( ClipboardManager ) getSystemService ( CLIPBOARD_SERVICE ); cliboardManager . addPrimaryClipChangedListener ( new OnPrimaryClipChangedListener () { @Override public void onPrimaryClipChanged () { ClipData clipData = cliboardManager . getPrimaryClip (); System . out . println ( "*****

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I am trying to get the IDispatch * of an open explorer window using IShellWindows::FindWindowSW ; however, I cannot seem to coax the method to return anything other than S_FALSE. The code I am using is basically: OleInitialize(nullptr); CComPtr<IShellWindows> spWindows; auto hr = spWindows.CoCreateInstance(CLSID_ShellWindows); auto pidl = ILCreateFromPath(L"C:\\temp"); VARIANT vtLoc; vtLoc.vt = VT_VARIANT | VT_BYREF; vtLoc.pbVal = (BYTE *) pidl; CComVariant vtEmpty; long lhwnd; CComPtr<IDispatch> spdisp; hr = spWindows->FindWindowSW(&vtLoc,

A colleague and I were trying to figure out a way of doing the equivalent of a "continue" statement within a VBScript "For/Next" loop. Everywhere we looked we found people had no way to do this in VBScript without having nasty nestings, which is not an option for us since it is a quite big loop. We came out with this idea. Would it work just as a "continue(to next iteration)"? Does anyone have any better workaround or improvement suggestion? For i=1 to N For workaroundloop = 1 to 1 [Code] If Condition1 Then Exit For End If [MoreCode] If Condition2 Then Exit For End If [MoreCode] If Condition2

前言：有时候，想要把自己的程序运行在别人的iphone手机上，但又不能通过本地真机调试的方法安装，这个时候我们就要打一个测试专用的ipa包给远方的测试小伙伴们测试。 步骤： 首先希望你的发布证书已经安装好了。 1.生成Ad Hoc描述文件 （1）进入证书管理后台网站 https://developer.apple.com/account/ios/certificate/ （2）找到"Provisioning Profiles"下的"Distribution" （3）点击“创建”铵钮 （4）选中"Ad Hoc"，然后点击"continue"按钮 （5）选择你项目所对应的 App ID，然后点击"continue"按钮 （6）选择项目对应的发布证书，然后点击"continue"按钮 （7）选择项目可以运行的测试机（如果没有加入测试机，需要提前通过udid加入测试机） （8）在此处输入Ad Hoc描述文件的名称，然后continue，就完成了描述文件的生成 （9）下载生成的Ad Hoc描述文件，并双击安装 2.在工程中配置证书 （1）打开工程，进入xcode，将此处的“Automatically manage signing”去掉勾选 （2）找到"Bulid Settings"->"Signing","Code Signing Identity"选择对应的发布证书,

Boolean is = false; for (String str : QiniuConstant.pictureMap.values()) { if(data.contains(str)){ is = true ; break;/continue;/return; } } System.out.println("is:"+is); break;直接跳出整个循环，打印结果。 continue;跳出本次循环，继续遍历下一个，遍历不结束不打印结果。 return;结束整个程序，不打印结果。 文章来源: break;continue;return;三者的区别

Ŀ¼ 给定年龄，用户可以猜三次年龄 年龄猜对，让用户选择两次奖励 用户选择两次奖励后可以退出 age = 18 count = 0 prize_dict = { 0 : '布娃娃' , 1 : '变形金刚' , 2 : '高达' , 3 : '零食大礼包' , 4 : '<墨菲定律>' } while count < 3 : inp_age = input ( '请输入您的名字: ' ) if not inp_age . isdigit (): #判断输入是否为数字.isdigit---字符串的内置函数 print ( '傻逼，你的年龄输入错误！！' ) continue inp_age_int = int ( inp_age ) if inp_age_int == age : print ( '你太聪明啦！！猜对啦！' ) print ( prize_dict ) for i in range ( 2 ): prize_choice = input ( '请输入你想要的奖品,如果不想要,则输入"n"退出!!!' ) if prize_dict != 'n' : print ( f '恭喜你获得奖品： {prize_dict[int(prize_choice)]}' ) else : break break elif inp_age_int < age : print ( '抱歉

旅行(不是加强版) 加强版 加强版数据范围： 我们注意到 也就是说要么是个树，要么是个基环树 60pts 这60分是个树，可以简单的贪心想到每次都走子树中编号最小的那个，并且把1作为根 dfs练手题 还是贴个代码叭 void dfs1(int now,int fa) { if(vis[now])return ; ans1[++t]=now; vis[now]=1; vector<int> qwq;//为了不使存储的点被后面的子树覆盖，所以用vector for(int e=head[now];e;e=ed[e].nxt) { int v=ed[e].to; if(v==fa)continue; qwq.push_back(v); } sort(qwq.begin(),qwq.end());//vector的排序 int qaq=qwq.size(); for(int i=0;i<qaq;i++) dfs1(qwq[i],now); } 100pts 大多数的解法 当m==n时，它是一个基环树（即树上挂着一个环的树） 基环树只要删掉环上的一条边，它就是个树了。所以我们可以枚举删掉哪条边。（需要开个 \(O_2\) ） 由于开 \(O_2\) 会让你的评测记录显得不优雅，我们要考虑考虑怎么不开 \(O_2\) 过掉这道题。 由于我们要删去环上的边，所以要先找个环