pop

I want to pop to the third view on the navigation stack back to the first view. I know how to pop one view at once: [self.navigationController popViewControllerAnimated:YES]; But how do I do two at once? Thanks... Meet You can try this to Jump between the navigation controller stack as well NSMutableArray *allViewControllers = [NSMutableArray arrayWithArray:[self.navigationController viewControllers]]; for (UIViewController *aViewController in allViewControllers) { if ([aViewController isKindOfClass:[RequiredViewController class]]) { [self.navigationController popToViewController

An error occurred in script '/usr/local/apache2/htdocs/read.php' on line 197: Only variables should be passed by reference (line 196 is $ext = strtolower(array_pop(explode('.',$filename))); ) if(!function_exists('mime_content_type')) { function mime_content_type($filename) { $mime_types = array( 'txt' => 'text/plain', 'htm' => 'text/html', 'html' => 'text/html', //ETC ); $ext = strtolower(array_pop(explode('.',$filename))); if (array_key_exists($ext, $mime_types)) { return $mime_types[$ext]; } elseif (function_exists('finfo_open')) { $finfo = finfo_open(FILEINFO_MIME); $mimetype = finfo_file(

I am new to Python and experimenting with lists I am using Python 3.2.3 (default, Oct 19 2012, 20:13:42), [GCC 4.6.3] on linux2 Here is my samplecode >>> l=[1,2,3,4,5,6] >>> for i in l: ... l.pop(0) ... print(l) ... I would expect the following output 1 [2, 3, 4, 5, 6] 2 [3, 4, 5, 6] 3 [4, 5, 6] 4 [5, 6] 5 [6] 6 [] Instead I am getting this 1 [2, 3, 4, 5, 6] 2 [3, 4, 5, 6] 3 [4, 5, 6] The for-loop stops iterating after 3 turns. Can somebody explain why? mgilson Unrolling a bit (the caret ( ^ ) is at the loop "index"): your_list = [1,2,3,4,5,6] ^ after popping off the first item: your_list = [2

https://leetcode-cn.com/problems/min-stack/ 设计一个支持 push，pop，top 操作，并能在常数时间内检索到最小元素的栈。 push(x) -- 将元素 x 推入栈中。 pop() -- 删除栈顶的元素。 top() -- 获取栈顶元素。 getMin() -- 检索栈中的最小元素。 示例: MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); --> 返回 -3. minStack.pop(); minStack.top(); --> 返回 0. minStack.getMin(); --> 返回 -2. 小菜鸡的尝试： 不是很能懂什么叫常数时间（猜测O(1)？）于是抱着必超时的心态开始写最普通的办法。 1 class MinStack { 2 public: 3 /** initialize your data structure here. */ 4 stack<int> main; 5 stack<int> b; 6 MinStack() { 7 8 } 9 10 void push(int x) { 11 main.push(x); 12 } 13 14

Is there any jQuery plugin or CSS technique to achieve this pop effect like google images? Thanks #images{ padding:30px; } #images img{ position:relative; float:left; height:100px; margin:5px; transition:0.3s; box-shadow: 0 0 0 10px #fff; } #images img:hover{ z-index:2; transform: scale(1.2); } #images img:hover:after{ content: attr(title); } <div id="images"> <img src="http://dummyimage.com/180x120/000/fff" alt=""/> <img src="http://dummyimage.com/175x104/f0f/fff" alt=""/> <img src="http://dummyimage.com/150x100/a3d/fff" alt=""/> <img src="http://dummyimage.com/278x125/cf5/fff" alt=""/> <img

This question already has an answer here: How can I concatenate str and int objects? 4 answers I am currently learning Python so I have no idea what is going on. num1 = int(input("What is your first number? ")) num2 = int(input("What is your second number? ")) num3 = int(input("What is your third number? ")) numlist = [num1, num2, num3] print(numlist) print("Now I will remove the 3rd number") print(numlist.pop(2) + " has been removed") print("The list now looks like " + str(numlist)) When I run the program, entering in numbers for num1, num2 and num3, it returns this: Traceback (most recent

设计一个支持 push，pop，top 操作，并能在常数时间内检索到最小元素的栈。 push(x) -- 将元素 x 推入栈中。 pop() -- 删除栈顶的元素。 top() -- 获取栈顶元素。 getMin() -- 检索栈中的最小元素。 示例: MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); --> 返回 -3. minStack.pop(); minStack.top(); --> 返回 0. minStack.getMin(); --> 返回 -2. class MinStack { public: /** initialize your data structure here. */ MinStack() { } void push(int x) { s1.push(x); if (s2.empty() || x <= s2.top()) s2.push(x); } void pop() { if (s1.top() == s2.top()) s2.pop(); s1.pop(); } int top() { return s1.top(); } int getMin() { return s2

Is there a way to delete an item from a dictionary in Python? Additionally, how can I delete an item from a dictionary to return a copy (i.e., not modifying the original)? Greg Hewgill The del statement removes an element: del d[key] However, this mutates the existing dictionary so the contents of the dictionary changes for anybody else who has a reference to the same instance. To return a new dictionary, make a copy of the dictionary: def removekey(d, key): r = dict(d) del r[key] return r The dict() constructor makes a shallow copy . To make a deep copy, see the copy module . Note that making