peek

I have a blocking queue of objects. I want to write a thread that blocks till there is a object on the queue. Similar to the functionality provided by BlockingQueue.take(). However, since I do not know if I will be able to process the object successfully, I want to just peek() and not remove the object. I want to remove the object only if I am able to process it successfully. So, I would like a blocking peek() function. Currently, peek() just returns if the queue is empty as per the javadocs. Am I missing something? Is there another way to achieve this functionality? EDIT: Any thoughts on if I

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 using Python 3.4.1 on Windows 8. I would like to read a file with a buffered interface that allows me to peek a certain number of bytes ahead as well reading bytes. io.BufferedReader seems like the right choice. Unfortunately, io.BufferReader.peek seems useless because it appears to just return all the bytes stored in the buffer, rather than the number requested. In fact, this is allowed by the documentation of this function (emphasis mine): peek([size]) Return bytes from the stream without advancing the position. At most one single

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 can I manually trigger a single click-like vibration like the ones that happen when using peek&pop, or application shortcuts? 回答1: It seems that these guys have found a way to do it for iOS 9 for phones with Taptic Engine (including the iPhone 6S). http://unifiedsense.com/development/using-taptic-engine-on-ios.html I'm not sure if it applies to iOS 8 or any other phones. They've essentially retrieved the Taptic Engine interface from UIDevice and called actuateFeedback with an integer corresponding to Peek or Pop. Unfortunately these apps

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'm searching for a simple way to evaluate a simple math expression from an string, like this: 3*2+4*1+(4+9)*6 I just want + and * operations plus ( and ) signs. And * has more priority than + . 回答1: I think you're looking for a simple recursive descent parser . Here's a very simple example: const char * expressionToParse = "3*2+4*1+(4+9)*6"; char peek() { return *expressionToParse; } char get() { return *expressionToParse++; } int expression(); int number() { int result = get() - '0'; while (peek() >= '0' && peek() = '0' && peek() 回答2: One

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 tried the following snippet of Java 8 code with peek . List list = Arrays.asList("Bender", "Fry", "Leela"); list.stream().peek(System.out::println); However there is nothing printed out on the console. If I do this instead: list.stream().peek(System.out::println).forEach(System.out::println); I see the following which outputs both the peek as well as foreach invocation. Bender Bender Fry Fry Leela Leela Both foreach and peek take in a (Consumer super T> action) So why is the output different? 回答1: The Javadoc mentions the following:

I am aware of the specific function in golang from the bufio package. func (b *Reader) Peek(n int) ([]byte, error) Peek returns the next n bytes without advancing the reader . The bytes stop being valid at the next read call. If Peek returns fewer than n bytes, it also returns an error explaining why the read is short. The error is ErrBufferFull if n is larger than b's buffer size. I need to be able to read a specific number of bytes from a Reader that will advance the reader . Basically, identical to the function above, but it advances the reader. Does anybody know how to accomplish this?

1. Stack.peek() peek()函数返回栈顶的元素，但不弹出该栈顶元素。 2. Stack.pop() pop()函数返回栈顶的元素，并且将该栈顶元素出栈。 文章来源: Stack.peek()和Stack.pop()的区别

使用栈实现队列的下列操作： push(x) – 将一个元素放入队列的尾部。 pop() – 从队列首部移除元素。 peek() – 返回队列首部的元素。 empty() – 返回队列是否为空。 示例: MyQueue queue = new MyQueue(); queue.push(1); queue.push(2); queue.peek(); // 返回 1 queue.pop(); // 返回 1 queue.empty(); // 返回 false 说明: 你只能使用标准的栈操作 – 也就是只有 push to top, peek/pop from top, size, 和 is empty 操作是合法的。 你所使用的语言也许不支持栈。你可以使用 list 或者 deque（双端队列）来模拟一个栈，只要是标准的栈操作即可。 假设所有操作都是有效的 （例如，一个空的队列不会调用 pop 或者 peek 操作）。 class MyQueue { public: /** Initialize your data structure here. */ stack<int> a; stack<int> b; MyQueue() { } /** Push element x to the back of queue. */ void push(int x) { a.push(x);

sudo add-apt-repository ppa:peek-developers/stable sudo apt update sudo apt install peek 来源： https://www.cnblogs.com/WaterGe/p/11726773.html