yield

5.并发编程协程

℡╲_俬逩灬. 提交于 2019-11-28 04:47:00
一 引子   本节的主题是基于单线程来实现并发,即只用一个主线程(很明显可利用的cpu只有一个)情况下实现并发,为此我们需要先回顾下并发的本质:切换+保存状态   cpu正在运行一个任务,会在两种情况下切走去执行其他的任务(切换由操作系统强制控制),一种情况是该任务发生了阻塞,另外一种情况是该任务计算的时间过长或有一个优先级更高的程序替代了它   协程本质上就是一个线程,以前线程任务的切换是由操作系统控制的,遇到I/O自动切换,现在我们用协程的目的就是较少操作系统切换的开销(开关线程,创建寄存器、堆栈等,在他们之间进行切换等),在我们自己的程序里面来控制任务的切换。     ps:在介绍进程理论时,提及进程的三种执行状态,而线程才是执行单位,所以也可以将上图理解为线程的三种状态   一:其中第二种情况并不能提升效率,只是为了让cpu能够雨露均沾,实现看起来所有任务都被“同时”执行的效果,如果多个任务都是纯计算的,这种切换反而会降低效率。为此我们可以基于yield来验证。yield本身就是一种在单线程下可以保存任务运行状态的方法,我们来简单复习一下: #1 yiled可以保存状态,yield的状态保存与操作系统的保存线程状态很像,但是yield是代码级别控制的,更轻量级 #2 send可以把一个函数的结果传给另外一个函数,以此实现单线程内程序之间的切换 import time def

python3 迭代器与生成器

淺唱寂寞╮ 提交于 2019-11-28 04:12:34
迭代器 迭代是python最强大的功能之一,是访问集合元素的一种方式。 迭代器是一个可以记住遍历的位置的对象 迭代器对象从集合的第一个元素开始访问,直到所有的元素被访问完结束。迭代器只能往前不会后退。 迭代器有两个基本的方法:iter()和next(). 字符串,列表或元组对象都可以用于创建迭代器 生成器 在 python中, 使用了yield的函数被称为生成器(generator) 跟普通函数不同的是,生成器是一个返回迭代器的函数,只能用于迭代操作,更简单的理解生成器就是一个迭代器。 在调用生成器运行的过程中,每次遇到yield时函数会暂停并保存当前所有的运行信息,返回yield的值。并在下一次执行next()方法从当前位置继续运行。 来源: https://www.cnblogs.com/fanx-1995/p/11391470.html

What is the equivalent syntax in VB.NET for “yield return”? [duplicate]

[亡魂溺海] 提交于 2019-11-28 03:56:35
问题 This question already has answers here : Yield in VB.NET (8 answers) Closed 6 years ago . Using the C# code below, how would you write it in Visual Basic? What is it trying to say? using System; using System.Collections.Generic; using System.IO; using System.Windows; using System.Windows.Media; using System.Windows.Media.Imaging; namespace Microsoft.LiveLabs.Pivot { /// <summary> /// Tile Builder class /// </summary> public static class TileBuilder { /// <summary> /// Specifies which images

yield函数的执行顺序

落爺英雄遲暮 提交于 2019-11-28 01:38:03
例子: 上图中标明了 行号出现的顺序 从顺序中可以看到 1.开始先执行for循环,执行到93行yield_test(1)时,会调用函数yield_test(),所以打印了79行内容 2.到80行时,遇到了yield,yield可以理解为return,所以93行yield_test(1)得到返回值0 3.接着开始执行94-96行代码 4.当再一次执行93行(for循环)时,从上图顺序中可以看到,程序执行了81-83行。因为n是1,所以这边程序也就停止执行了。 执行结果 从上面的结果中可以看到,yield函数被再次调用时,并不是从函数第一行代码开始执行,而是从yield语句后面一行开始的。 把上面例子中n改成3 1 def yield_test(n): 2 for i in range(n): 3 print(str(sys._getframe().f_lineno) + " before.") 4 yield call(i) 5 print(str(sys._getframe().f_lineno) + " after.") 6 print('i====', i) # 输出的是当前的i值 7 print(str(sys._getframe().f_lineno) + " afterPrint.") 8 9 10 def call(i): 11 return i * 2 12 13

python_0基础开始_day12

女生的网名这么多〃 提交于 2019-11-27 22:09:29
第十二节 一,生成器 生成器的核心:生成器的本质就是迭代器 迭代器是python自带的 生成器是程序员自己写的一种迭代器 在python中有三种方式来创建生成器: 基于函数编写 推导式方式编写 python内置函数或者模块提供(其实1,3两种本质上差不多,都是通过函数的形式生成,只不过1是自己写的生成器函数,3是python提供的生成器函数而已) # def func():# print("这是一个函数")# return "函数"# func()​def func(): print("这是一个生成器") yield "生成器"# 坑func() # 生成一个生成器print(func().__next__()) # 启动生成器print(func().__next__()) yield与return: 相同点: 都是返回内容 都可以返回多次(但是return写多个只会执行一个) 不同点: return是终止函数,yield是暂停生成器 yield能够记录当前执行位置 一个yield对应一个next def func(): msg = input("请输入内容") yield msg print("这是第二次启动") yield "生成器2" yield "生成器3" yield "生成器4"​g = func()print(next(g))print(next(g))print

Does thread.yield() lose the lock on object if called inside a synchronized method?

∥☆過路亽.° 提交于 2019-11-27 21:25:29
问题 I understand that Thread.currentThread().yield() is a notification to thread scheduler that it may assign cpu cycle to some other thread of same priority if any such is present. My question is: If current thread has got lock on some object and calls yield() , will it loses that lock right away? And when thread scheduler finds out there is no such thread to assign cpu cycle, then the thread which has called yield() will again be in fight to get lock on the object which it has lost earlier?? I

Why is using a sequence so much slower than using a list in this example

☆樱花仙子☆ 提交于 2019-11-27 21:15:07
Background: I have a sequence of contiguous, time-stamped data. The data-sequence has holes in it, some large, others just a single missing value. Whenever the hole is just a single missing value, I want to patch the holes using a dummy-value (larger holes will be ignored). I would like to use lazy generation of the patched sequence, and I am thus using Seq.unfold. I have made two versions of the method to patch the holes in the data. The first consumes the sequence of data with holes in it and produces the patched sequence . This is what i want, but the methods runs horribly slow when the

Passing multiple code blocks as arguments in Ruby

让人想犯罪 __ 提交于 2019-11-27 21:01:31
I have a method which takes a code block. def opportunity @opportunities += 1 if yield @performances +=1 end end and I call it like this: opportunity { @some_array.empty? } But how do I pass it more than one code block so that I could use yield twice, something like this: def opportunity if yield_1 @opportunities += 1 end if yield_2 @performances +=1 end end and: opportunity {@some_other_array.empty?} { @some_array.empty? } I am aware that this example could be done without yield, but it's just to illustrate. horseyguy You can't pass multiple blocks, per se, but you can pass multiple procs or

yield

我们两清 提交于 2019-11-27 20:53:24
https://blog.csdn.net/mieleizhi0522/article/details/82142856 先把yield理解为return #!/usr/bin/python3.5 # -*- coding=utf-8 -*- def foo(): print('starting...') while True: res = yield 4 print('res:',res) g=foo() print(next(g)) print('*'*20) print(next(g))    root@ubuntu:/home/vigossr/exercise# ./yield_ex.py starting... 4 ******************** res: None 4 我直接解释代码运行顺序,相当于代码单步调试: 1.程序开始执行以后,因为foo函数中有yield关键字,所以foo函数并不会真的执行,而是先得到一个生成器g(相当于一个对象) 2.直到调用next方法,foo函数正式开始执行,先执行foo函数中的print方法,然后进入while循环 3.程序遇到yield关键字,然后把yield想想成return,return了一个4之后,程序停止,并没有执行赋值给res操作,此时next(g)语句执行完成,所以输出的前两行

How does this function with a “yield” work in detail?

半世苍凉 提交于 2019-11-27 20:43:26
I got this method (inside a Unity C# Script), but I do not understand how the "yield" part actually works. I know from the MSDN that the function will return an IEnumerator which I could iterate throught, but this code waits 1,5 seconds and does not get iterated because this would mean, the objects created inside were created multiple times. Anyone here who can explain me how this code works? IEnumerator DestroyShip() { // create new gameobject Instantiate(ExplosionPrefab, transform.position, transform.rotation); // make current gameobject invisible gameObject.renderer.enabled = false; // set