All functions are actually variables, so it’s actually pretty easy to store all your functions in array (by referencing them without the ()): // Create your functions, in a variety of manners… // (The second method is preferable, but I show the first for reference.) function fun1() { alert(“Message 1”); }; var fun2 = function() … Read more
priority_queue doesn’t allow iteration through all the members, presumably because it would be too easy in invalidate the priority ordering of the queue (by modifying the elements you traverse) or maybe it’s a “not my job” rationale. The official work-around is to use a vector instead and manage the priority-ness yourself with make_heap, push_heap and … Read more
You can implement a stack with O(1) pop(), push() and get_min(): just store the current minimum together with each element. So, for example, the stack [4,2,5,1] (1 on top) becomes [(4,4), (2,2), (5,2), (1,1)]. Then you can use two stacks to implement the queue. Push to one stack, pop from another one; if the second … Read more
As of Python >=3.2, os.makedirs() can take a third optional argument exist_ok: os.makedirs(mydir, exist_ok=True)
Use: Queue<Object> queue = new LinkedList<>(); You can use .offer(E e) to append an element to the end of the queue and .poll() to dequeue and retrieve the head (first element) of the queue. Java defined the interface Queue, the LinkedList provided an implementation. It also maintains references to the Head and Tail elements, which … Read more
Well, let’s take a closer look at the scenario you are describing above. I think it’s important to paste the documentation immediately prior to the snippet in your question to provide context: Section 4.7 of the AMQP 0-9-1 core specification explains the conditions under which ordering is guaranteed: messages published in one channel, passing through … Read more
It is best to make the condition (monitored by your condition variable) the inverse condition of a while-loop: while(!some_condition). Inside this loop, you go to sleep if your condition fails, triggering the body of the loop. This way, if your thread is awoken–possibly spuriously–your loop will still check the condition before proceeding. Think of the … Read more
Did you try these? Pop from queue x, a = a[0], a[1:] Pop from stack x, a = a[len(a)-1], a[:len(a)-1] Push a = append(a, x) From: https://code.google.com/p/go-wiki/wiki/SliceTricks
Yeah. Queue LinkedList being the most trivial concrete implementation.
Use KVO to observe the operations property of your queue, then you can tell if your queue has completed by checking for [queue.operations count] == 0. Somewhere in the file you’re doing the KVO in, declare a context for KVO like this (more info): static NSString *kQueueOperationsChanged = @”kQueueOperationsChanged”; When you setup your queue, do … Read more