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CS2 in C++ Peer Instruction Materials by Cynthia Bailey Lee is licensed under a Creative Commons Attribution- NonCommercial - ShareAlike 4.0 International License . Permissions beyond the scope of this license may be available at http://peerinstruction4cs.org .
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CS2 in C++ Peer Instruction Materials by Cynthia Bailey Lee is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.Permissions beyond the scope of this license may be available at http://peerinstruction4cs.org. CS 106X – Programming Abstractions in C++ Cynthia Bailey Lee
Today’s Topics • Queues contrasted with Stacks • Queues example: Mouse events • Queue example: Josephus Survivor Puzzle • Iterating using range-based for loop
Stacks vs. Queues • Stack is referred to as a LIFO data structure • “Last in, first out” • Queue is referred to as a FIFO data structure • “First in, first out” • How many of these operate as FIFO/Queue? • Patients waiting in a hospital Emergency Room • Passengers boarding and exiting an airplane • Passengers boarding and exiting an elevator (assume all board on ground floor and exit at the same destination) • People waiting in line for an amusement park ride (A) None (B) 1 (C) 2 (D) All 3
LIFO: FIFO: Image is modified by Cynthia Lee based on original from Wikimedia Commons “A seatmap over the Airbus A380” Source=self-made |Date=24 September 2007 |Author= S. Solberg J.Used under Creative Commons License
Queue Applications Mouse events
Event queues • While your code executes, a separate program is constantly listening for events and recording them • Mouse moved, mouse clicked, mouse dragged, keyboard key pressed, etc • Every once in a while, your code can call getNextEvent() to see what has happened • getNextEvent returns the events one at a time in the same order they happened • In other words, returns them in FIFO order! • When it is “recording” events, it is enqueuingevents in an event QUEUE • Very common use of the Queue ADT
Queue Applications Josephus Survivor Puzzle
Josephus Survivor Puzzle • N people seated in a circle • Skip M living people and kill the (M+1)th, repeatedly around the circle, until two remain • Question: Where do you sit? 1 2 12 3 11 4 10 5 9 6 8 7
Josephus Survivor Puzzle • Question: (you will need paper & pencil) • N = 12, M = 3 Where do you sit? • 1 or 2 • 1 or 6 • 5 or 10 • 5 or 6 • Other/none/more 1 2 12 3 11 4 10 5 9 6 8 7
Queues and Josephus Puzzle • Enqueue everybody • Dequeue and immediately re-enqueueM people. • Dequeue somebody forever. If more than two alive, repeat steps 2&3 • N = 5, M = 2 1. 2. 3. 2. 3. 2. 3.
Queues and Josephus Puzzle • Enqueue everybody • Dequeue and immediately re-enqueueM people. • Dequeue somebody forever. • If more than two alive, repeat steps 2&3 • N = 5, M = 2 Contents of the queue, reading from top down: • 1, 2, 4 • 2, 4, 5 • 2, 4, 5, 1 • 4, 5, 1, 2 • Other 1. 2. 3. 2. 3. 2. 3.
Queues and Josephus Puzzle:What does that look like in code? • Enqueue everybody • Dequeue and immediately re-enqueueMpeople. • Dequeuesomebody forever • If more than two alive, repeat steps 2&3
C++11 has a nice for loop just for collections ADTs • Range-based for loop gives you each item in a collection, one at a time • For ADTs with a clear linear order (Vector, Stack, Queue), they are given to you in that order • [Spoiler for Wednesday!] Some ADTs are not linear. In theory, we often consider the ordering of iteration on these to be “undefined,” but in practice some implementations allow you to rely on a certain order
C++11 range-based for loop • One way to iterate over a Vector: for (inti=0; i<classlist.size(); i++){ cout << classlist[i] << endl; } • Range-based for loop makes the code cleaner: for (string name : classlist){ cout << name << endl; }
Other iteration techniques in C++ (NOT recommended for this class) • STL vector and iterators for (vector<int>::iterator it = classlist.begin(); it != classlist.end(); ++it){ cout << *it << endl; } • Stanford library “foreach” loop (obsoleted by range-based for loop in C++11): foreach (string name in classlist){ cout << name << endl; }