CSE 14 Lecture Notes - Lecture 12: Standard Streams

52 views5 pages

coralox416

10 May 2018

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

CMPS12A Lecture 12: More Arrays & Introduction to Input/Output Streams

There is a little bit more about how arrays work. One thing for sure is how one array can be

opied to aothe aa sie ou a’t just ake a aa aiale eual aothe aa

variable. Thus, what you have to do is:

Example:

class copyArray {

public static void main(String[] args) {

int [] x = {0, 1, 2, 3};

int [] y = new int[x.length];

y = x; //Does not work so take this line off

for (int i=0; i<x.length; i++)

y[i] = x[i];

“ste.out.pitlFo :  + [] +   + [] +   + [] +   + [];

“ste.out.pitlFo :  + [] +   + [] +   + [] +   + [];

y[2] = 10;

y[1] = 20;

“ste.out.pitlFo :  + [] +   + [] +   + [] +   + [];

“ste.out.pitlFo :  + [] +   + [] +   + [] +   + [];

}

You’ll see that  doig the fo loop ad opig eah eleet of  dietl ito , e get a

op of it hee oth aiales do’t efeee the sae aa. “o i the fist ouput, it pits

out:

OUTPUT

0 1 2 3

find more resources at oneclass.com

Unlock document

This preview shows pages 1-2 of the document.
Unlock all 5 pages and 3 million more documents.

Already have an account? Log in

In the second output:

OUTPUT

0 1 2 3

0 20 10 3

Thus, we did succeed in making a unique copy of the array. Each array is unique and does not

reference each other at the same time. Notice that for the y array, we need to initiate how

much space there should be to store items. Since we want another copy of the x array, we

ake the ’s aa stoage e the sae length as x, so x.length.

Another example with arrays but using functions:

class arrayFunctions {

p s v m() { //puli stati…

int [] x = {1, 4, 6, 10};

int [] y = new int[x.length];

copyArray(x, 0, y, x.length);

printArray(x);

printArray(y);

}

static void printArray(int[] a) {

for (int i = 0; i < a.length; i++) {

“ste.out.pita[i] +  ;

System.out.println();

}

static int [] copyArray(int[] a, int n, int[] b, int m) {

for (int i = n; i < m; i++)

b[i] = a[i];

find more resources at oneclass.com

Unlock document

This preview shows pages 1-2 of the document.
Unlock all 5 pages and 3 million more documents.

Already have an account? Log in

Document Summary

Cmps12a lecture 12: more arrays & introduction to input/output streams. There is a little bit more about how arrays work. One thing for sure is how one array can be (cid:272)opied to a(cid:374)othe(cid:396) a(cid:396)(cid:396)a(cid:455) si(cid:374)(cid:272)e (cid:455)ou (cid:272)a(cid:374)"t just (cid:373)ake a(cid:374) a(cid:396)(cid:396)a(cid:455) (cid:448)a(cid:396)ia(cid:271)le e(cid:395)ual a(cid:374)othe(cid:396) a(cid:396)(cid:396)a(cid:455) variable. (cid:455)ste(cid:373). out. p(cid:396)i(cid:374)tl(cid:374)(cid:894)(cid:862)fo(cid:396) (cid:454): (cid:862) + (cid:455)[(cid:1004)] + (cid:862) (cid:862) + (cid:455)[(cid:1005)] + (cid:862) (cid:862) + (cid:455)[(cid:1006)] + (cid:862) (cid:862) + (cid:455)[(cid:1007)](cid:895); y[2] = 10; y[1] = 20; (cid:455)ste(cid:373). out. p(cid:396)i(cid:374)tl(cid:374)(cid:894)(cid:862)fo(cid:396) (cid:454): (cid:862) + (cid:454)[(cid:1004)] + (cid:862) (cid:862) + (cid:454)[(cid:1005)] + (cid:862) (cid:862) + (cid:454)[(cid:1006)] + (cid:862) (cid:862) + (cid:454)[(cid:1007)](cid:895); (cid:455)ste(cid:373). out. p(cid:396)i(cid:374)tl(cid:374)(cid:894)(cid:862)fo(cid:396) (cid:454): (cid:862) + (cid:455)[(cid:1004)] + (cid:862) (cid:862) + (cid:455)[(cid:1005)] + (cid:862) (cid:862) + (cid:455)[(cid:1006)] + (cid:862) (cid:862) + (cid:455)[(cid:1007)](cid:895); You"ll see that (cid:271)(cid:455) doi(cid:374)g the fo(cid:396) loop a(cid:374)d (cid:272)op(cid:455)i(cid:374)g ea(cid:272)h ele(cid:373)e(cid:374)t of (cid:454) di(cid:396)e(cid:272)tl(cid:455) i(cid:374)to (cid:455), (cid:449)e get a (cid:272)op(cid:455) of it (cid:449)he(cid:396)e (cid:271)oth (cid:448)a(cid:396)ia(cid:271)les do(cid:374)"t (cid:396)efe(cid:396)e(cid:374)(cid:272)e the sa(cid:373)e a(cid:396)(cid:396)a(cid:455). O i(cid:374) the fi(cid:396)st ouput, it p(cid:396)i(cid:374)ts out:

Related Questions

(0) public class ReferenceTest 6 public static void methodSix (int) 1 x){ x[0] = new int [2]; x[0][0] = 0; x(01 (1) = 1; x = new int [10] [5); x[0] (3) = 3; public static void main(String[] args) { int[] O Y = [ {1,2,3}, {2,4}}; methodSix (y); //WHAT ARE THE VALUES IN THE ARRAY OF ARRAYS //REFERRED TO BY Y AT THIS POINT? (ANSWER IN CHART)

2. In this section you will be shown several classes. Inside the code for each class will be a question. You should provide your answer to the question inside the chart. Many of the questions will ask you whether a variable has changed or not. In cases that the variable being asked about is a primitive type, you should provide your answer based on the data stored in the primitive type. In cases that the variable being asked about is a reference type, you should provide your answer based on the data stored at the Object linked to by the reference type. For example, in the first snippet, the question asks about the variable x defined in the main method. Since x is a primitive type, you should answer YES if you think the variable x has had its value changed after the method methodOne() returns. All your answers should go in the chart below: (9 points) Part Answer to question x changed (circle one)? YES NO y changed (circle one)? a changed (circle one)? b changed (circle one)? d1 a changed (circle one)? b changed (circle one)? YES e words changed (circle one)? YES NO Values in array of array referred to by y: What is printed to the screen? (a) public class ReferenceTesti public static void methodOne (int x){ x++; public static void main(String[args) { int x = 0; methodOne (x); //IS THE DATA STORED OR REFERRED TO BY X THE SAME? (ANSWER IN CHART)

(g) Suppose I define a type Point which contains two properties, x and y. The Point class has both getters and setters defined on it which do the obvious thing of setting and getting x and y. Remember that the get (int index) method defined on an ArrayList will return a reference to the index element of the ArrayList. import java.util.ArrayList; public class ReferenceTest7 public static void methodSevenA (ArrayList coordinates) Point first - coordinates.get(0); first.set X (200.0); public static void methodsevenB (ArrayList coordinates) Point second = coordinates.get (1); second = new Point (5.0,100.0); second. setX (100); public static void main(String[] args) { ArrayList coordinates = new ArrayList(); coodinates.add(new Point (1.0,2.0)); coordinates.add(new Point(3.0, 4.0)); methodSevenA (coordinates); methodSeven (coordinates); System.out.println (coordinates.get (0).getX()); System.out.println (coordinates.get (1).getX()); 1/WHAT IS PRINTED TO THE SCREEN? (ANSWER IN CHART)

CSE 14 Lecture Notes - Lecture 12: Standard Streams

Document Summary

Get access

Related Documents

CSE 14 Lecture Notes - Lecture 14: Bubble Sort

CSE 14 Lecture Notes - Lecture 10: 5,6,7,8

CSE 14 Lecture Notes - Lecture 11: Dynamic Array

Related Questions