Consider the following statements and tell the difference ? • int *p; p=(int *)malloc(sizeof(int)); p=NULL; free(p); • int *p; p=(int *)malloc(sizeof(int)); free(p); p=NULL;
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Consider the following statements and tell the difference ?
• int *p; p=(int *)malloc(sizeof(int)); p=NULL; free(p);
• int *p; p=(int *)malloc(sizeof(int)); free(p); p=NULL;
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- Consider the following two statements and tell the difference ? • int *p; p=(int *)malloc(sizeof(int)); p=NULL; free(p); • int *p; p=(int *)malloc(sizeof(int)); free(p); p=NULL; (C language)JAVA Implement the equals method of the Arrow class. Two arrows are equal when they have the same starting point and direction. public class Arrow{ private Point start; private String direction; /** Constructs an arrow. @param x the x-position @param y the y-position @param direction a compass direction N E S W NE NW SE SW */ public Arrow(int x, int y, String direction) { start = new Point(); start.move(x, y); this.direction = direction; } /** Checks whether this arrow is equal to another. @param otherObject another arrow @return true if this arrow and otherObject have the same position and direction. */ public boolean equals(Object otherObject) { /* code goes here */ }}public class Soru32{ // Returns length of LCS for X[0..m-1], Y[0..n-1] static void lcs(String X, String Y, int m, int n) { int[][] L = new int[m+1][n+1]; // Following steps build L[m+1][n+1] in bottom up fashion. Note // that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] for (int i=0; i<=m; i++) { for (int j=0; j<=n; j++) { if (i == 0 || j == 0) L[i][j] = 0; else if (X.charAt(i-1) == Y.charAt(j-1)) L[i][j] = L[i-1][j-1] + 1; else L[i][j] = Math.max(L[i-1][j], L[i][j-1]); } } // Following code is used to print LCS int index = L[m][n]; int temp = index; // Create a character array to store the lcs string char[] lcs = new char[index+1]; lcs[index] = '\u0000'; // Set the terminating character // Start from the right-most-bottom-most…
- public class Soru32{ // Returns length of LCS for X[0..m-1], Y[0..n-1] static void lcs(String X, String Y, int m, int n) { int[][] L = new int[m+1][n+1]; // Following steps build L[m+1][n+1] in bottom up fashion. Note // that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] for (int i=0; i<=m; i++) { for (int j=0; j<=n; j++) { if (i == 0 || j == 0) L[i][j] = 0; else if (X.charAt(i-1) == Y.charAt(j-1)) L[i][j] = L[i-1][j-1] + 1; else L[i][j] = Math.max(L[i-1][j], L[i][j-1]); } } // Following code is used to print LCS int index = L[m][n]; int temp = index; // Create a character array to store the lcs string char[] lcs = new char[index+1]; lcs[index] = '\u0000'; // Set the terminating character // Start from the right-most-bottom-most…Using Jcreator, open, compile, and run java and observe its runtime. Write your observation regarding how encapsulation works for the program and the data it processed. Limit your paper to a single paragraph with at least 50 words or five (5) sentences. Source code: public class EncapTest { private String name; private String idNum; private int age; public int getAge() { return age; } public String getName() { return name; } public String getIdNum() { return idNum; } public void setAge( int newAge) { age = newAge; } public void setName(String newName) { name = newName; } public void setIdNum( String newId) { idNum = newId; }} public class RunEncap { public static void main(String args[]) { EncapTest encap = new EncapTest(); encap.setName("James"); encap.setAge(20); encap.setIdNum("12343ms"); System.out.print("Name : " + encap.getName() + " Age : " + encap.getAge()); }}Please help debug this: Debugging Exercise 4-3, Farrell, Joyce, Java Programming, 9th edition, Cengage Learning public class DebugBox { private int width; private int length; private int height; private FixDebugBox() { length = 1; width = 1; height = 1; } public DebugBox(int width, int length, int height) { width = width; length = length; height = height; } public void showData() { System.out.println("Width: " + width + " Length: " + length + " Height: " + height); } public double getVolume() { double vol = length - width - height; return vol; } }
- WRITE THE BODY OF Java program - THE checkWinner METHOD BELOW SO THAT THE GAME STOPS WHEN SOMEONE WINS, OR WHEN THE BOARD IS FULL. AS SHOWN IN THE SAMPLE OUTPUT AT THE BOTTOM. Code: import java.util.Scanner; public class TicTacBoard { private char[][] board; // 2-D array of characters private char curPlayer; // the player whose turn it is (X or O) // Constructor: board will be size x size public TicTacBoard(int size) { board = new char[size][size]; // initialize the board with all spaces: for(int row=0; row < board.length; row++) for(int col=0; col < board[row].length; col++) board[row][col] = ' '; curPlayer = 'X'; // X gets the first move } public void playGame() { display(); do { takeTurn(); display(); }while(!checkWinner()); } /////// display //////// // Display the current status of the board on the // screen, using hyphens (-) for horizontal lines // and pipes (|) for vertical lines. public void…I need a help in making a method that will Input a text that will be converted to Huffman code and Input a Huffman code that will be converted to its text equivalent.You can try this sample Character input: HAAHAABAAACACAAAABABABEBEBBAABEEBBCCCCCAAAAAAACCCDDDDDDAAADDDDDEEEEEEDDADDDADDEEEEEEEECCCEEEEEEEEEEDDEEEEEEEEEEBBEEEEEFFGGHAHHEEHHA Huffman.java import java.util.PriorityQueue;import java.util.Scanner;import java.util.Comparator;import java.util.ArrayList; class HuffmanNode { int data; char c; HuffmanNode left; HuffmanNode right;} class MyComparator implements Comparator<HuffmanNode> { public int compare(HuffmanNode x, HuffmanNode y) { return x.data - y.data; }} public class Huffman { public static void printCode(HuffmanNode root, String s) { if (root.left == null && root.right == null && Character.isLetter(root.c)) { System.out.println(s + "tttt"+ s.length()); return; } printCode(root.left, s + "0");…Please implement in Java implement a keyed bag in which the items to be stored are strings (perhaps people’s names) and the keys are numbers (perhaps Social Security or other identification numbers). So, the in- sertion method has this specification: public void insert(String entry, int key);// Precondition: size( ) < CAPACITY, and the // bag does not yet contain any item// with the given key.// Postcondition: A new copy of entry has// been added to the bag, with the given key. When the programmer wants to remove or retrieve an item from a keyed bag, the key of the item must be specified rather than the item itself. The keyed bag should also have a boolean method that can be used to determine whether the bag has an item with a specified key. In a keyed bag, the pro- grammer using the class specifies a particular key when an item is inserted. Here’s an implementation idea: A keyed bag can have two private arrays, one that holds the string data and one that holds the corresponding…
- Write a program Calculator.java. Instead of using integers, use rationals, as shown in Figure .You will need to use the split method in the String class, introduced inSection10.10.3, Replacing and Splitting Strings, to retrieve the numerator stringand denominator string, and convert strings into integers using the Integer.parseIntmethod.PROGRAM SIMULATION. Understand the given JAVA program and write the output. b. public class ChangeIt { public void doIt( int[] z ) { z[0] = 0; } } public class TestIt { public static void main ( String[] args ) { int[] myArray = {1, 2, 3, 4, 5} ; ChangeIt.doIt(myArray ); for (int j=0; j<myArray.length; j++ ) System.out.print( myArray[j] + " " ) ; } }Design a java program that simulates a checkout line at a supermarket. The line is a queue object. Customers (i.e., customer objects) arrive in random integer intervals of from 1 to 4 minutes. Also, each customer is serviced in random integer intervals of from 1 to 4 minutes. Obviously, the rates need to be balanced. If the average arrival rate is larger than the average service rate, the queue will grow infinitely. Even with “balanced” rates, randomness can still cause long lines. Run the supermarket simulation for a 12-hour day (720 minutes), using the following algorithm: Choose a random integer between 1 and 4 to determine the minute at which the first customer arrives .At the first customer’s arrival time, do the following: Determine customer’s service time (random integer from 1 to 4). Begin servicing the customer. Schedule the arrival time of the next customer (random integer 1 to 4 added to the current time). For each minute of the day, consider the following: If the next…