92 B 19 C 27 D E 3
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- . What will be the output of the following code, consider the memory address of variable x is 0x0044c and pointer variable ptr is 0x0066f, and consider int data type take 2 byte in the memory void main() { int x[5] = { 1,2,3,4,5 }; int *ptr = x; ptr = ptr + 1; cout << ptr << endl; cout << *ptr << endl; cout << &ptr << endl; cout << &x[0] << endl; cout << *x << endl; x += 2; cout << x << endl; }What is the value of the following expressions (or ERROR)?1. __________ s.length()2. __________ t.length()3. __________ 1 + a4. __________ a.toUpperCase()5. __________ "Tomorrow".indexOf("r")6. __________ "Tomorrow".lastIndexOf('o')7. __________ "Tomorrow".substring(2,4)8. __________ (a.length() + a).startsWith("a")9. __________ s = = a10. __________ a.substring(1,3).equals("bc")Please help me make memory diagrams for points one, two and three in the program. I dont know how to do so and I need to understand, also please explain what youre doing as youre solving#include <stdio.h> int foo (int);int jupiter(int, int);int mercury(int, int); int main(void){ int x = 2, y =4, z = 10; y = foo(y); y = -4 + jupiter(x, z) * 2 ; // Point 3 return 0;} int mercury(int m, int n){ int t; t = m + 2 * n; // Point 1 return t;} int jupiter(int i, int j){ int d = j; d = mercury(i % 400, i); // Point 2 return d;} int foo(int x){ int y = x; x = y * 2; return x;}
- Create the generated machine code from snippet code as follow: int i=10, j, x;for (a=10; a>0; a--) {if (a % 2 == 1) i += a;else i /= 2;j = i*2;}x = i+j; please solve max in 30 minutes thank uWhat is the output of the following code int a = 10, b = -12, c = 0, d; d = ++a + b ++ / --c; cout<<"\n"<<a<<" "<<b<<" "<<c<<" "<<d<<endl; 11 11 -1 23 12 -11 -1 23 10 -11 -1 24 11 -11 -1 23 11 -11 -1 22 No hand written and fast answer with explanationWrite the C equivalent “train” function #void train(int*A, int*B, int*C, int k) train: add $t4, $0, $0bus: add $t5, $t4, $a1 lw $t6, 0($t5) add $t5, $t4, $a2 lw $t7, 0($t5) add $t6, $t6, $t7 add $t5, $t4, $a0 sw $t6, 0($t5) addi $t4, $t4, 4 slt $t5, $t4, a3 bne $t5, $0, bus jr ra
- Identify all private, shared variables in the following code snippet. What is the output of the program (e.g. the value of IS)? int A[100]; int count; void work(int index[]) { float B[10]; …… } void main(){ int IS = 10; int i=0; #pragma omp parallel private(i) { int index[10]; work(index); #pragma omp for reduction(+:IS) for(i=0;i<100;i++) { IS = IS + i; } } printf(“%i\n”, IS); }I need to create an UML digram from the following code: import java.util.Scanner;//Code Provided For Project Two.class Main {public static void main(String[] args) {Scanner input = new Scanner(System.in);System.out.print("Enter the first complex number: ");double a = input.nextDouble();double b = input.nextDouble();Complex c1 = new Complex(a, b);System.out.print("Enter the second complex number: ");double c = input.nextDouble();double d = input.nextDouble();Complex c2 = new Complex(c, d);System.out.println("(" + c1 + ")" + " + " + "(" + c2 + ")" + " = " + c1.add(c2));System.out.println("(" + c1 + ")" + " - " + "(" + c2 + ")" + " = " + c1.subtract(c2));System.out.println("(" + c1 + ")" + " * " + "(" + c2 + ")" + " = " + c1.multiply(c2));System.out.println("(" + c1 + ")" + " / " + "(" + c2 + ")" + " = " + c1.divide(c2));System.out.println("|" + c1 + "| = " + c1.abs()); Complex c3 = null;try {c3 = (Complex)c1.clone();}catch (CloneNotSupportedException e)…need main class in java with user input that prints like the following in a (y/n) loop. reads from a txt file and second class
- When the code below is run, it produces undefined and unpredictable results. Why?Can you suggest an intervention to resolve this issue without removing or changingany of the existing lines of code? (hint: add a line of code) char *myPtr;myPtr = malloc(10*sizeof(char));for (int i = 0; i < 10; i++){myPtr[i] = 'a';}myPtr[10] = '\0';Java Program ASAP Modify this program so it passes the test cases in Hypergrade becauses it says 5 out of 7 passed. Also I need one one program and please correct and change the program so that the numbers are in the correct places as shown in the correct test case. import java.io.BufferedReader;import java.io.FileReader;import java.io.IOException;import java.util.ArrayList;import java.util.Arrays;import java.util.InputMismatchException;import java.util.Scanner;public class FileSorting { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); while (true) { System.out.println("Please enter the file name or type QUIT to exit:"); String fileName = scanner.nextLine(); if (fileName.equalsIgnoreCase("QUIT")) { break; } try { ArrayList<String> lines = readFile(fileName); if (lines.isEmpty()) { System.out.println("File " +…In python, Problem Description:Sheldon and Leonard are physicists who are fixated on the BIG BANG theory. In order to exchange secret insights they have devised a code that encodes UPPERCASE words by shifting their letters forward. Shifting a letter by S positions means to go forward S letters in the alphabet. For example, shifting B by S = 3 positions gives E. However, sometimes this makes us go past Z, the last letter of the alphabet. Whenever this happens we wrap around, treating A as the letter that follows Z. For example, shifting Z by S = 2 positions gives B. Sheldon and Leonard’s code depends on a parameter K and also varies depending on the position of each letter in the word. For the letter at position P, they use the shift value of S = 3P + K. For example, here is how ZOOM is encoded when K = 3. The first letter Z has a shift valueof S = 3 × 1 + 3 = 6; it wraps around and becomes the letter F. The second letter, O, hasS = 3 × 2 + 3 = 9 and becomes X. The last two letters…
