Explain Relative Addressing mode?
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- Java Program ASAP ************This program must work in hypergrade and pass all the test cases.********** This program down below does not pass the test cases in Hypergrade as shown in the scheenshot. Please modify it. Also I have provided the code instructions. Thank you import java.io.BufferedReader;import java.io.FileReader;import java.io.IOException;import java.io.InputStreamReader;import java.util.regex.Pattern;import java.util.regex.Matcher;public class WordSeparator { public static void main(String[] args) { try (BufferedReader reader = new BufferedReader(new InputStreamReader(System.in))) { String inputFileName; boolean fileNotFound; System.out.print("Please enter the file name or type QUIT to exit:\n"); do { fileNotFound = false; // Reset fileNotFound flag for each iteration inputFileName = reader.readLine(); if (inputFileName.equalsIgnoreCase("QUIT")) {…305. A program that runs in parts on several computers is said to be a. delegated b. spread c. distributed d. recursiveComputer Science Write in Javascript please and use high order functions liek map or filter or reduce let obj = [ { mode: "hot" }, { temp: "low" }, { mode: "cold } ] I want to create a function that get object by key such as: myFunc("mode"); returns { mode: "hot" } next call of myFunc("mode") returns { mode: "cold
- Code to shows how you can implement the steps using a PyCommand for Immunity Debugger. When you call the program, you pass it the DGA function's starting address.Write Implicit and Explicit code blocksComputer Science Course: Reconfigurable Computing Content: Make an Experimental report on Implement a 2-D convolution using HLS C and Optimize the HLS with directives. Please include images of the task in the report.
- Computer Science using java The program has to evaluate arithmetic expressions using a BST for thatpurpose. The rules are the following:The program asks the user to enter an arithmetic expression in the infixformat. Then the program builds a BST for that expression. After buildingthe BST, a menu will be present to the user allowing him to: traverse the BST in inorder, traverse the BST in postorder, traverse the BST in preorder.Users will choose among the options present. The result will be thedisplay of the original arithmetic expression and the new one thatdepends on the option chosen. The program should repeat as long as theuser wants.Use the C programming language, not C++. #include “driverlib.h” #include <stdint.h> #include <stdbool.h> #define MAXCOUNT 500 Bool isPrime(uint16_t x); // Main Function Void main(void){ volatile uint16_t ii; // declare local loop variable // Halt the WDT and unlock PMM for Port I/O Configuration WDT_A_hold(WDT_A_BASE); PMM_unlockLPM5(); //Set P1.0 (RED LED) as Output GPIO_setAsOutputPin(GPIO_PORT_P1, GPIO_PIN0); //Set P1.0 (RED LED) Low -- Turns RED LED OFF GPIO_setOutputLowOnPin(GPIO_PORT_P1, GPIO_PIN0); for(ii=0; ii<MAXCOUNT; ii++){ if(isPrime(ii)) GPIO_toggleOutputOnPin(GPIO_PORT_P1, GPIO_PIN0); } while(1); // Spin Loop } // Function isPrime // Input x -- number to check if prime // Returns: True, if x is prime else FALSE // Bool isPrime(uint16_t x){ volatile uint16_t ii; uint16_t maxValue; if((x==0) || (x==1)) return false; if((x==2) || (x==3)) return true; maxValue = (x/2) +…C++ ProgrammingTopic: Stack, Ques and DequesBelow is the initial program of the main file, only modify the main file, sllstack file also provided. See attached photo for instructions. main.cpp #include <iostream> #include <cstring> #include "sllstack.h" using namespace std; int main(int argc, char** argv) { SLLStack* stack = new SLLStack(); int test; string str; cin >> test; switch (test) { case 0: getline(cin, str); // PERFORM SOLUTION TO BRACKETS PROBLEM HERE // FYI: Place your variable declarations, if any, before switch. break; case 1: stack->push('a'); stack->push('b'); stack->push('c'); cout << stack->pop() << endl; cout << stack->pop() << endl; cout << stack->pop() << endl; cout << stack->isEmpty() << endl; break;…
- Modify the following insert functions of this code and implement a logic that user will not be able to add same value again for exam one he enter 5 he will not be able to add 5 again #include<iostream>#include<cstdio>#include<cstdlib>using namespace std;/* * Node Declaration */class node{ public: int info; node *next;}*start; /* * Class Declaration */class single_llist{ public: node* create_node(int); void insert_begin(); void insert_pos(); void insert_last(); void delete_pos(); void sort(); void search(); void update(); void reverse(); void display(); single_llist() { start = NULL; }}; /* * Main :contains menu */main(){ int choice, nodes, element, position, i; single_llist sl; start = NULL; while (1) { cout<<endl<<"Operations on singly linked list"<<endl;…C Programming Language (Code With C Programming Language) Problem Title : Visible Trees There is a legend about a magical park with N × N trees. The trees are positioned in a square grid with N rows (numbered from 1 to N from north to south) and N columns (numbered from 1 to N from west to east). The height (in metres) of each tree is an integer between 1 and N × N, inclusive. Magically, the height of all trees is unique. Bunga is standing on the northmost point of the park and wants to count the number of visible trees for each Column. Similarly, Lestari is standing on the westmost point of the park and wants to count the number of visible trees for each Row. A tree X is visible if all other trees in front of the tree X are shorter than the tree X. For example, let N = 3 and the height (in metres) of the trees are as follows6 1 87 5 32 9 4 On the first column, Bunga can see two trees, as the tree on the third row is obstructed by the other trees. On the second column, Bunga can see…Modify the Calculator program to include the following functions. 1. Display a menu of the following options at startup Enter a 1 to 3 digit integer for push D - pop and display top element P - display top element but do not pop L - print all elements on the stack without any pop H - display number of elements currently on the stack C - clear stack X - Halt M - display menu + - pop top two elements, add them, push result - - negate the top element * - multiply top two elements and push result
