Answer:
Create two classes, PlayingCard and Main, as depicted in the attachments.
Explanation:
The getters/setters are not actually used in this code, but creating them is pretty standard. Although, you could argue that a PlayingCard object should never change value/suit, and is what it is as long as it exists. Such objects would be called "immutable".
Answer:
Let’s identify variables needed for this program.
First variables will be the one which will hold the values present in the Given Numbers in Array list and key of 16-bit and it will be array ARR and KEY. variables will be holding the Messages MSG1 “KEY IS FOUND AT “, RES ” POSITION”, 13, 10,” $” and MSG2 ‘KEY NOT FOUND!!!.$’ to be printed for the User. Other variables will be holding Length of the Array and it will be LEN, So in all Six variables.
The identified variables are ARR, KEY, LEN, RES, MSG1 and MSG2.
DATA SEGMENT
ARR DW 0000H,1111H,2222H,3333H,4444H,5555H,6666H,7777H,8888H,9999H
LEN DW ($-ARR)/2
KEY EQU 7777H
MSG1 DB "KEY IS FOUND AT "
RES DB " POSITION",13,10," $"
MSG2 DB 'KEY NOT FOUND!!!.$'
DATA ENDS
CODE SEGMENT
ASSUME DS:DATA CS:CODE
START:
MOV AX,DATA
MOV DS,AX
MOV BX,00
MOV DX,LEN
MOV CX,KEY
AGAIN: CMP BX,DX
JA FAIL
MOV AX,BX
ADD AX,DX
SHR AX,1
MOV SI,AX
ADD SI,SI
CMP CX,ARR[SI]
JAE BIG
DEC AX
MOV DX,AX
JMP AGAIN
BIG: JE SUCCESS
INC AX
MOV BX,AX
JMP AGAIN
SUCCESS: ADD AL,01
ADD AL,'0'
MOV RES,AL
LEA DX,MSG1
JMP DISP
FAIL: LEA DX,MSG2
DISP: MOV AH,09H
INT 21H
MOV AH,4CH
INT 21H
CODE ENDS
END START
In this Assembly Language Programming, A single program is divided into four Segments which are 1. Data Segment, 2. Code Segment, 3. Stack Segment, and 4. Extra Segment. Now, from these one is compulsory i.e. Code Segment if at all you don’t need variable(s) for your program.if you need variable(s) for your program you will need two Segments i.e. Code Segment and Data Segment.
Explanation:
The attached Logic is a C like Program to conduct a binary search we need small Algorithm Shown above in a very simple way, So Just we will covert the logic into Assembly There are many things uncommon in the programing Language. There are No While Loops or Modules but this things are to be implemented in different ways.
Answer:
Explanation:
The following code is written in Java, it prompts the user to enter 10 integers and saves them to an ArrayList. Then it prompts for 5 strings and saves them to another ArrayList. Then, it calls the sort method and adds the lists as parameters. Finally, it prints out both lists completely sorted in ascending order.
import java.util.ArrayList;
import java.util.Scanner;
class Division{
public static double division(double a, double b) throws Exception {
if(b == 0)
//throw new Exception("Invalid number.");
return (a / b);
return a / b;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
ArrayList<Integer> mylist2 = new ArrayList<>();
for (int x = 0; x < 10; x++) {
System.out.println("Please Enter a Number: ");
int number = in.nextInt();
mylist2.add(number);
}
ArrayList<String> mylist = new ArrayList<>();
for (int x = 0; x < 5; x++) {
System.out.println("Please Enter a Word: ");
String word = in.nextLine();
mylist.add(word);
}
sort(mylist);
sort(mylist2);
for (String x: mylist) {
System.out.print(x + ", ");
}
System.out.println("");
for (int x: mylist2) {
System.out.print(x + ", ");
}
}
public static <E extends Comparable<E>> ArrayList<E> sort(ArrayList<E> list) {
E temp;
if (list.size()>1) // check if the number of orders is larger than 1
{
for (int x=0; x<list.size(); x++) // bubble sort outer loop
{
for (int i=0; i < list.size() - x - 1; i++) {
if (list.get(i).compareTo(list.get(i+1)) > 0)
{
temp = list.get(i);
list.set(i,list.get(i+1) );
list.set(i+1, temp);
}
}
}
}
return list;
}}
