Answer:
Hello attached is the Java program written to solve the problem
The Pet.java and Dog.java files are unaltered
Explanation:
The input and output codes are attached as well i.e the second image is the input while the third image is the output code
Answer:
I will write the code in C++ and JAVA
Explanation:
<h2>
JAVA CODE</h2>
public class Main
{ public static void main(String[] args) {
// displays Gershwin,George
System.out.println("Gershwin,George"); } }
<h2>
C++ Code:</h2>
#include <iostream>
using namespace std;
int main()
{ cout<<"Gershwin,George";
}
// displays last name Gershwin followed by , followed by first name George
//displays Gershwin,George as output.
The answer is the Quick Access Toolbar. However, it is not only for saving files or undoing your work. Containing a set of commands that are independent, this toolbar is actually customizable wherein you could change these icons to the ones you really need and frequently use. By tweaking the settings, you can even add commands to the Quick Access Toolbar that are not in the ribbon (like New, Open and Print).
Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
