3. Problem 5. A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. a. How many bits are needed for the opcode? b. How many bits are left for the address part of the instruction? c. What is the maximum allowable size for memory? d. What is the largest unsigned binary number that can be accommodated in one word of memory?
Answers
The number of bits that are needed for this operation code (opcode) is 8 bits.
What is binary encoding?
Binary encoding can be defined as a standard technique that is designed and developed for converting data in plain text (source alphabets) to a form that is easily used by different operating systems (OS), especially through the use of a binary digit (bit) or two-symbol system.
How to calculate the number of bits.
The number of bits that are needed for this operation code (opcode) is given by this mathematical expression:
n = 8 bits.
b. To determine the number of bits that are left for the address part of the instruction:
- Opcode, O, I = 8 bits.
- Instruction size = 24 bits.
Address = 16 bits.
c. To determine the maximum allowable size for memory:
Based on the memory unit, this is given by
d. The largest unsigned binary number that can be accommodated in one word of memory is given by:
Read more on binary encoding here: brainly.com/question/5381889
Answer:
a) 8 bits b) 16 bits. c) 2¹⁶ * 24 bits d) 2²⁴ -1
Explanation:
a) In order to be able to accommodate 150 different instructions, the number of bits needed must be equal to the minimum power of 2 that satisfies this equation:
2n > 150, i.e. n=8.
b) If the total number of bits for a word is 24, and 8 are used for the op code, there are 16 bits left for the address part.
c) If the address part has 16 bits, this means that the total addressable space is just 2¹⁶, so the maximum allowable size is 2¹⁶ * 24 bits.
d) As we have 24 bits to be filled, the largest unsigned binary number is just 2²⁴ – 1.
(As we need to leave a position for all zeros).
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void f1(int array, int size);
void f1(int& array, int size);
void f1(int array[100], int size);
void f1(float array[], int size);
All of the above
C and D
A and B
Answers
Answer:
Only
Option: void f1(float array[], int size);
is valid.
Explanation:
To pass an array as argument in a function, the syntax should be as follows:
functionName (type arrayName[ ] )
We can't place the size of the array inside the array bracket (arrayName[100]) as this will give a syntax error. The empty bracket [] is required to tell the program that the value that passed as the argument is an array and differentiate it from other type of value.
Answers
Answer:
Direct access storage device.
Explanation:
DASD stands for direct access storage device.
Answers
Answer:
Following are the code to this question:
public class Pig //Defining class Pig
{
private String name; //Defining string variable name
private int age; // Defining integer variable age
private double weight; // Defining double variable weight
Pig (String name, int age, double weight) //Defining parameterized constructor
{
super(); //using super key
this.name = name; //holding value in name variable
this.age = age; // holding value in age variable
this.weight = weight; // holding value in weight variable
}
String getName() //Defining method getName
{
return name; //return name value
}
void setName(String name) // Defining method setName
{
this.name = name; //hold name value
}
int getAge() // Defining method getAge
{
return age; //return value
}
void setAge(int age) // Defining method setAge
{
this.age = age; // hold age value
}
double getWeight() //Defining method getWeight
{
return weight; //return weight value
}
void setWeight(double weight) //Defining method setWeight
{
this.weight = weight; //hold weight value
}
void display() //Defining method display
{
System.out.println("Name:" + name + " Age:" + age + " Weight:" + weight); //print value
}
public static void main(String[] ar) //Defining main method
{
Pig onc = new Pig("Jig",5,14.5); //creating class object and called parameterized constructor
onc.display();//calling display method
}
}
Output:
please find the attachment.
Explanation:
In the given java program, a class "Pig" is declared, in which three name, age, and weight is defined which differs in datatypes, in the next step, parameterized constructor, get and set method, and display method declared, which can be described as follows:
- In the parameterized constructor, uses super and this keyword to call and holds parameter value.
- In the get method three methods "getName, getAge, and getWeight" are defined, that return method values, and the set method "setName, setAge, and setWeight" uses this keyword to hold value in its variables.
- The display method is used to print all method store values by its variables name.
- Inside the main method, class object "onc" is created, which stores the value in it and calls the display method that print value with a message.
Answers
Answer:
Its A Dynamic Web Page
Explanation:
until the price of goods reflects equal supply and demand.
Answers
o True
o False
Answers
Answer:
False
Explanation:
The answer is False. In Stacks, we can access only the top element present in the stack. Stack is the collection of elements which follow LIFO ( Last In First Out ) which means the last element inserted in the stack is the first element to out. Stack has restriction that only the element which is present at the top called as top element is only accessible. That means only the top element can be inserted and deleted.