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maw [93]
3 years ago
9

Prompt the user to input an integer, a double, a character, and a string, storing each into separate variables. Then, output tho

se four values on a single line separated by a space.
Computers and Technology
1 answer:
cupoosta [38]3 years ago
7 0

Answer:

Following are the program in java language:

import java.util.*; // import package

public class Main

{

public static void main(String[] args) // main method

{

 int num; // integer variable declaration

 double r; //double varaible declaration

 char ch; // char variable declaration

Scanner scnr=new Scanner(System.in); // creating the instance of scanner

 System.out.println("Enter the string:");

 String str=scnr.nextLine(); // reda string by user

System.out.println("Enter the integer value:");

 num=scnr.nextInt(); // Read integer value by the user

 System.out.println("Enter the Double value:");

 r=scnr.nextDouble(); // read double value by user

 System.out.println("Enter the character value:");

 ch= scnr.next().charAt(0); // read character by user

 System.out.print(num + " ,"); // display them in a single line

 System.out.print( " " +r +" ,");//dislpay in a single line  

  System.out.print( " " +ch +",");//dislpay in a single line

 System.out.print(" " +str);//dislpay in a single line

}

}

Output:

Enter the string:

ram

Enter the integer value:

34

Enter the Double value:

45.789

Enter the character value:

r

34 , 45.789 , r, ram

Explanation:

Following are the description of program

  • Read the integer ,string,char,double value by the user by using scanner class in "num","str","ch" and "r" variable respectively
  • Print that value in a single line separated by comma by System.out.println statement.

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Computer system with a 32-bit logical address and 4k-byte page size. assume that each entry of a page table consists of 4bytes.
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The number of bits in the physical address is 26 bits. The number of entries in a page table is \mathbf{2^{20}} entries. The size of the page table in a one-level paging scheme is 4MB.

<h3>
What is paging in Operating System?</h3>

Paging is a storage method used in operating systems to recover activities as pages from secondary storage and place them in primary memory. The basic purpose of pagination is to separate each procedure into pages.

We are given the following parameters:

  • 32-bit logical address
  • Page size = 4 KB = \mathbf{2^{12} \ bytes }
  • Size of each Page table entry = 4 bytes

Suppose the system supports physical memory size = 64 MB = \mathbf{2^{26} \ bytes}

Thus, the number of bits in the physical address is computed as:

= \mathbf{log_2 \{Physical-memory-size\}}

=\mathbf{log_2(2^{26})}}

= 26 bits

The number of entries in a page table = logical address space size/page size

The number of entries in a page table \mathbf{= \dfrac{2^{32}}{2^{12}}}

\mathbf{=2^{20}} entries

In a one-level paging scheme, the size of the table is:

= entire no. of page entries × page table size

= \mathbf{2^{20}\times 4 \ bytes}

= 4 MB

b.

suppose that this system supports up to 2^30 bytes of physical memory.

  • The size of the page table will be the same as 4 MB due to the fact that the number of entries, as well as, the page table entry size is the same.

Since the size of the page table surpasses that of a single page. A page cannot include a whole page table. Therefore, the page table must be broken into parts to fit onto numerous pages, and an additional level of the page table is required to access this page table.

  • This is called the Multi-Level Paging system.

Therefore, we can conclude that the number of bits in the physical address is 26 bits, the number of entries in a page table is \mathbf{2^{20}} entries, and the size of the page table in a one-level paging scheme is 4MB.

Learn more about Paging in Operating System here:

brainly.com/question/17004314

#SPJ1

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Explanation:

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