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Mice21 [21]
1 year ago
9

onsider a file system on a disk that has both logical and physical block sizes of 512 bytes. assume that the information about e

ach file is already in memory. for each of the three allocation strategies (contiguous, linked, and indexed), answer the following question: if we are currently at logical block 10 (the last block accessed was block 10) and want to access logical block 4, how many physical blocks must be read from the disk?
Computers and Technology
1 answer:
anygoal [31]1 year ago
4 0

For three disk allocation strategies (contiguous, linked, and indexed),

in this case, the  physical blocks must be read from the disk are 1,4 and 2 respectively.

In the disk, every block has a sequential number. The majority of disks today do logical I/O in hardware. Physical I/O is seldom ever required of operating systems today.

File access uses virtual I/O. Similar to disk logical I/O is disk virtual I/O. The distinction is that with virtual I/O, as opposed to physical I/O, the blocks that make up a file are

You could want to transfer from virtual block 10 to virtual block 4 in a contiguous allocation. Since the file is contiguous and you are at logical block 10 + Z, block four can be accessed by reading logical block 4 + Z. Without any intermediaries, that can be done directly.

Each block in the linked structure has 511 bytes of data and one byte that serves as an offset to the following block.

Read the first block, determine the offset to the second block, read the second block, determine the offset to the fourth block, and then read the fourth block in order to gain access to the fourth block.

If blocks are numbered from 1 to 10, then then is the "4" correct. The remainder of the solutions, however, make the assumption that block numbering begins at 0, using division to obtain offsets.

Assume there is a contiguous index for the indexed way of allocating files. Once more, it is assumed that each offset in the index is one byte in size.

Let's say that Z serves as the index's origin. Then, the byte at block Z + 3 DIV 512 and the offset at 3 MOD 512 (3 = 4 - 1) make up the entry for block 4 (numbered 1 to 4).

You must read the index and the data block in order to locate the block (2 reads).

To learn more about disk allocation strategies click here:

brainly.com/question/15083707

#SPJ4

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The intention of this problem is to analyze a user input word, and display the letter that it starts with (book → B).
yulyashka [42]

Answer:

Here is the C++ program:

#include<iostream>  // to use input output functions

using namespace std;     //to identify objects like cin cout

void StartChar(string str)  {  // function that takes a word string as parameter and returns the first letter of that word in capital

   int i;  

   char c;

   do  //start of do while loop

   {

   for (int i = 0; i < str.length(); i++) {  //iterates through each character of the word str

       if(str.length()>25){  //checks if the length of input word is greater than 25

           cout<<"limit exceeded"; }  //displays this message if a word is more than 25 characters long

      c = str.at(i);   // returns the character at position i

       if (! ( ( c >= 'a' && c <= 'z' ) || ( c >= 'A' && c <= 'Z' ) ) ) {  //checks if the input word is contains characters other than the alphabet

            cout<<str<<" is not a word!"<<endl; break;}  //displays this message if user enters characters other than the alphabet

        if (i == 0) {  //check the first character of the input word

           str[i]=toupper(str[i]);  //converts the first character of the input word to uppercase

           cout<<str<<" starts with letter "<<str[i]<<endl;  }  // prints the letter that the word starts with in capital letter

   }   cout<<"Enter a word: ";  //prompts user to enter a word

      cin>>str;   //reads input word from user

}while(str!="#");   }    //keeps prompting user to enter a word until the user enters #

int main()   //start of the main() function body

{ string str;  //declares a variable to hold a word

cout<<"Enter a word: "; //prompts user to enter a word

cin>>str; //reads input word from user

   StartChar(str); }  //calls function passing the word to it

     

Explanation:

The program prompts the user to enter a word and then calls StartChar(str) method by passing the word to the function.

The function StartChar() takes that word string as argument.

do while loop starts. The for loop inside do while loop iterates through each character of the word string.

First if condition checks if the length of the input word is greater than 25 using length() method which returns the length of the str. For example user inputs "aaaaaaaaaaaaaaaaaaaaaaaaaaa". Then the message limit exceeded is printed on the screen if this if condition evaluates to true.

second if condition checks if the user enters anything except the alphabet character to form a  word. For example user $500. If this condition evaluates to true then the message $500 is not a word! is printed on the screen.

Third if condition checks if the first character of input word, convert that character to capital using toupper() method and prints the first letter of the str word in capital. For example if input word is "Computer" then this prints the message: Computer starts with the letter C in the output screen.

The program keeps prompting the user to enter the word until the user enters a hash # to end the program.

The output of the program is attached.

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