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Monica [59]
3 years ago
13

Typically, at the completion of a device I/O, a single interrupt is raised and appropriately handled by the host processor. In c

ertain settings, however, the code that is to be executed at the completion of the I/O can be broken into two separate pieces. The first piece executes immediately after the I/O completes and schedules a second piece of code (sometimes called a "Deferred Procedure Call") to be executed at a later time. What is the purpose of using this strategy in the design of interrupt handlers?
Computers and Technology
1 answer:
inessss [21]3 years ago
5 0
No idea sorrrrrrryyyyyyyy
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Compare and contrast CD and DVD?
anzhelika [568]

Answer:

Both Flat, round discs.

A DVD can hold six times as much as compacity than a disc.

A CD is a Compact Disc.

4 0
2 years ago
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Design a program takes as input, X, an unsorted list of numbers, and returns the sorted list of numbers in X. The program must b
Lelechka [254]

Answer:

The program in python is as follows:

def split(X):

   L = []; G = []

   for i in range(len(X)):

       if X[i]>=X[0]:

           G.append(X[i])

       else:

           L.append(X[i])

   L.sort(); G.sort()

   return L,G

X = []

n = int(input("Length of X: "))

for i in range(n):

   inp = int(input(": "))

   X.append(inp)

   

if len(X) == 0 or len(X) == 1:

   print(X)

else:

   X1,X2=split(X)

   newList = sorted(X1 + X2)

   print(newList)

Explanation:

The following represents the split function in the previous problem

def split(X):

This initializes L and G to empty lists

   L = []; G = []

This iterates through X

   for i in range(len(X)):

All elements of X greater than 0 equal to the first element are appended to G

      <em> if X[i]>=X[0]:</em>

<em>            G.append(X[i])</em>

Others are appended to L

<em>        else:</em>

<em>            L.append(X[i])</em>

This sorts L and G

   L.sort(); G.sort()

This returns sorted lists L and G

   return L,G

The main function begins here

This initializes X

X = []

This gets the length of list X

n = int(input("Length of X: "))

This gets input for list X

<em>for i in range(n):</em>

<em>    inp = int(input(": "))</em>

<em>    X.append(inp)</em>

This prints X is X is empty of has 1 element

<em>if len(X) == 0 or len(X) == 1:</em>

<em>    print(X)</em>

If otherwise

else:

This calls the split function to split X into 2

   X1,X2=split(X)

This merges the two lists returned (sorted)

   newList = sorted(X1 + X2)

This prints the new list

   print(newList)

7 0
2 years ago
Is it more costly to Andrew to go to graduate business school full time or part time if the tuition is the same for each? A. Par
andrezito [222]

Answer:

A. Part time

Explanation:

Since tuition is the same whether he goes part time or full time, in the long run he will end up paying tuition a greater number of times if he is a part time student.

4 0
3 years ago
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Write a function wordcount() that takes the name of a text file as input and prints the number of occurrences of every word in t
artcher [175]

Answer:

Explanation:

The following Python program uses a combination of dictionary, list, regex, and loops to accomplish what was requested. The function takes a file name as input, reads the file, and saves the individual words in a list. Then it loops through the list, adding each word into a dictionary with the number of times it appears. If the word is already in the dictionary it adds 1 to its count value. The program was tested with a file named great_expectations.txt and the output can be seen below.

import re

def wordCount(fileName):

   file = open(fileName, 'r')

   wordList = file.read().lower()

   wordList = re.split('\s', wordList)

   wordDict = {}

   for word in wordList:

       if word in wordDict:

           wordDict[word] = wordDict.get(word) + 1

       else:

           wordDict[word] = 1

   print(wordDict)

wordCount('great_expectations.txt')

4 0
3 years ago
Using MARS/MIPS
slega [8]

Answer:

Explanation:

MIPS program which increments from 0 to 15 and display results in Decimal on the console

In this program the user defined procedures print_int and print_eot were used to print the integer values and new line characters(\n) respectively.the mechanisam of the loop is explaine in the comment section of the program.

     addi $s0, $0, 0

     addi $s1, $0, 15

print_int:

              li $v0, 1                # system call to print integer

              syscall                

              jr $ra                     # return

print_eol:                      # prints "\n"

            li $v0, 4            

              la $a0, linebrk      

              syscall              

              jr $ra                  # return

main: . . .          

              li $a0, 0                # print 0

              jal print_int         # print value in $a0

loop:   move $a0, $s0           # print loop count

       jal print_int        

       jal print_eol           # print "\n" character

       addi $s0, $s0, 1        # increment loop count by 1

       ble $s1, $s0, loop      # exit if $s1<$s0

beq $s0, $0, end

end:

MIPS progam to increment from 0 to 15 and display results in Hexadecimal on the console

this program is slightly differed from the previous program in this program the system call issued in print_int is implemented with a system call that prints numbers in hex.

addi $s0, $0, 15

     addi $s1, $0, 0

print_int:

   li      $v0,34                  # syscall number for "print hex"

   syscall                         # issue the syscall

              jr $ra                     # return

print_eol:                      # prints "\n"

            li $v0, 4            

              la $a0, linebrk      

              syscall              

              jr $ra                  # return

main: . . .          

              li $a0, 0                # print 0

              jal print_int         # print value in $a0

loop:   move $a0, $s0           # print loop count

       jal print_int        

       jal print_eol           # print "\n" character

       addi $s0, $s0, 1        # increment loop count by 1

       ble $s1, $s0, loop      # exit if $s0>$s1

beq $s0, $0, end

end:

5 0
3 years ago
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