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worty [1.4K]
3 years ago
11

The "persistent interrupt" is a bug in the code that exhibits a behavior that’s similar to an infinite loop. The ISR exits and g

ets called back immediately an infinite amount of times even though the interrupt event occurred once. The persistent interrupt occurs when three conditions are met. What are these conditions? How do we avoid a persistent interrupt from occurring?
Computers and Technology
1 answer:
Mumz [18]3 years ago
4 0

Answer:

1. the 3 conditions have been listed below

2. to avoid persistent interrupt, we break any of the 3 conditions.

Explanation:

The persistent interrupt occurs when these three conditions below are met.

They are as follows:

1. GIE=1

2. xIE=1

3. xIFT=1

in avoidance of the persistent interrupt from occurring, we break any of these 3 conditions that I have listed above.

in order To break any one of the 3 conditions, we clear the flag that corresponds to the associated

we can also avoid the persistent interrupt occurrence by having the Interrupt Service Routine (ISR) clear maybe one or all of the three bits mentioned above before the ISR terminates, so it wouldn't get called back again. clearing any of the bits will not allow the ISR to be called again.

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Write a loop statement to count and display the number of positive integers in the array. That is, your loop should display the
dlinn [17]

Answer:

int count =0;

    for(int i=0;i<10;i++)

    {

        if(myArray[i]>=0)

        {

            count++;

        }

    }

    cout<<"Number of positive integers is "<<count<<endl;

Explanation:

The above written loop is for counting positive integers in the myArray[].

For counting we have taken a count integer initialized with 0.On iterating over the array if the element is greater than or equal to 0 we consider it as positive and increasing the count.At the end printing the count.

3 0
3 years ago
A palindrome is a string that reads the same from left to right and from right to left. Design an algorithm to find the minimum
stepladder [879]

Answer:

Explanation:

The following code is written in Python. It is a recursive function that tests the first and last character of the word and keeps checking to see if each change would create the palindrome. Finally, printing out the minimum number needed to create the palindrome.

import sys

def numOfSwitches(word, start, end):

   if (start > end):

       return sys.maxsize

   if (start == end):

       return 0

   if (start == end - 1):

       if (word[start] == word[end]):

           return 0

       else:

           return 1

   if (word[start] == word[end]):

       return numOfSwitches(word, start + 1, end - 1)

   else:

       return (min(numOfSwitches(word, start, end - 1),

                   numOfSwitches(word, start + 1, end)) + 1)

word = input("Enter a Word: ")

start = 0

end = len(word)-1

print("Number of switches required for palindrome: " + str(numOfSwitches(word, start, end)))

3 0
3 years ago
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
3 years ago
List the three control problems associated with competingprocesses and briefly define each.
hjlf

Answer and explanation :

the three control problems associated with competing process are

  • MUTUAL EXCLUSION : We know that some resources are shareable and some are not shareable. which means only one process can access the resource at a time this type of resources are called critical resources this code can be access at only one process at a time. the other process if required to access should not be allowed
  • DEADLOCK: this hold the process without complete for example suppose there are two resources R1 and R2 and two process P1 and P2 and P1 use R1 and P2 use R2 but after some time when P1 needs R2 but R2 is not available as it is used by P2 so the all process will be on hold
  • STARVATION : when priorities are given to the process as high priorities and low priorities. And high priorities process always competing  then low priorities process have to wait for very long time this is called starvation
3 0
3 years ago
Computers that are close to one another are connected to form a LAN
Archy [21]

Explanation:

different computer are connected to a LAN by a cable and an interface card

6 0
3 years ago
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