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IgorLugansk [536]
2 years ago
9

5.11 Of these two types of programs:a. I/O-boundb. CPU-boundwhich is more likely to have voluntary context switches, and which i

s more likely to have nonvoluntary context switches? Explain your answer
Computers and Technology
1 answer:
Leno4ka [110]2 years ago
7 0

Answer:

The answer is "both voluntary and non-voluntary context switch".

Explanation:

The description to this question can be described as follows:

Whenever processing requires resource for participant contextual switch, it is used if it is more in the situation of I/O tied. In which semi-voluntary background change can be used when time slice ends or even when processes of greater priority enter.

  • In option a, It requires voluntary context switches in I /O bound.
  • In option b, it requires a non-voluntary context switch for CPU bound.
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Write a program that repeatedly reads in integers until a negative integer is read. The program keeps track of the largest integ
vfiekz [6]

Answer:

In Python:

num = int(input("Enter number: "))

maxn = num

while num >=0:

   if num>maxn:

       maxn = num

   num = int(input("Enter number: "))

print("Largest: "+str(maxn))

Explanation:

Get input from the user

num = int(input("Enter number: "))

Initialize the largest to the first input

maxn = num

This loop is repeated until a negative input is recorded

while num >=0:

If the current input is greater than the previous largest

   if num>maxn:

Set largest to the current input

       maxn = num

Get another input from the user

   num = int(input("Enter number: "))

Print the largest

print("Largest: "+str(maxn))

5 0
2 years ago
For any element in keysList with a value greater than 50, print the corresponding value in itemsList, followed by a space. Ex: I
VARVARA [1.3K]

Answer:

import java.util.Scanner;  //mentioned in the question.

public class ArraysKeyValue {  //mentioned in the question.

public static void main (String [] args) {  //mentioned in the question.

final int SIZE_LIST = 4; //mentioned in the question.

int[] keysList = new int[SIZE_LIST]; //mentioned in the question.

int[] itemsList = new int[SIZE_LIST]; //mentioned in the question.

int i; //mentioned in the question.

keysList[0] = 13; //mentioned in the question.                                

keysList[1] = 47; //mentioned in the question.

keysList[2] = 71; //mentioned in the question.

keysList[3] = 59; //mentioned in the question.

itemsList[0] = 12; //mentioned in the question.

itemsList[1] = 36; //mentioned in the question.

itemsList[2] = 72; //mentioned in the question.

itemsList[3] = 54;//mentioned in the question.

// other line to complete the solution is as follows--

for(i=0;i<(keysList.length);i++) //Loop to access all the element of keysList array variable.

    {// open for loop braces

        if(keysList[i]>50) // compare the value of keysList array variable with 50.

        System.out.println(itemsList[i]); // print the value

   }// close for loop braces.

}//  close main function braces.

} //close the class braces.

Output:

72

54

Explanation:

In the solution part--

  1. One loop is placed which moves from 0 to (size-1) of the 'keyslist' array.
  2. Then the 'if' statement is used to check the element of 'keyslist' array, which is greater than 50 or not.
  3. If it is greater, then print the element of item_list array of 'i' index which is also the index of greater value of keyslist array element.

5 0
2 years ago
Write a program that uses the function isPalindrome given in Example 6-6 (Palindrome). Test your program on the following string
Artist 52 [7]

Answer:

#include <bits/stdc++.h>

using namespace std;

bool isPalindrome(string str)

{

   char a,b;

int length = str.length();

for (int i = 0; i < length / 2; i++)

{

   a=tolower(str[i]);//Converting both first characters to lowercase..

   b=tolower(str[length-1-i]);

   if (b != a )

return false;

}

return true;

   

}

int main() {

   string t1;

   cin>>t1;

   if(isPalindrome(t1))

   cout<<"The string is Palindrome"<<endl;

   else

   cout<<"The string is not Palindrome"<<endl;

return 0;

}

Output:-

Enter the string

madam

The string is Palindrome

Enter the string

abba

The string is Palindrome

Enter the string

22

The string is Palindrome

Enter the string

67876

The string is Palindrome

Enter the string

444244

The string is not Palindrome

Explanation:

To ignore the cases of uppercase and lower case i have converted every character to lowercase then checking each character.You can convert to uppercase also that will also work.

6 0
2 years ago
Complete the program below that takes in one positive, odd, integer, n (at least 3), and prints a "diamond" shape of stars with
avanturin [10]

Answer:

n = int(input("Enter the n (positive odd integer): "))

for i in range(1, n+1, 2):

   print(i*"*")

for i in range(n-2, 0, -2):

   print(i*"*")

Explanation:

*The code is in Python.

Ask the user to enter the n

Create a for loop that iterates from 1 to n, incrementing by 2 in each iteration. Print the corresponding number of asterisks in each iteration

Create another for loop that iterates from n-2 to 1, decrementing by 2 in each iteration. Print the corresponding number of asterisks in each iteration

5 0
2 years ago
Problem 1a. Write a function named hasFinalLetter that takes two parameters 1. strList, a list of non-empty strings 2. letters,
mash [69]

Answer:

The answer is the programming in Python language has strings and characters that has to be declared in the method.

Explanation:

#method

def hasFinalLetter(strList,letters):

output = #output list

#for every string in the strList

for string in strList:

#findout the length of each string in strList

length = len(string)

#endLetter is last letter in each string

endLetter = string[length-1]

#for each letter in the letters list

for letter in letters:

#compare with endLetter

#if we found any such string

#add it to output list

if(letter == endLetter):

output.append(string)

#return the output list

return output

#TestCase 1 that will lead to return empty list

strList1 = ["user","expert","login","compile","Execute","stock"]

letters1 = ["a","b","y"]

print hasFinalLetter(strList1,letters1)

#TestCse2

strList2 = ["user","expert","login","compile","Execute","stock"]

letters2 = ["g","t","y"]

print hasFinalLetter(strList2,letters2)

#TestCase3

strList3 = ["user","expert","login","compile","Execute","stock"]

letters3 = ["k","e","n","t"]

print hasFinalLetter(strList3,letters3)

8 0
2 years ago
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