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4 years ago

Consider the partially-filled array named a. What does the following loop do? (cin is a Scanner object)

Computers and Technology

1 answer:

Keith_Richards [23]4 years ago

4 0

Answer:

Option (1)

Explanation:

value is read outside the loop so it is read only once. After that the loop has begun and in the loop condition it is checked if size < capacity and as size is initialized as 3 so it is true and value if entered more than 0 then loop starts and value is assigned to a[3] and size is incremented and it continues till size is 6 and condition becomes false. So loop executes 3 times but same value is assigned.

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Which of the following occurs when the amount of money earned is greater than the

swat32

Answer:

When this happens, there is a surplus of money and you can pay off your debts.

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4 years ago

Lance is at a bus station. His friend is using the ATM machine to withdraw some money. Lance notices a stranger deceptively watc

Blababa [14]

Answer:

The person watching Lances friend typing the ATM pin is an example of "shoulder browsing" and Lances friend should "Change the ATM pin."

I hope this is the answer you were looking for!

8 0

3 years ago

Read 2 more answers

Write a code that calculates the Greatest Common Divisor (GCD) of two positive integers (user-defined inputs). Include an except

STALIN [3.7K]

Answer:

This program is written using Java programming language

import java.util.*;

public class calcGcd

{

public static void main (String [] args)

{

int num1, num2;

Scanner input = new Scanner(System.in);

//Input two integers

num1 = input.nextInt();

num2 = input.nextInt();

//Get least of the two integers

int least = num1;

if(num1 > num2)

{

least = num2;

}

//Initialize gcd to 1

int gcd = 1;

//Calculate gcd using for loop

for(int i=1;i<=least;i++)

{

if(num1%i == 0 && num2%i == 0)

{

gcd = i;

}

if(gcd == 1)

{

System.out.print("GCD is 1");

}

else

{

System.out.print("GCD is "+gcd);

}

Explanation:

To calculate the GCD, the program uses a for loop that iterates from 1 to the smaller number of the user input.

Within this iteration, the program checks for a common divisor of the two user inputs by the iterating element

The GCD is then displayed afterwards;

However, if the GCD is 1; the program prints the message "GCD is 1"

Line by Line Explanation

This line declares two integer numbers

int num1, num2;

This line allows user the program to accept user defined inputs

Scanner input = new Scanner(System.in);

The next two line allows gets inputs from the user

num1 = input.nextInt();

num2 = input.nextInt();

To calculate the GCD, the smaller of the two numbers is needed. The smaller number is derived using the following if statement

int least = num1;

if(num1 > num2)

{

least = num2;

}

The next line initializes GCD to 1

int gcd = 1;

The GCD is calculated using the following for loop

The GCD is the highest number that can divide both numbers

for(int i=1;i<=least;i++)

{

if(num1%i == 0 && num2%i == 0)

{

gcd = i;

}

The following is printed if the calculated GCD is 1

if(gcd == 1)

{

System.out.print("GCD is 1");

}

Otherwise, the following is printed

else

{

System.out.print("GCD is "+gcd);

}

8 0

3 years ago

To make it easier to enter names into your code, NetBeans provides a feature known as

Alexus [3.1K]

Answer:

Code completion.

Explanation:

The NetBeans is an integrated development environment framework of java language. Code completion is the features of NetBeans which is easier to complete the code This feature is known as smart code features because it completes the code in a very easy manner. The Code completion features are needful when we have to fill the missing code or program in the java language.

Hence Code completion. is used to enter the name into the code in the NetBeans.

3 0

3 years ago

and assuming main memory is initially unloaded, show the page faulting behavior using the following page replacement policies. h

Svet_ta [14]

FIFO

// C++ implementation of FIFO page replacement

// in Operating Systems.

#include<bits/stdc++.h>

using namespace std;

// Function to find page faults using FIFO

int pageFaults(int pages[], int n, int capacity)

{

// To represent set of current pages. We use

// an unordered_set so that we quickly check

// if a page is present in set or not

unordered_set<int> s;

// To store the pages in FIFO manner

queue<int> indexes;

// Start from initial page

int page_faults = 0;

for (int i=0; i<n; i++)

{

// Check if the set can hold more pages

if (s.size() < capacity)

{

// Insert it into set if not present

// already which represents page fault

if (s.find(pages[i])==s.end())

{

// Insert the current page into the set

s.insert(pages[i]);

// increment page fault

page_faults++;

// Push the current page into the queue

indexes.push(pages[i]);

}

// If the set is full then need to perform FIFO

// i.e. remove the first page of the queue from

// set and queue both and insert the current page

else

{

// Check if current page is not already

// present in the set

if (s.find(pages[i]) == s.end())

{

// Store the first page in the

// queue to be used to find and

// erase the page from the set

int val = indexes.front();

// Pop the first page from the queue

indexes.pop();

// Remove the indexes page from the set

s.erase(val);

// insert the current page in the set

s.insert(pages[i]);

// push the current page into

// the queue

indexes.push(pages[i]);

// Increment page faults

page_faults++;

}

return page_faults;

}

// Driver code

int main()

{

int pages[] = {7, 0, 1, 2, 0, 3, 0, 4,

2, 3, 0, 3, 2};

int n = sizeof(pages)/sizeof(pages[0]);

int capacity = 4;

cout << pageFaults(pages, n, capacity);

return 0;

}

LRU

//C++ implementation of above algorithm

#include<bits/stdc++.h>

using namespace std;

// Function to find page faults using indexes

int pageFaults(int pages[], int n, int capacity)

{

// To represent set of current pages. We use

// an unordered_set so that we quickly check

// if a page is present in set or not

unordered_set<int> s;

// To store least recently used indexes

// of pages.

unordered_map<int, int> indexes;

// Start from initial page

int page_faults = 0;

for (int i=0; i<n; i++)

{

// Check if the set can hold more pages

if (s.size() < capacity)

{

// Insert it into set if not present

// already which represents page fault

if (s.find(pages[i])==s.end())

{

s.insert(pages[i]);

// increment page fault

page_faults++;

}

// Store the recently used index of

// each page

indexes[pages[i]] = i;

}

// If the set is full then need to perform lru

// i.e. remove the least recently used page

// and insert the current page

else

{

// Check if current page is not already

// present in the set

if (s.find(pages[i]) == s.end())

{

// Find the least recently used pages

// that is present in the set

int lru = INT_MAX, val;

for (auto it=s.begin(); it!=s.end(); it++)

{

if (indexes[*it] < lru)

{

lru = indexes[*it];

val = *it;

}

// Remove the indexes page

s.erase(val);

// insert the current page

s.insert(pages[i]);

// Increment page faults

page_faults++;

}

// Update the current page index

indexes[pages[i]] = i;

}

return page_faults;

}

// Driver code

int main()

{

int pages[] = {7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2};

int n = sizeof(pages)/sizeof(pages[0]);

int capacity = 4;

cout << pageFaults(pages, n, capacity);

return 0;

}

You can learn more about this at:

brainly.com/question/13013958#SPJ4

4 0

1 year ago