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jeka94
4 years ago
8

In this challenge, write a function to add two floating point numbers. Determine the integer floor of the sum. The floor is the

truncated float value, i.e. anything after the decimal point is dropped. For instance, floor(1.1 + 3.05) = floor(4.15) = 4.
Computers and Technology
1 answer:
Wewaii [24]4 years ago
5 0

Answer:

def floor_sum(number1, number2):

   total = number1 + number2

   if total >= 0:

       return int(total)

   else:

       return int(total) - 1

   

print(floor_sum(1.1, 3.05))

Explanation:

*The code is in Python.

Create a function called floor_sum that takes two parameters, number1 and number2

Sum them and set it to the total

Check if the total is positive number or 0, return the integer part of the total. Otherwise, subtract 1 from the integer part of the total and return it.

Call the function with given parameters in the question and print the result

Note that if the result is a negative value like -5.17, the floor of the result is -6 not -5.

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4 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

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