Please help change the program below as follows:

a. Use pthreads instead of C++ version 11 threads.

b. Instead of using three threads (for Students A, B and C), make the program work with n threads. Use a symbolic constant value: NSTUDENTS to define this value. You must use arrays to keep track of the different threads. Odd id values are to be associated with expired library cards.

c. Change the program to ensure that all students have found their books (returned from findBooks) before allowing any students (threads) to continue with checking out their books (invoking checkOut). Add an output message (all books found) that is to be issued just once when all students have returned from findBooks. Solve it without using the pthread_barrier-functions

Code:

#include #include #include using namespace std; mutex Out; void println (string s) { Out.lock(); cout

private: int maxCustomers; int numberOfCustomers; int numberOK; int numberExpired; int numberOfBooks; mutex shared_var;

public: Library (int); void close (); int findBooks (int); bool checkOut (int, bool, int); }; Library::Library (int mc) { maxCustomers = mc; numberOfCustomers = 0; numberOK = 0; numberExpired = 0; numberOfBooks = 0; }

void Library::close () { cout

int Library::findBooks (int id) { return ( id * id + id + 1 ); }

bool Library::checkOut(int id, bool expired, int books) { shared_var.lock(); numberOfCustomers++; if (expired) numberExpired++; else { numberOK++; numberOfBooks += books; } shared_var.unlock();

return (!expired); } class Student { private: int id; bool cardExpired; Library *library;

public: Student (); Student (int, bool, Library*); void operator () () ; void run (); }; Student::Student () { id = 0; cardExpired = false; library = NULL; } Student::Student ( int i, bool x, Library* lib ) { id = i; cardExpired = x; library = lib; } void Student::operator () () { run (); }

void Student::run () { int books;

books = library->findBooks (id); if ( library->checkOut(id, cardExpired, books)) { string sOut = "Student " + to_string(id) + " leaves library with " + to_string(books) + " books"; println (sOut); } } int main(int argc, const char * argv[]) { Library* library = new Library (3);

thread studentA(Student (0, false, library)); thread studentB{Student (1, true, library)}; thread studentC{Student (2, false, library)};

println ("All students are visiting the library.");

studentA.join(); studentB.join(); studentC.join();

library->close();

return 0; }

Output with 5 Students with solution to parts a - c: Student O found 1 All students are visiting the library. Student 4 found 21 Student 2 found 7 Student 3 found 13 Student i found 3 all books found Student O leaves library with 1 books Student 4 leaves library with 21 books Student 2 leaves library with 7 books Number of customers: 5 Number of books checked out: 29 Output with 5 Students with solution to parts a - c: Student O found 1 All students are visiting the library. Student 4 found 21 Student 2 found 7 Student 3 found 13 Student i found 3 all books found Student O leaves library with 1 books Student 4 leaves library with 21 books Student 2 leaves library with 7 books Number of customers: 5 Number of books checked out: 29