5779 a

#include <iostream>
using namespace std;

using namespace std;
class A {
public:
    A() { cout << "A(0)" << endl; }
    A(const A& a) { cout << "A(1)" << endl; }
};

class B {
public:
    B() : a() { cout << "B(0)" << endl; }
    B(const B& b) { cout << "B(1)" << endl; }
private:
    A a;
 };

int main() {
    B object1;
    B object2 = object1;
    return 0;
}

What is the output?

B object1;

will result in, since a is a member field of B it will initialized before and the constructor will be called before Bs ctor

A(0)
B(0)

and

B object2 = object1;

will output, since the copy ctor B(const B& b) will be called

A(0)
B(1)

Remember!

An object is copied when:

1. Constructing new object from existing Copy ctor

2. Passing parameter by value Copy ctor

3. Returning by value. Copy ctor

4. Assigning existing to existing. Assignment ctor

Cases 1-3 are handled by copy constructor;

Case 4 is handled by assignment operator.

part b

remove this line

B(const B& b) { cout << "B(1)" << endl; }

And now what will be the output?

#include <iostream>
using namespace std;

using namespace std;
class A {
public:
    A() { cout << "A(0)" << endl; }
    A(const A& a) { cout << "A(1)" << endl; }
};

class B {
public:
    B() : a() { cout << "B(0)" << endl; }
private:
    A a;
 };

int main() {
    B object1;
    B object2 = object1;
    return 0;
}

Solution

A(0)
B(0)
A(1)

first B object1; will result in A(0) B(0) as above.

Now the copy ctor is called since it is the creation of the object2. Since a is initialized before any object it uses the default.

part c

now delete this line and return the previous line

A(const A& a) { cout << "A(1)" << endl; }

#include <iostream>
using namespace std;

class A {
public:
    A() { cout << "A(0)" << endl; }
};

class B {
public:
    B() : a() { cout << "B(0)" << endl; }
    B(const B& b) { cout << "B(1)" << endl; }
private:
    A a;
};

int main() {
    B object1;
    B object2 = object1;
    return 0;
}

What is the output?

Solution

A(0) B(0) A(0) B(1)

first B object1; will result in A(0) B(0) as above. Then initialize a A(0) and finally call the copy ctor B(1)

Question 2 (9pts)

#include <iostream>
using namespace std;

class A;

class B {
public:
  explicit B (A& o) { cout << "B (A)" << endl; }
};

class A {
public:
  A() { cout << "A()" << endl; }
};

void printit(B arg) {}

int main() {
      A a;
      B b(a);
      printit(a);
      return 0;
 }

The following cause a compile time error

printit(a);

Explain the error and why does it happen

Solution

printit(B arg) receives a B object but was given an A object, it can't convert a since the ctor is explicit

Part b

fix the code above, only fix this line

B b(a);
printit(a); // <======= this line
return 0;

Solution

We can call converting ctor explicitly

printit(B (a))

or just give a B object

printit(b)

Part c

After the fix what will be the output?

Solution

If we used the 1st answer above printit(B (a)) then the output would be like this

int main() {
      A a;            // A()
      B b(a);         // B (A)
      printit(B (a)); // B (A)
      return 0;
 }

If we used the 2nd answer above printit(b) then the output would be like this

int main() {
      A a;            // A()
      B b(a);         // B (A)
      printit(b);     
      return 0;
 }

printit(B arg) will call the copy ctor, which hasn't been defined

Question 3: (9pts)

#include <iostream>
using namespace std;

class Shape {
public:
    virtual void draw() = 0;
    virtual ~Shape() { cout << "~Shape()" << endl; }
};


class Circle : public Shape {
public:
    void draw() { cout << "Circle::draw()" << endl; }
    ~Circle() { cout << "~Circle()" << endl; }
};


void drawTheShape(Shape s) {
     s.draw();
}

int main() {
    Circle c;
    drawTheShape(c);
    return 0;
}

Part a:

On this the following line there is a compile time error

void drawTheShape(Shape s) {

Explain the error, and why is it happening?

Solution

void drawTheShape(Shape s) Here a new instance of Shape is called but since Shape is pure virtual as seen here

class Shape {
public:
    virtual void draw() = 0;

Objects can't be created.

Part b:

fix the line void drawTheShape(Shape s) { and only this line

Solution 1:

receive s by reference.

void drawTheShape(Shape& s) {

Solution 2:

receive an implemented class type

void drawTheShape(Circle s) {

Part c:

what is the output?

using the previous Solution 1:

Circle::draw()
~Circle()
~Shape()

This is because we first call the derived dtor and only after we call the bases dtor.

using the previous Solution 2:

Circle::draw()
~Circle()
~Shape()
~Circle()
~Shape()

Question 4: (9pts)

#include <iostream>
#include <string>
using namespace std;

class string {
    const char* p;
public:
    string(const char* p) {this->p = p;}
};

int main() { // a demo program
    string s = "abc";
}

main.cpp:12:2: error: reference to 'string' is ambiguous
string s = "abc";
^

Part a:

Why type of error is this:

• compilation

• linker

• runtime

• logic

• memory leak

Additionally, what caused the error?

Solution

This is a compile-time error

• We included #include <iostream> which includes std::string

• Also, we used using namespace std;

• Finally, we created a class by the name of string

  class string {
      ...
  };

Now the compiler doesn't know which string to call, our string or std::string

Read more here and here about different types of errors

This make sense that here is a compile error since it has nothing to do typing object files together or using a library rather ambiguity.

Part b:

Fix the code by removing only 1 line of code, don't alter the main

Solution

remove this line

using namespace std;

now the compiler will knows who to call since string refers to our class and std::string refers to the std

Part c

class string {
    const char* p;
public:
    string(const char* p) {this->p = p;}
};

Here string has pointer p but string doesn't have a dtor.

Will this cause an error? If so, why type of error, if not why won't it cause an error?

Solution

This is no error, notice the following

string(const char* p) {this->p = p;}

string doesn't allocate memory there is no new or malloc it only receives a pointer to allocated memory

Question 5-6: (22pts)

Question 7: (22pts)

#include <iostream>
using namespace std;
#include "Money.hpp"
using ariel::Money;

int main() {
// Set the exchange rates in shekels:
    Money::set_rate("ILS", 1);
    Money::set_rate("USD", 4);
    Money::set_rate("EUR", 6);
// Define amounts and do calculations:
    Money shekel1(1, "ILS");
    Money dollar3(3, "USD");
    Money sum = shekel1 + dollar3;
    cout << sum << endl; // 13 ILS
    cout << (dollar3 + Money(1, "EUR")) << endl; // 4.5 USD
    (sum += Money(2, "EUR")) += Money(0.5, "ILS");
    cout << sum << endl; // 25.5 ILS
// Money(1, "XXX"); // throws an exception
}

• Write Money.cpp and Money.hpp

• Allow addition of different sums with automatic conversion of units

• The rate of each coin is set in the beginning of the code

Solution

Money.hpp

#include <string>
#include <map>
#include <iostream>
using std::string, std::map, std::endl, std::ostream, std::ifstream;
namespace ariel{
    class Money {
        double amount;
        string currency;
        double rate;
        static map<string,double> map_currency_to_rate;
    public:
        Money(double amount, string currency): amount(amount), currency(currency) {
            rate = map_currency_to_rate.at(currency); // throws exception if not found
        }
        friend Money operator+(const Money& a, const Money& b);
        friend Money& operator+=(Money& a, const Money& b);
        friend ostream& operator<< (ostream& out, const Money& m);
        static void set_rate(string currency, double rate) {
            map_currency_to_rate[currency] = rate;
        }
    };
}

Money.cpp

#include "Money.hpp"
namespace ariel {
    Money operator+(const Money& a, const Money& b) {
        double amount = a.amount + b.amount*b.rate/a.rate;
        return Money(amount, a.currency);
    }

    Money& operator+=(Money& a, const Money& b) {
        a.amount += b.amount*b.rate/a.rate;
        return a;
    }

    ostream& operator<< (ostream& out,  const Money& m) { //on test forgot const and &
        return (out << m.amount << " " << m.currency) << endl;
    }

    map<string,double> Money::map_currency_to_rate{};
}

13 ILS

4.5 USD

25.5 ILS