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9.x — Chapter 9 comprehensive quiz

In this chapter, we explored topics related to operator overloading, as well as overloaded typecasts, and topics related to the copy constructor.

Summary

Operator overloading is a variant of function overloading that lets you overload operators for your classes. When operators are overloaded, the intent of the operators should be kept as close to the original intent of the operators as possible. If the meaning of an operator when applied to a custom class is not clear and intuitive, use a named function instead.

Operators can be overloaded as a normal function, a friend function, or a member function. The following rules of thumb can help you determine which form is best for a given situation:

  • If you’re overloading assignment (=), subscript ([]), function call (()), or member selection (->), do so as a member function.
  • If you’re overloading a unary operator, do so as a member function.
  • If you’re overloading a binary operator that modifies its left operand (e.g. operator+=), do so as a member function if you can.
  • If you’re overloading a binary operator that does not modify its left operand (e.g. operator+), do so as a normal function or friend function.

Typecasts can be overloaded to provide conversion functions, which can be used to explicitly or implicitly convert your class into another type.

A copy constructor is a special type of constructor used to initialize an object from another object of the same type. Copy constructors are used for direct/uniform initialization from an object of the same type, copy initialization (Fraction f = Fraction(5,3)), and when passing or returning a parameter by value.

If you do not supply a copy constructor, the compiler will create one for you. Compiler-provided copy constructors will use memberwise initialization, meaning each member of the copy is initialized from the original member. The copy constructor may be elided for optimization purposes, even if it has side-effects, so do not rely on your copy constructor actually executing.

Constructors are considered converting constructors by default, meaning that the compiler will use them to implicitly convert objects of other types into objects of your class. You can avoid this by using the explicit keyword in front of your constructor. You can also delete functions within your class, including the copy constructor and overloaded assignment operator if desired. This will cause a compiler error if a deleted function would be called.

The assignment operator can be overloaded to allow assignment to your class. If you do not provide an overloaded assignment operator, the compiler will create one for you. Overloaded assignment operators should always include a self-assignment check.

New programmers often mix up when the assignment operator vs copy constructor are used, but it’s fairly straightforward:

  • If a new object has to be created before the copying can occur, the copy constructor is used (note: this includes passing or returning objects by value).
  • If a new object does not have to be created before the copying can occur, the assignment operator is used.

By default, the copy constructor and assignment operators provided by the compiler do a memberwise initialization or assignment, which is a shallow copy. If your class dynamically allocates memory, this will likely lead to problems, as multiple objects will end up pointing to the same allocated memory. In this case, you’ll need to explicitly define these in order to do a deep copy. Even better, avoid doing your own memory management if you can and use classes from the standard library.

Quiz Time

1) Assuming Point is a class and point is an instance of that class, should you use a normal/friend or member function overload for the following operators?

1a) point + point
1b) -point
1c) std::cout << point
1d) point = 5;

Show Solution

2) Write a class named Average that will keep track of the average of all integers passed to it. Use two members: The first one should be type int32_t, and used to keep track of the sum of all the numbers you’ve seen so far. The second should be of type int8_t, and used to keep track of how many numbers you’ve seen so far. You can divide them to find your average.

2a) Write all of the functions necessary for the following program to run:

and produce the result:

4
6
12
6.5
7
7

Hint: Remember that int8_t is usually typedef’d as a char, so std::cout treats it accordingly.

Show Solution

2b) Does this class need an explicit copy constructor or assignment operator?

Show Solution

3) Write your own integer array class named IntArray from scratch (do not use std::array or std::vector). Users should pass in the size of the array when it is created, and the array should be dynamically allocated. Use assert statements to guard against bad data. Create any constructors or overloaded operators needed to make the following program operate correctly:

This programs should print:

5 8 2 3 6
5 8 2 3 6

Show Solution

4) Extra credit: This one is a little more tricky. A floating point number is a number with a decimal where the number of digits after the decimal can be variable. A fixed point number is a number with a fractional component where the number of digits in the fractional portion is fixed.

In this quiz, we’re going to write a class to implement a fixed point number with two fractional digits (e.g. 12.34, 3.00, or 1278.99). Assume that the range of the class should be -32768.99 to 32767.99, that the fractional component should hold any two digits, that we don’t want precision errors, and that we want to conserve space.

4a) What type of member variable(s) do you think we should use to implement our fixed point number with 2 digits after the decimal? (Make sure you read the answer before proceeding with the next questions)

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4b) Write a class named FixedPoint2 that implements the recommended solution from the previous question. If either (or both) of the non-fractional and fractional part of the number are negative, the number should be treated as negative. Provide the overloaded operators and constructors required for the following program to run:

This program should produce the result:

34.56
-2.08
-2.08
-2.08
-0.05
-0.05

Hint: Although it may initially seem like more work initially, it’s helpful to store both the non-fractional and fractional parts of the number with the same sign (e.g. both positive if the number is positive, and both negative if the number is negative). This makes doing math much easier later.
Hint: To output your number, first cast it to a double.

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4c) Now add a constructor that takes a double. You can round a number (on the left of the decimal) by using the round() function (included in header cmath).

Hint: You can get the non-fractional part of a double by static casting the double to an integer
Hint: To get the fractional part of a double, you’ll first need to zero-out the non-fractional part. Use the integer value to do this.
Hint: You can move a digit from the right of the decimal to the left of the decimal by multiplying by 10. You can move it two digits by multiplying by 100.

The follow program should run:

This program should produce the result

0.01
-0.01
5.01
-5.01

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4d) Overload operator==, operator >>, operator- (unary), and operator+ (binary).

The following program should run:

And produce the output:

true
true
true
true
true
true
true
true
-0.48
0.48
Enter a number: 5.678
You entered: 5.68

Hint: Add your two FixedPoint2 together by leveraging the double cast, adding the results, and converting back to a FixedPoint2.
Hint: For operator>>, use your double constructor to create an anonymous object of type FixedPoint2, and assign it to your FixedPoint2 function parameter

Show Solution

10.1 -- Object relationships
Index
9.15 -- Shallow vs. deep copying

186 comments to 9.x — Chapter 9 comprehensive quiz

  • Satti

    Hello, I have a small question that I can't seem to answer.
    For question number 1, I have written everything identical to the solution code except the following:

    When I compiled and ran the above code, I got 6.4 at the end, which means that it calculated only (ans += 6)/5 and not the other +=10 also.
    I then noticed that in the solution, there was a '&' after Average and I didn't. So I added the '&' to my original code, and it worked perfectly.
    I don't understand why this would work and mine wouldn't. Why does it make a difference?

    • avg += 6 executes first. It increases @avg.m_total and @avg.m_numbers. It then returns a copy(!) of @avg.
      copy += 10 executes next, this doesn't affect @avg.

      Lesson 6.11 and 7.4a

  • AlexR

    Hi. In quiz 3 I was playing around with the code in order to understand the copy constructor a bit more but there's something I just can't understand:

    Consider this code (I'm pasting the relevant code only).

    Note that in this code I removed the reference to the second parameter of operator<<().

    The copy constructor should be called every time I want to print with cout because the object will be passed by value (I know this is discouraged and a reference should be used). If I remove the copy constructor the compiler will use the default one which does a shallow copy and that means I should expect a bad output in the console.
    However in my machine I'm receiving this:

    Why is only the second one incorrect and not the first one if in both calls to cout the default copy constructor is doing shallow copies of the object?

    • Hi Alex!

      Line 29: You don't have a copy constructor, this line alone will crash your program. It doesn't, because of compiler optimizations.
      Assuming your compiler optimizes line 29 and your program continues with @a in a valid state.
      Line 30: A shallow copy of @a is created for @operator<<. The clone's @m_numbers pointer is valid, there's no problem.
      @operator<< finishes, calling the clone's destructor. The destructor deletes the array (!).

      Since the array was shared between @a and it's clone, @a's @m_numbers is no longer valid.

      Line 33 uses a reference parameter and doesn't access @a's @m_numbers. This is fine.
      Line 34 passes @a by reference, but also accesses @a's @m_numbers. Behavior is undefined.
      Line 36: @b's array is valid, but was initialized with undefined values (In your case, zeroes). When @operator<< is called, what happened to @a in line 30 will now happen to @b.

      At the end of @main, both @a's and @b's @m_numbers have already been deleted. Since they go out of scope now, their destructors will be called and another delete will be performed. Behavior is undefined (Or a crash is guaranteed, I'm not sure).

  • Dean Cutillar

    Hello all,

    I have been slowly but surely working through these chapters learning C++.  I have a lingering confusion over distinguishing a pointer vs a reference to a variable.

    Here is an example:

    So this function to overload operator=, does it return an address to a fraction, or a reference to a fraction?

    by returning *this at the end of the function, is that a give away that means it's an address?  Also, I thought "*this" was a de-reference and would result in the actual data being pointed to.  So does it return an address, or the actual data pointed to?  

    Lastly, in the "self assignment guard" if statement (this==&fraction), so I always read "&" in the parameter as "a reference to" the variable.  But that statement is comparing it to "this" which is a pointer variable.  So is it a pointer?  Does including the "&" in the body of the function mean the data being pointed to?

    Please clarify these issues.  It really would help me a lot.  Thanks.

    Dean.

    • Alex

      This function returns a reference.

      If & is applied to an object, it means address of.
      If & is applied to a type, it means reference (in the case of function parameters, this case applies)

      Since this is a pointer, *this is the object being pointed to by "this". In the self-assignment guard, &fraction means "address-of" in this context, which is a pointer.

  • Silviu

    Hello, i was reading the code and something slipped and noticed just now.
    ->operator double()
    usually is operator()(),operator+ and so on ...

    There is a lesson that i missed or something ? can you explain ? Thank you.

  • Jörg

    Hi, wouldn't it be better to make the operator- overload return a const reference instead of a copy, like that:

    Is there a reason not to?

  • Danang

    this is my code for number 4a - 4c :  i use int for member variable to increase size of decimal number can hold to handle the case where decimal is > 99 or < -99 here.

    is this better way to handle the case where decimal is > 99 or < -99 than using int16_t or int8_t?

  • MikiBG

    In the last question, operators +, >> and << do not need to access member variables. Is there any reason why one should prefer to define them as friend functions over just normal functions?

    • Alex

      Most seasoned devs would tell you it's better to define overloaded operators as non-members if you can, because it "improves encapsulation" by ensuring you don't access internal members directly, and reduces the number of functions you have to consider updating when making a change to the class.

      I've updated the solution to define those outside the class, as per best practice.

  • kjm

    Hello,
    Why not to use a single int32_t field for 4a? (fixed-point behaves essentially like an integer while memory is aligned anyway)
    Thank you.

  • Hi, worked through 4(b) and then moved on to 4(c) (it's only another function after all) and ended up with the following:

    Hopefully this is good enough to move on to the next?

    • * Line 17
      * Line 19: Unnecessary check. You're checking again in the following lines
      *-Line 24, 36, 38: Can be written without repetition of variable names
      *-@FixedPoint2::FixedPoint2(double): Favor initialization over assignment
      * Line 46: @round should be @std::round
      * Line 46, 51: Use double numbers when calculating with doubles (100.0 instead of 100)
      * Line 64-76: Way to go

      > Hopefully this is good enough to move on to the next?
      See if you can work out the *- suggestions first. Other than that you code looks fine.

  • For number 3 I give you:

    • * Line 16, 22, 24, 36, 53, 65, 68
      * Line 19: That 0 only initializes the first element. Empty curly brackets mean "initialize the entire array to default values (0)".
      * Line 43: Compilation error. Should be

      or for large types

  • Well, here's my attempt at number 2.  It produces the correct result so I suppose it is ok

    • Hi Nigel!

      * Line 11, 12: Thank goodness!
      * Line 11, 12: Should be @std::int32_t and std::int8_t, but those shouldn't be used at all, because they're optional. Alex is to blame for this, you did what you were asked for.
      * @Average::Average is unused, you might as well remove it.
      * Line 37, 54

  • TheBendeee

    Hi! I have problems with understanding this part of the code:
    (line 31:)

    Why does this work, and how?

    • Alex

      Consider a number like 2.56. 2 would be stored in m_base, and 56 would be stored in m_decimal. To reconstruct the actual value, we first need to convert m_base from 56 back to 0.56. We do this by dividing it by 100. We static_cast(m_decimal) to make sure we're doing floating point division instead of integer division. Then we add m_base to finish reconstructing the entire number.

      • Silviu

        Hello,
        Alex great example and great lessons, but i think you should explain this in your code, it's still a key part in the code, that is not really explained and I think it's a good explanation and very helpful.
        Thank you.

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