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3. Week 2 Tuesday: Working with Numbers

I had planned to discuss integer division and the modulo operator last Thursday, but I did not get to discuss them in full detail. So we’ll reiterate some things that I didn’t get to describe fully, and we’ll also describe some messy details involving numbers in C++.

Integer Division and the Modulo Operator

Whenever one performs an arithmetic operation in C++ with two integers, one will always end up with another integer. This is not a problem most of the time: the sum, difference, and product of two integers is always an integer anyways. However, the quotient of two integers is not always an integer! Consider the following code snippet:

1cout << 18 / 3 << endl; 
2cout << 19 / 3 << endl; 
3cout << -20 / 3 << endl; 

The output of this code consists of three numbers: 6, 6, and -6, each on its own line. C++’s method of converting the mathematical answers — \(6\), \(6.3333\ldots\), and \(-6.6666\ldots\) — into integers is simply to cut off everything from the decimal point onwards.

One serious ramification of this problem is that multiplication and division are no longer true inverse operations, i.e. division and multiplication no longer cancel each other out all the time. For instance, if (a) and (b) are any two integers, one would (mathematically) expect the following equations to be true: \[\begin{align*} (a\times b) \div b = a && \textrm{and} && (a\div b)\times b = a.\end{align*}\] The first one will pretty much always be true, but the second may not.

1cout << (19 / 3) * 3 << endl; 

This line of code will output 18, not 19!

This is where the modulo operator enters the fray: it represents the discrepancy in the equation \((a\div b) \times b = a\) in C++. More specifically, the expression a % b will always be equal to a - (a / b) * b for any integers a and b, so that the equation

a = (a / b) * b + a % b

is always true when a and b are integers in C++. When a and b are positive integers, a % b coincides with the remainder of the integer division a / b! But when a or b is negative, this above equation unambiguously determines the value of a % b whereas the notion of a “remainder” may not be so clear.

Double Trouble

So what happens if you want \(6.333\ldots\) instead of \(6\) when you perform 19 / 3?

Recall that decimal numbers are represented using the double variable type in C++.

The most important thing to know about a double is that it has finite precision; we will see some examples of this later on.

Arithmetic operations involving at least one double will produce doubles:

1cout << 19 / 3 << endl; 
2cout << 19.0 / 3 << endl; 
3cout << 19 / 3.0 << endl; 
4cout << 19.0 / 3.0 << endl; 

Only the first expression is an integer; the latter 3 all were “promoted” to doubles, despite the presence of an integer.

While one can always force a literal number to be a double by adding a .0 to the end of it, one has no such luxury with integer variables. Instead, one needs to “cast” integers into doubles. To do so, if a is an integer variable, then static_cast<double>(a) will turn a into a double.

1int a = 19; 
2int b = 3; 
3
4cout << a / b << endl;  // does integer division
5
6cout << static_cast<double>(a / b) << endl; // ??? 
7cout << static_cast<double>(a) / b << endl; 

Note the output of the second line! You need to cast to a double first, then perform the division. The second line prints out 6.0 because it does the division first (which produces the integer 6), then converts that into a double.

The problem of converting a double back into an integer is a similarly slippery problem, one that is perhaps best demonstrated with an example.

SAMPLE RUNS: 
Enter a time in seconds: 124.556
124.556s = 2min, 4s, and 556ms. 

Let’s first write some pseudocode for this:

1. Take a decimal input time from the user.
2. Define the integer ms = time * 1000.
3. Define the integer seconds = ms / 1000 and replace ms with ms % 1000.
4. Define the integer minutes = seconds / 60 and replace seconds with seconds % 60.
5. Print out the results to the screen.

We might naïvely cobble together the following program:

 1#include <iostream> 
 2
 3using namespace std; 
 4
 5int main() {
 6    // 1. take input from user 
 7    double time; 
 8    cout << "Enter a time in seconds: "; 
 9    cin >> time; 
10
11    // 2-4. compute ms, seconds, minutes 
12    int ms = time * 1000; 
13
14    int seconds = ms / 1000; 
15    ms = ms % 1000; 
16
17    int minutes = seconds / 60; 
18    seconds = seconds % 60; 
19
20    // 5. print everything out. 
21    cout << time << "s = " 
22         << minutes << "min, " 
23         << seconds << "s, and "
24         << ms << "ms." << endl; 
25
26    return 0; 
27} 

However, consider the following run of the code:

Enter a time in seconds: 16.002
16.002s = 0min, 16s, and 1ms.

Something went wrong, and it’s that C++ thinks 16.002 * 1000 is just barely less than 16002 due to the limited precision of a double! Thus it rounds it down to 16001 instead.

 1#include <cmath> // needed for floor, round, ceil 
 2
 3int main() {
 4    double d = 20.15; 
 5    int i1 = d * 100;        // truncates d * 100
 6    int i2 = floor(d * 100); 
 7    int i3 = round(d * 100); 
 8    int i4 = ceil(d * 100); 
 9
10    return 0; 
11} 

With that out of the way, try to fix the code we wrote above yourself before peeking at the best way to proceed!

 1#include <iostream> 
 2#include <cmath>    // needed for round 
 3
 4using namespace std; 
 5
 6int main() {
 7    // 1. take input from user 
 8    double time; 
 9    cout << "Enter a time in seconds: "; 
10    cin >> time; 
11
12    // 2-4. compute ms, seconds, minutes 
13    int ms = round(time * 1000); 
14
15    int seconds = ms / 1000; 
16    ms = ms % 1000; 
17
18    int minutes = seconds / 60; 
19    seconds = seconds % 60; 
20
21    // 5. print everything out. 
22    cout << time << "s = " 
23         << minutes << "min, " 
24         << seconds << "s, and "
25         << ms << "ms." << endl; 
26
27    return 0; 
28} 

In general, it’s a good idea to use the round function from the cmath library to convert from doubles to integers over using the default integer conversion in C++.

Sample run: 
INPUT   30.3571 
OUTPUT  7 

INPUT   -39.82113
OUTPUT  1

INPUT   5
OUTPUT  0