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The issue is that the solution always assumes that there will be two solutions from the quadratic formula, yet this is not always the case. We identified a few circumstances where this program would fail:

1. If (b^2-4ac<0), then there are no real solutions. You cannot take the squareroot of a negative number, so the program outputs nan twice.
2. If (b^2-4ac=0), both solutions from the quadratic formula are identical. We should not repeat this solution in the output.
3. If (a=0) but (b\neq 0), we have a linear equation (bx+c=0). In this case, the quadratic formula fails because it’ll divide by zero. There’s also just one solution: (x=-\frac{c}{b} ).
4. If (a=b=0), then we have the equation (c=0). If (c\neq 0), then this equation is false no matter what (x) is, and there are no solutions. If (c=0), then every value of (x) is a solution, and we should indicate as much.

This was the solution we produced together as a class:

 1#include <iostream>
 2#include <cmath>
 3
 4using namespace std;
 5
 6int main() {
 7    double a, b, c, d;
 8
 9    // read in the variables a, b, and c.
10    // they are the coefficients of the equation
11    // ax^2 + bx + c = 0
12    cin >> a >> b >> c;
13
14    if(a == 0 && b == 0 && c == 0) {
15        cout << "Infinitely many solutions." << endl;
16    } else if(a == 0 && b == 0) {
17        cout << "No solutions." << endl;
18    } else if(a == 0) {
19        cout << "Solution: " << -1 * c / b << endl;
20    } else {
21        // use the quadratic formula
22        // x = (-b +/- sqrt(b^2 - 4ac)) / 2a
23        // d = sqrt(b^2-4ac)/ 2a
24        d = b * b - 4 * a * c;
25        if(d > 0) {
26            cout << "Solutions: "
27                 << (-1 * b + sqrt(d)) / (2 * a)
28                 << ", " << (-1 * b - sqrt(d)) / (2 * a) << endl;
29        } else if(d < 0) {
30            cout << "No solutions." << endl;
31        } else {
32            cout << "Solution: " << -1 * b / (2 * a) << endl;
33        }
34    }
35
36    return 0;
37}

This time, the issue was that the if statements were “independent”. That is to say, if both the year and the month were invalid, then the if statements on lines 11 and 16 would both run, producing (at least!) two “Invalid” outputs. One way to fix this is by wrapping things in else statements. However, one should stay away from this when more complex logic is involved (such as finding the number of days in a month). Having many nested if/else-if/else statements produces code that is simultaneously hard to read and easy to mistype, making errors painful to chase down.

The solution we ultimately settled on in class was to use a bool variable to keep track of whether or not the date was invalid. If the year is invalid, then the boolean is set to false; if the month is invalid, it is also set to false; if the day is invalid, it too is set to false. This way, even if we detect multiple sources of invalid input, it doesn’t make a difference. After we check all three numbers, we output “Valid” or “Invalid” according to the contents of this boolean variable. Here is the code:

 1#include <iostream>
 2
 3using namespace std;
 4
 5int main() {
 6    // read year, month, and day from user
 7    int year, month, day;
 8    cin >> year >> month >> day;
 9    bool valid = true;
10
11    // if year is negative, it's invalid.
12    if(year < 0) {
13        valid = false;
14    }
15
16    // if month < 1 or month > 12, it's invalid
17    if(month < 1 || month > 12) {
18        valid = false;
19    }
20
21    // months 1, 3, 5, 7, 8, 10, 12 have 31 days.
22    // months 4, 6, 9, 11 have 30 days.
23    // month 2 has 28 days most of the time, 
24    // but 29 days on leap years
25    int last_day = 31;
26    if(month == 4 || month == 6 || month == 9 || month == 11) {
27        last_day = 30;
28    } else if(month == 2) {
29        last_day = 28;
30        if(year % 4 == 0 && year % 100 != 0 || year % 400 == 0) {
31            last_day = 29;
32        }
33    }
34
35    if(day < 1 || day > last_day) {
36        valid = false;
37    }
38
39    if(valid) {
40        cout << "Valid" << endl;
41    } else {
42        cout << "Invalid" << endl;
43    }
44
45    return 0;
46}

Once again, regardless of how many of the if statements on lines 11, 17, and 35 get triggered, the final result will still only output one “Invalid”.

I mentioned an alternative solution briefly during class. Rather than using a boolean variable to “remember” if the date is invalid, we could realise that the moment we determine that a date is invalid, the rest of the program has no bearing on the output. We should just print “Invalid” and exit the program!

To do so, we can actually use the return 0; to our benefit. This line tells the program, “We have finished executing the main function.” The 0 is a status indicator, and it represents a status of “everything is okay”. As such, putting a return 0; earlier in the main function will just terminate the program early. This is the corresponding code:

 1#include <iostream>
 2
 3using namespace std;
 4
 5int main() {
 6    // read year, month, and day from user
 7    int year, month, day;
 8    cin >> year >> month >> day;
 9
10    // if year is negative, it's invalid.
11    if(year < 0) {
12        cout << "Invalid" << endl; 
13        return 0; // early return - exits the program 
14    }
15
16    // if month < 1 or month > 12, it's invalid
17    if(month < 1 || month > 12) {
18        cout << "Invalid" << endl; 
19        return 0; // early return - exits the program
20    }
21
22    // months 1, 3, 5, 7, 8, 10, 12 have 31 days.
23    // months 4, 6, 9, 11 have 30 days.
24    // month 2 has 28 days most of the time, 
25    // but 29 days on leap years
26    int last_day = 31;
27    if(month == 4 || month == 6 || month == 9 || month == 11) {
28        last_day = 30;
29    } else if(month == 2) {
30        last_day = 28;
31        if(year % 4 == 0 && year % 100 != 0 || year % 400 == 0) {
32            last_day = 29;
33        }
34    }
35
36    if(day < 1 || day > last_day) {
37        cout << "Invalid" << endl; 
38    } else {
39        cout << "Valid" << endl; 
40    } 
41
42    return 0;
43}

This is a programming pattern that will appear again in the future. Often, you will encounter situations where your program’s behaviour is very simple and predictable for a very specific set of inputs or edge cases. Rather than handling these edge cases in if statements and wrapping the rest of your entire program in an else statement, it’s better to just exit the program early immediately after handling the edge cases. This produces clean, efficient, and maintainable code!