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7. Week 4 Tuesday: Practise with If Statements

Our very first example of control flow is the if statement. This allows us to tell C++ to run certain lines of code and avoid running other lines of code, perhaps depending on the user’s input or on the status of our computer.

The general syntax is as follows:

1if( [condition] ) {
2    // code 
3}

The [condition] is an expression that is true or false, and the code within the curly braces runs if the condition is true. Typically, it will be a comparison of numbers, such as x <= 15, y > 0, etc. More on conditional expressions below!

You can add an else block at the end:

1if( [condition] ) {
2    // code 
3} else {
4    // more code 
5}

This time, if the condition is true, the first block of code will run; otherwise, the second block will run. Finally, you can chain multiple conditions together:

1if( [condition 1] ) {
2    // code block 1 
3} else if( [condition 2] ) {
4    // code block 2
5} else if( [condition 3] ) {
6    // code block 3 
7} // etc. 

It will go through the conditions in order, one by one, until it finds a true statement. Then, it will run the corresponding code block before exiting. Here is an example:

 1int x = 73; 
 2if(x >= 90) {
 3    cout << "You got an A" << endl; 
 4} else if(x >= 80) {
 5    cout << "You got a B" << endl; 
 6} else if(x >= 70) {
 7    cout << "You got a C" << endl; 
 8} else if(x >= 60) {
 9    cout << "You got a D" << endl; 
10} else {
11    cout << "See me after class." << endl; 
12}

This snippet of code will output You got a C. It will check the condition x >= 90, which is false. Then, it checks x >= 80, which is also false. It finally checks x >= 70, which is true, and so it runs the following code block. It then skips the rest of the else if/else blocks!

There are several things to remark:

• Note where the semicolons are. None of the curly braces are followed by semicolons, and the conditions inside the if statements don’t get semicolons!
• Else blocks are not mandatory. You can have a single if statement or an if statement followed by a few else-if’s without an else statement at the end.
• Curly braces are not mandatory. This will cause the if statement to run the first statement (i.e., up to the first semicolon) only. It is strongly recommended that you use curly braces all the time, but unfortunately programmers can be lazy and omit them. For instance, the output of the following code is pie; can you see why?

1int x = 50; 
2if(x < 0) 
3    cout << "I love "; 
4    cout << "pie" << endl; 

Conditionals

We have to put conditionals inside the if statements, and these will generally take the form of comparisons between numbers. You can compare integers, chars, and doubles with each other. If a and b are numbers, you can perform the following comparisons:

• Greater than (or equal to): a > b, a >= b.
• Less than (or eqaul to): a < b, a <= b.
• Equal to: a == b. There are two equal signs, not one. This is because a = b looks like you’re saying “overwrite the value of a with the value of b”. This is a very common source of error.
• Not equal to: a != b.

These should hopefully be self-explanatory.

Sometimes, however, we may have more than one condition that we want to be true, or several conditions of which we want at least one to be true. If p and q are two boolean expressions (i.e., true or false, such as the comparisons above), you can combine them in the following two ways:

• And: p && q, both p and q must be true.
• Or: p || q, either p or q must be true.

Of course, one can string together any number of conditions using a mix of && and ||; however, you should be careful with the order in which they are combined together. This StackOverflow answer provides a great explanation; in short, && is evaluated first in absence of parentheses.

 1#include <iostream>
 2
 3using namespace std;
 4
 5int main() {
 6    int a = 15;
 7    double b = 4.6;
 8    double c = -9.7;
 9    int d = 100;
10    double e = 2.718;
11    double f = 4.178e18;    // scientific notation - 4.178 * 10^18
12
13    if(a == 10 || b > 3 && c < 0 || d * e < f && f < 0 && e > 0) {
14        cout << "I love C++ so much, I want to marry C++." << endl;
15    } else {
16        cout << "C++ is as pleasant as a soggy sandwich." << endl;
17    }
18
19    return 0;
20}

Let’s analyse this particular problem together. A naïve but entirely plausible solution in pseudocode may look as follows:

1. Let the user enter three integers, \(a\), \(b\), and \(c\).
2. If \(b-a=c-b\), the numbers are evenly spaced.
3. Otherwise, the numbers are not evenly spaced.

This works perfectly well when the numbers are in increasing or decreasing order — we’re checking if the differences between consecutive numbers are the same.

However, this algorithm falls apart entirely when the three numbers are out of order: the user inputting 4 2 6 would cause a = 4, b = 2, c = 6 to be read in, and b - a = -2 while c - b = 4.

One immediate way to remedy this is to first put the numbers in order, then perform the above algorithm. In pseudocode,

1. Let the user enter three integers, \(a\), \(b\), and \(c\).
2. By swapping the contents of \(a\), \(b\), and/or \(c\) as necessary, put the three numbers in ascending or descending order.
3. If \(b-a=c-b\), the numbers are evenly spaced.
4. Otherwise, they are not evenly spaced.

Try to implement these steps algorithmically before peeking at the solution!

 1#include <iostream> 
 2
 3using namespace std; 
 4
 5int main() {
 6    // 1. read user input into 3 integers, a, b, and c 
 7    int a, b, c; 
 8    cout << "Please enter 3 integers: "; 
 9    cin >> a >> b >> c; 
10
11    // 2. rearrange the three integers so they're "in order". 
12    // cases 4 and 5, swapping a and b
13    if((b < a && a >= c) || (b > a && a >= c)) { 
14        int temp = b; 
15        b = a; 
16        a = temp; 
17    } 
18
19    // cases 3 and 6, swapping b and c
20    else if((b < c && c <= a) || (b > c && c >= a)) {
21        int temp = b; 
22        b = c; 
23        c = temp; 
24    } 
25
26    // 3. if b - a and c - b are equal, we are evenly spaced 
27    if( b - a == c - b ) {
28        cout << "Evenly spaced" << endl; 
29    } 
30
31    // 4. otherwise, they aren't evenly spaced. 
32    else {
33        cout << "Unevenly spaced" << endl; 
34    } 
35
36    return 0; 
37} 

Sample runs:
INPUT    1 -1 -1         (x^2 - x - 1 = 0)
OUTPUT   Solutions: 1.61803, -0.618034

INPUT    1 2 1           (x^2 + 2x + 1 = 0)
OUTPUT   Solution: -1

 1#include <iostream>
 2#include <cmath>
 3
 4using namespace std;
 5
 6int main() {
 7    double a, b, c, d;
 8    cin >> a >> b >> c;
 9    d = sqrt(b * b - 4 * a * c);
10
11    cout << "Solutions: "
12         << (-1 * b + d) / (2 * a)
13         << ", " << (-1 * b - d) / (2 * a) << endl;
14
15    return 0;
16}