Chapter 3 Exercises

Work through these after reading Chapter 3. Try each one yourself before revealing the solution — you learn far more from an honest attempt than from reading a finished answer. Type the code into CLion and run it; do not just read it.

When you open a solution it appears blurred — click it once more to reveal it, so you do not see the answer by accident.

Each exercise is a small program with its own main(). Now that you have read CMake, you can keep them in one project — one add_executable line per file — and pick which to run from the dropdown next to the green ▶ button.

---

1. Sensor statistics

Practises: C++ Standard Library

Put these readings in a std::vector<int>: 17, 42, 99, 8, 23. Then, using standard-library algorithms rather than hand-written loops, print three things: the readings sorted ascending, their sum, and the largest value.

Hint: <algorithm> has std::sort and std::max_element; <numeric> has std::accumulate. Each takes a .begin(), .end() range.

Show solution

#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>

int main() {
    std::vector<int> readings = {17, 42, 99, 8, 23};

    std::sort(readings.begin(), readings.end());

    int sum = std::accumulate(readings.begin(), readings.end(), 0);
    int largest = *std::max_element(readings.begin(), readings.end());

    std::cout << "Sorted:";
    for (int r : readings) {
        std::cout << " " << r;
    }
    std::cout << "\n";

    std::cout << "Sum: " << sum << "\n";
    std::cout << "Largest: " << largest << "\n";
}

Each algorithm works on the whole container via the .begin(), .end() range. std::max_element returns an iterator to the largest element, so the * in front reads the value it points at. Letting the library sort and sum for you is shorter and harder to get wrong than writing the loops by hand — the chapter's main point.

---

2. Count the colours

Practises: Data Structures

You are given a list of colour names, some repeated — {"red", "green", "red", "blue", "green", "red"}. Count how many times each colour appears, then print each colour with its count. Use the container the chapter recommends for counting by key.

Hint: looking up a missing key in a map creates it with the value 0, so ++counts[colour] does the right thing the first time too.

Show solution

#include <iostream>
#include <string>
#include <vector>
#include <map>

int main() {
    std::vector<std::string> colours = {"red", "green", "red", "blue", "green", "red"};

    std::map<std::string, int> counts;
    for (const std::string& colour : colours) {
        ++counts[colour];
    }

    for (const auto& [colour, count] : counts) {
        std::cout << colour << ": " << count << "\n";
    }
}

++counts[colour] works because the first lookup of a new key inserts it with a value-initialised 0, which the ++ then makes 1. Using std::map prints the colours in alphabetical order; std::unordered_map would count them just as well but in no particular order. The [colour, count] in the loop is a structured binding — it splits each key/value pair into two named pieces, exactly as the chapter showed.

---

3. Count the distinct IDs

Practises: Data Structures

A stream of sensor IDs arrives, with some repeats — {4, 8, 4, 15, 16, 8, 23, 42, 16}. Print how many distinct IDs there were. Use the container designed for the "have I seen this before?" question.

Show solution

#include <iostream>
#include <vector>
#include <unordered_set>

int main() {
    std::vector<int> ids = {4, 8, 4, 15, 16, 8, 23, 42, 16};

    std::unordered_set<int> distinct(ids.begin(), ids.end());

    std::cout << "Distinct IDs: " << distinct.size() << "\n";
}

A set silently discards duplicates, so once every ID has gone in, its size() is the count of distinct values. Building the set straight from the vector's begin(), end() range is the shortest way; inserting in a loop would work too. The chapter's decision table points to a set for exactly this "track which items I have seen" job.