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Print a literal string on standard output
std::cout << "Hello World" << std::endl;
Loop to execute some code a constant number of times
for (int i = 0; i < 10; ++i)
  cout << "Hello\n";
Like a function which doesn't return any value, thus has only side effects (e.g. Print to standard output)
void finish(char* name){
    cout << "My job here is done. Goodbye " << name << "\n";
Create a function which returns the square of an integer
int square(int x){
  return x*x;
Declare a container type for two floating-point numbers x and y
struct Point{
  double x;
  double y;
Do something with each item x of the list (or array) items, regardless indexes.
for(const auto &x : items) {
Print each index i with its value x from an array-like collection items
std::size_t i = 0;
for(const auto & x: items) {
  std::cout << "Item " << i++ << " = " << x << std::endl;
Alternative implementation:
for (std::size_t ix = 0; ix < items.size(); ++ix) {
  std::cout << "Item " << ix << " = " << items[ix] << std::endl;
Alternative implementation:
for (auto const [i, x] : std::views::enumerate(items)) {
  std::cout << i << ": " << x << '\n';
Create a new map object x, and provide some (key, value) pairs as initial content.
std::map<const char*, int> x;
x["one"] = 1;
x["two"] = 2;
Alternative implementation:
std::unordered_map<std::string, double> mymap = {
    {"mom", 5.4},
    {"dad", 6.1},
    {"bro", 5.9}
The structure must be recursive because left child and right child are binary trees too. A node has access to children nodes, but not to its parent.
struct binary_tree
 int data;
 binary_tree *left = nullptr, *right = nullptr;
Generate a random permutation of the elements of list x
std::random_shuffle(x.begin(), x.end());
The list x must be non-empty.
std::mt19937 gen;
std::uniform_int_distribution<size_t> uid (0, x.size () - 1);
x[uid (gen)];
Alternative implementation:
std::ranges::sample(x, &result, 1, std::mt19937{std::random_device{}()});
Check if the list contains the value x.
list is an iterable finite container.
bool Contains(const std::vector<int> &list, int x)
	return std::find(list.begin(), list.end(), x) != list.end();
Alternative implementation:
auto contains(auto list, auto x) -> bool {
  return std::ranges::find(list, x) != std::ranges::end(list);
Access each key k with its value x from an associative array mymap, and print them.
for (const auto& kx: mymap) {
	std::cout << "Key: " << kx.first << " Value: " << kx.second << std::endl;
Alternative implementation:
for (const auto& [k, x]: mymap) {
	std::cout << "Key: " << k << " Value: " << x << '\n';
Alternative implementation:
for (auto entry : mymap) {
  auto k = entry.first;
  auto x = entry.second;
  std::cout << k << ": " << x << "\n";
Pick a random number greater than or equals to a, strictly inferior to b. Precondition : a < b.
float pick(float a, float b)
	std::default_random_engine generator;
	std::uniform_real_distribution distribution(a, b);
	return distribution(generator);
Pick a random integer greater than or equals to a, inferior or equals to b. Precondition : a < b.
std::mt19937 gen;
std::uniform_int_distribution<size_t> uid (a, b);
uid (gen);
The structure must be recursive. A node may have zero or more children. A node has access to its children nodes, but not to its parent.
template<typename T>
struct Node{
  T value;
  std::vector<Node<T>> children;
Call a function f on every node of a tree, in depth-first prefix order
void dfs(const auto &tree, const auto &root)

	for (auto child : tree)
		dfs(tree, child);
Reverse the order of the elements of the list x.
This may reverse "in-place" and destroy the original ordering.
std::reverse(begin(x), end(x));
Implement a function search which looks for item x in a 2D matrix m.
Return indices i, j of the matching cell.
Think of the most idiomatic way in the language to return the two values at the same time.
template<typename T, size_t len_x, size_t len_y>
std::pair<size_t, size_t> search (const T (&m)[len_x][len_y], const T &x) {
    for(size_t pos_x = 0; pos_x < len_x; ++pos_x) {
        for(size_t pos_y = 0; pos_y < len_y; ++pos_y) {
            if(m[pos_x][pos_y] == x) {
                return std::pair<size_t, size_t>(pos_x, pos_y);

    // return an invalid value if not found
    return std::pair<size_t, size_t>(len_x, len_y);
Alternative implementation:
bool _search(const Matrix& _m, const float _x, size_t* const out_i, size_t* const out_j) {
  for (size_t j = 0; j < _m.rows; j++) {
    for (size_t i = 0; i < _m.cols; i++) {
      if (, j) == _x) {
         *out_i = i;
         *out_j = j;
         return true;
  return false;
Swap the values of the variables a and b
swap(a, b);
Extract the integer value i from its string representation s (in radix 10)
int i = std::atoi(s);
Alternative implementation:
int i = std::stoi(s);
Alternative implementation:
std::stringstream ss(str);
int i;
ss >> i;
Alternative implementation:
std::string s("123");
int i;
std::from_chars(, + s.size(), i, 10);
Given a real number x, create its string representation s with 2 decimal digits following the dot.
std::stringstream ss;
ss << std::fixed << std::setprecision(2) << x;
s = ss.str();
Alternative implementation:
std::array<char, 23> buffer;
std::string s{};
if (auto [ptr, ec] = std::to_chars(, + buffer.size(), x, std::chars_format::fixed, 2); ec == std::errc{}) {
  s = std::string(, ptr);
} else {
  s = std::make_error_code(ec).message();
Declare a new string s and initialize it with the literal value "ネコ" (which means "cat" in japanese)
std::string s = u8"ネコ";
Share the string value "Alan" with an existing running process which will then display "Hello, Alan"
auto mutex = std::mutex{};
auto cv = std::condition_variable{};
auto variable = std::optional<std::string>{};
auto future = std::async([&]() {
	auto lock = std::unique_lock{mutex, std::defer_lock};
	cv.wait(lock, [&variable](){ return variable.has_value(); });
	std::cout << "Hello, " << *variable << "!" << std::endl;

//passing value in main thread (or it can be done in others as well)
	auto lock = std::unique_lock{mutex};
	variable = "Alan";
Declare and initialize a matrix x having m rows and n columns, containing real numbers.
std::vector<std::vector<double>> x (m, std::vector<double>(n));
Declare and initialize a 3D array x, having dimensions boundaries m, n, p, and containing real numbers.
std::vector<std::vector<std::vector<double>>> x (m, std::vector<std::vector<double>> (n, std::vector<double> (p)));
Sort the elements of the list (or array-like collection) items in ascending order of x.p, where p is a field of the type Item of the objects in items.
std::sort(begin(items), end(items), [](const auto& a, const auto& b) { return a.p < b.p; });
Alternative implementation:
struct {
    bool operator()(const Item &lhs, const Item &rhs) const { return lhs.p < rhs.p; }
} custom_sort;

std::sort(begin(items), end(items), custom_sort);
Alternative implementation:
std::ranges::sort(items, {}, &Item::p);
Remove i-th item from list items.
This will alter the original list or return a new list, depending on which is more idiomatic.
Note that in most languages, the smallest valid value for i is 0.
items.erase (items.begin () + i);
Launch the concurrent execution of procedure f with parameter i from 1 to 1000.
Tasks are independent and f(i) doesn't return any value.
Tasks need not run all at the same time, so you may use a pool.
auto futures = std::vector<std::future<void>>{};
for (auto i = 1; i <= 1000; ++i)
	futures.emplace_back(std::async(f, i));
Create the recursive function f which returns the factorial of the non-negative integer i, calculated from f(i-1)
unsigned int f( unsigned int i ) {
	if ( i == 0 ) return 1;
	return i * f( i - 1 );
Create function exp which calculates (fast) the value x power n.
x and n are non-negative integers.
int pow(int x, int p)
  if (p == 0) return 1;
  if (p == 1) return x;

  int tmp = pow(x, p/2);
  if (p%2 == 0) return tmp * tmp;
  else return x * tmp * tmp;
Assign to the variable x the new value f(x), making sure that no other thread may modify x between the read and the write.
std::atomic<int> x{};

auto local_x = x.load();
while(!x.compare_exchange_strong(local_x, f(local_x))) {
	local_x = x.load();
Alternative implementation:
auto mutex = std::mutex{};
auto x = someValue();

	auto lock = std::unique_lock{mutex};
	x = f(x);
Declare and initialize a set x containing unique objects of type T.
std::unordered_set<T> x;
Alternative implementation:
std::unordered_set<T, hasher, eq> x;
Implement a function compose (A -> C) with parameters f (A -> B) and g (B -> C), which returns the composition function g ∘ f
std::function<C (A)> compose(B (&f)(A), C (&g)(B))
    return [&f, &g](A a){return g(f(a));};

auto fn = compose(f, g);
Implement a function compose which returns composition function g ∘ f for any functions f and g having exactly 1 parameter.
template<typename Pred1, typename Pred2>
auto compose(Pred1 f, Pred2 g){
  return [f,g](auto x){ return g(f(x)); };	
Transform a function that takes multiple arguments into a function for which some of the arguments are preset.
// function taking many parameters
int add(int a, int b)
    return a + b;

// define a new function preseting the first parameter
std::function<int (int)> add_def(int a)
    return [a](int b){return add(a, b);};

int result = add_def(4)(6);
Alternative implementation:
auto add(int a, int b) -> int {
	return a + b;

//curry with std::bind
using namespace std::placeholders;
auto add5 = std::bind(add, _1, 5);

//curry with lambda
auto add5 = [](int x) { return add(x, 5); };

auto result = add5(1);
assert(result == 6);
Find substring t consisting in characters i (included) to j (excluded) of string s.
Character indices start at 0 unless specified otherwise.
Make sure that multibyte characters are properly handled.
auto t = s.substr(i, j-i);
Set the boolean ok to true if the string word is contained in string s as a substring, or to false otherwise.
bool ok = s.find(word) != std::string::npos;
Print each item v of list a which is not contained in list b.
For this, write an outer loop to iterate on a and an inner loop to iterate on b.
auto a = {1,2,3,4,5};
auto b = {3,5};

for (auto va: a){
    for (auto vb: b){
        if (va==vb) goto OUTER;
    std::cout << va << '\n';
    OUTER: continue;
Look for a negative value v in 2D integer matrix m. Print it and stop searching.
auto indices = findNegativeValue (m, 10, 20);
std::cout << m[indices.first][indices.second] << '\n';

std::pair<int, int> findNegativeValue (int **m, int rows, int columns) {
  for (int i = 0; i < rows; ++i) {
    for (int j = 0; j < columns; ++j) {
      if (m[i][j] < 0) return make_pair (i, j);
  throw "No negative value!";
Insert the element x at position i in the list s. Further elements must be shifted to the right.
s.insert (s.begin () + i, x);
Sleep for 5 seconds in current thread, before proceeding with the next instructions.
Create the string t consisting of the 5 first characters of the string s.
Make sure that multibyte characters are properly handled.
auto t = s.substr(0, 5);
Create string t consisting in the 5 last characters of string s.
Make sure that multibyte characters are properly handled.
std::string t = s.substr(s.length() - 5);
Assign to variable s a string literal consisting in several lines of text, including newlines.
std::string s = R"( Earth is a planet.
So is the Jupiter)";
Build list chunks consisting in substrings of the string s, separated by one or more space characters.
std::istringstream x(s);
std::list<std::string> chunks;
std::copy(std::istream_iterator<std::string>(x), std::istream_iterator<std::string>(), std::back_inserter(chunks));
Write a loop that has no end clause.
for (;;) {
	/// Do something
Alternative implementation:
while (true) {
	// Do something
Determine whether the map m contains an entry for the key k
bool key_exists = m.find(k) != m.end();
Alternative implementation:
bool key_exists = m.count(k) != 0;
Alternative implementation:
bool key_exists = m.contains(k);
Determine whether the map m contains an entry with the value v, for some key.
std::find_if(m.begin(), m.end(), [v](const auto& mo) {return mo.second == v; }) != m.end();
Concatenate elements of string list x joined by the separator ", " to create a single string y.
std::vector<std::string> x;
std::string y;
const char* const delim = ", ";

switch (x.size())
	case 0: y = "";   break;
	case 1: y = x[0]; break;
		std::ostringstream os;
		std::copy(x.begin(), x.end() - 1,
			std::ostream_iterator<std::string>(os, delim));
		os << *x.rbegin();
		y = os.str();
Alternative implementation:
y = x | views::join_with(',') | to<std::string>();
Calculate the sum s of the integer list or array x.
int s = std::accumulate(x.begin(), x.end(), 0, std::plus<int>());
Alternative implementation:
auto s = std::ranges::fold_left(x, 0, std::plus<int>{});
Create the string representation s (in radix 10) of the integer value i.
auto s = std::to_string(i);
Fork-join : launch the concurrent execution of procedure f with parameter i from 1 to 1000.
Tasks are independent and f(i) doesn't return any value.
Tasks need not run all at the same time, so you may use a pool.
Wait for the completion of the 1000 tasks and then print "Finished".
size_t taskCount = 1000;
auto remainingTasks = std::latch(static_cast<std::ptrdiff_t>(taskCount));
auto f = [&remainingTasks](int i) {
  // do work

std::vector<std::jthread> threads{ 10 };
for (auto i = 0; i < taskCount; i++) {
  auto& workerThread = threads[i % 10];
  if (workerThread.joinable()) {
  workerThread = std::jthread(f, i);

std::cout << "Finished";
Create the list y containing the items from the list x that satisfy the predicate p. Respect the original ordering. Don't modify x in-place.
std::copy_if (x.begin (), x.end (), std::back_inserter(y), p);
Alternative implementation:
auto y = x | std::views::filter(p);
Create the string lines from the content of the file with filename f.
std::string fromFile(std::string _f)
    std::ifstream t(_f);
    t.seekg(0, std::ios::end);
    size_t size = t.tellg();
    std::string buffer(size, ' ');
    t.seekg(0);[0], size); 
Print the message "x is negative" to standard error (stderr), with integer x value substitution (e.g. "-2 is negative").
std::cerr << x << " is negative\n";
Alternative implementation:
int main(){
	int x = -2;
	std::cerr << x <<" is negative\n";
Assign to x the string value of the first command line parameter, after the program name.
int main(int argc, char *argv[])
 vector<string> args(1 + argv, argc + argv);

 string x =;
Assign to the variable d the current date/time value, in the most standard type.
std::time_t d = std::time(nullptr);
Set i to the first position of string y inside string x, if exists.

Specify if i should be regarded as a character index or as a byte index.

Explain the behavior when y is not contained in x.
int i = x.find(y);
Assign to x2 the value of string x with all occurrences of y replaced by z.
Assume occurrences of y are not overlapping.
std::stringstream s;
std::regex_replace(std::ostreambuf_iterator<char>(s), x.begin(), x.end(), y, z);
auto x2 = s.str()
Alternative implementation:
constexpr std::string ReplaceAllCopy(const std::string& x, const std::string& y, const std::string& z) {
	auto x2 = x;
	size_t i = 0;

	while (true) {
		i = x2.find(y, i);
		if (i == std::string::npos) break;
		x2.replace(i, y.length(), z);
	return x2;
Create an object x to store n bits (n being potentially large).
std::vector<bool> x(n, 0);
Alternative implementation:
std::bitset<n> x;
Use seed s to initialize a random generator.

If s is constant, the generator output will be the same each time the program runs. If s is based on the current value of the system clock, the generator output will be different each time.
Basic implementation of the Echo program: Print all arguments except the program name, separated by space, followed by newline.
The idiom demonstrates how to skip the first argument if necessary, concatenate arguments as strings, append newline and print it to stdout.
int main(int argc, char * argv[])
	for(int i = 1; i < argc; ++i)
		std::cout <<(1==i ? "" : " ") << argv[i];
        std::cout <<std::endl;
Create a factory named fact for any sub class of Parent and taking exactly one string str as constructor parameter.
template <typename T,
            std::is_base_of_v<Parent, T>, bool> = true>
T fact(const std::string& str) {
  return T(str);
Compute the greatest common divisor x of big integers a and b. Use an integer type able to handle huge numbers.
unsigned long long int GCD(unsigned long long int a, unsigned long long int b)
    unsigned long long int c=a%b;
        return b;
    return GCD(b, c);
Alternative implementation:
auto x = std::gcd(a, b);
Compute the least common multiple x of big integers a and b. Use an integer type able to handle huge numbers.
auto x = std::lcm(a, b);
Create the string s of integer x written in base 2.

E.g. 13 -> "1101"
std::bitset<sizeof(x)*8> y(x);
auto s = y.to_string();
Alternative implementation:
std::string ToBinary(int x) {
  std::array<char, 64> buf;
  auto[ptr, ec] = std::to_chars(, + buf.size(), x, 2);
  auto s = std::string(, ptr);
  return s;
Declare a complex x and initialize it with value (3i - 2). Then multiply it by i.
using namespace std::complex_literals;
auto x = 3i - 2.;
x *= 1i;
Execute a block once, then execute it again as long as boolean condition c is true.
do {
} while(c);
Declare the floating point number y and initialize it with the value of the integer x .
float y = static_cast<float>(x);
Declare integer y and initialize it with the value of floating point number x . Ignore non-integer digits of x .
Make sure to truncate towards zero: a negative x must yield the closest greater integer (not lesser).
int y = static_cast<int>(x);
Declare the integer y and initialize it with the rounded value of the floating point number x .
Ties (when the fractional part of x is exactly .5) must be rounded up (to positive infinity).
int y = static_cast<int>(std::floor(x + 0.5f));
Declare regular expression r matching strings "http", "htttp", "httttp", etc.
 auto r = regex("htt+p");
Count number c of 1s in the integer i in base 2.

E.g. i=6 → c=2
uint32_t c = 0;
for (; i; i &= i - 1, ++c);
Write boolean function addingWillOverflow which takes two integers x, y and return true if (x+y) overflows.

An overflow may be above the max positive value, or below the min negative value.
bool addingWillOverflow(int x, int y) {
  return ((x > 0) && (y > std::numeric_limits<int>::max() - x)) ||
         ((x < 0) && (y < std::numeric_limits<int>::min() - x));
Exit immediately.
If some extra cleanup work is executed by the program runtime (not by the OS itself), describe it.
Create a new bytes buffer buf of size 1,000,000.
vector<byte> _buf(1'000'000);
You've detected that the integer value of argument x passed to the current function is invalid. Write the idiomatic way to abort the function execution and signal the problem.
throw domain_error("oops!");
Expose a read-only integer x to the outside world while being writable inside a structure or a class Foo.
class Foo final {
  int mX = 0;

  const auto& X() const { return mX; }
Implement the procedure control which receives one parameter f, and runs f.
void control(invocable auto&& f)
Print the name of the type of x. Explain if it is a static type or dynamic type.

This may not make sense in all languages.
Assign to variable x the length (number of bytes) of the local file at path.
std::ifstream f(path, std::ios::in | std::ios::binary);
f.seekg(0, std::ios::end);
size_t x = f.tellg();
Set the boolean b to true if string s starts with prefix prefix, false otherwise.
bool b = s.starts_with(prefix);
Alternative implementation:
std::string prefix = "something";
bool b =, prefix.size(), prefix) == 0;
Set boolean b to true if string s ends with string suffix, false otherwise.
return suffix.size() == s.size() - s.rfind(suffix);
Alternative implementation:
bool b = s.ends_with(suffix);
Convert a timestamp ts (number of seconds in epoch-time) to a date with time d. E.g. 0 -> 1970-01-01 00:00:00
std::time_t d = std::chrono::system_clock::to_time_t(ts);
Assign to the string x the value of the fields (year, month, day) of the date d, in format YYYY-MM-DD.
int main()
	char x[32]{};
	time_t a = time(nullptr);
	struct tm d;
	if (localtime_s(&d, &a) == 0) {
		strftime(x, sizeof(x), "%F", &d);
		std::cout << x << std::endl;

	return 0;

Sort elements of array-like collection items, using a comparator c.
struct is_less {
    bool operator () (int a, int b){
        return a < b;

int main(){
    std::vector<int> items = {1337, 666, -666, 0, 0, 666, -666};
    std::sort(items.begin(), items.end(), is_less());

    std::vector<int> expected = {-666, -666, 0, 0, 666, 666, 1337};
    assert(items.size() == expected.size());
    for (size_t i = 0; i < items.size(); i++){
        assert(items[i] == expected[i]);
    return 0;
Assign to the string s the name of the currently executing program (but not its full path).

int main(int argc, char* argv[])
    std::cout << std::filesystem::path(argv[0]).filename() << '\n';
Assign to string dir the path of the working directory.
(This is not necessarily the folder containing the executable itself)
dir = std::filesystem::current_path();
Set n to the number of bytes of a variable t (of type T).
std::size_t n = sizeof(t);
Set the boolean blank to true if the string s is empty, or null, or contains only whitespace ; false otherwise.
bool blank = false;
if (s.empty() || std::all_of(s.begin(), s.end(), [](char c){return std::isspace(c);})) {
      blank = true;
Print each key k with its value x from an associative array mymap, in ascending order of k.
std::map< K, V > _mymap;
for (const auto& pair : _mymap) {
    std::cout << pair.first << ": " << pair.second << "\n";
Alternative implementation:
auto print(auto&... args) {
  // c++17 fold expression
  (std::cout << ... << args) << std::endl;

auto print_map_contents(auto mymap) {
  // c++17 structured binding
  for (auto [k, x] : mymap) {
    print("mymap[", k, "] = ", x);

auto main() -> int {
  // map is ordered map, iteration is ascending by key value
  auto map = std::map<std::string, int> {
    {"first entry", 12},
    {"second entry", 42},
    {"third entry", 3},

Print each key k with its value x from an associative array mymap, in ascending order of x.
Multiple entries may exist for the same value x.
for_each(begin(mymap), end(mymap),
    [&s](const auto& kv) { s.insert(kv.second); });
Set n to the number of elements of the list x.
size_t n = x.size();
Create the set y from the list x.
x may contain duplicates. y is unordered and has no repeated values.
std::unordered_set<T> y (x.begin (), x.end ());
Alternative implementation:
std::set<T> y (x.begin (), x.end ());
Remove duplicates from the list x.
Explain if the original order is preserved.
std::sort(x.begin(), x.end());
auto last = std::unique(x.begin(), x.end());
x.erase(last, x.end());
Alternative implementation:
std::vector<std::string> x = {"one", "two", "two", "one", "three"};
std::unordered_set<std::string> t;
for (auto e : x)
Read an integer value from the standard input into the variable n
std::cin >> n;
Create an enumerated type Suit with 4 possible values SPADES, HEARTS, DIAMONDS, CLUBS.
enum class Suit {
Alternative implementation:
enum class Color : char{
	Red, Black, Green
Verify that predicate isConsistent returns true, otherwise report assertion violation.
Explain if the assertion is executed even in production environment or not.
Write the function binarySearch which returns the index of an element having the value x in the sorted array a, or -1 if no such element exists.
template<typename T>
int binarySearch(const std::vector<T> &a, const T &x)
    if(a.size() == 0) return -1;

    size_t lower = 0;
    size_t upper = a.size() - 1;

    while(lower <= upper)
        auto mid = lower + (upper-lower) / 2;

        if(x == a[mid])
            return (int)mid;
        else if(x > a[mid])
            lower = mid + 1;
            upper = mid - 1;

    return -1;
measure the duration t, in nanoseconds, of a call to the function foo. Print this duration.
auto start = std::chrono::steady_clock::now();
auto end = std::chrono::steady_clock::now();
auto t = std::chrono::duration_cast<std::chrono::nanoseconds>(end-start).count();
Write a function foo that returns a string and a boolean value.
std::tuple<std::string, bool> foo() {
  return std::make_tuple("someString", true);
Alternative implementation:
auto foo()
    return std::make_tuple("Hello", true);
Import the source code for the function foo body from a file "foobody.txt".
void _foo() {
#include "foobody.txt"
Call a function f on every node of a tree, in breadth-first prefix order
void bfs(Node& root, std::function<void(Node*)> f) {
  std::deque<Node*> node_queue;
  while (!node_queue.empty()) {
    Node* const node = node_queue.front();
    for (Node* const child : node->children) {
Call th function f on every vertex accessible from the vertex v, in depth-first prefix order
void dfs(Node& root, std::function<void(Node*)> f) {
  std::stack<Node*> queue;
  std::unordered_set<const Node*> visited;
  while (!queue.empty()) {
    Node* const node =;
    for (Node* const neighbor : node->neighbors) {
      if (visited.find(neighbor) == visited.end()) {
Execute f1 if condition c1 is true, or else f2 if condition c2 is true, or else f3 if condition c3 is true.
Don't evaluate a condition when a previous condition was true.
if (c1)
else if (c2)
else if (c3)
Alternative implementation:
if (c1)
else if (c2)
else if (c3)
Alternative implementation:
(c1 && (f1(), true))
 || (c2 && (f2(), true))
 || (c3 && (f3(), true));
Run the procedure f, and return the duration of the execution of f.
std::clock_t start = std::clock();
double duration = (std::clock() - start) / (double) CLOCKS_PER_SEC;
Alternative implementation:
#include <chrono>

double measure() {
    using namespace std::chrono;
    const auto start{ high_resolution_clock::now() };
    const auto elapsed{ high_resolution_clock::now() - start };
    const double seconds{ duration_cast<duration<double>>(elapsed).count() };
    return seconds;
Set boolean ok to true if string word is contained in string s as a substring, even if the case doesn't match, or to false otherwise.
auto ok = std::search(std::begin(s), std::end(s), std::begin(word), std::end(word),
    [](auto c, auto d){
        return std::tolower(c) == std::tolower(d);
) != std::end(s);
Declare and initialize a new list items, containing 3 elements a, b, c.
std::vector<T> items = {a, b, c};
Alternative implementation:
list items { a, b, c };
Remove at most 1 item from list items, having the value x.
This will alter the original list or return a new list, depending on which is more idiomatic.
If there are several occurrences of x in items, remove only one of them. If x is absent, keep items unchanged.
erase(items, x);
Remove all occurrences of the value x from list items.
This will alter the original list or return a new list, depending on which is more idiomatic.
Set the boolean b to true if the string s contains only characters in the range '0'..'9', false otherwise.
bool b = false;
if (! s.empty() && std::all_of(s.begin(), s.end(), [](char c){return std::isdigit(c);})) {
    b = true;
Delete from map m the entry having key k.

Explain what happens if k is not an existing key in m.
Iterate in sequence over the elements of the list items1 then items2. For each iteration print the element.
void print_seq(const auto& ...xs)
 (std::for_each(std::begin(xs), std::end(xs), 
           [](const auto& x) { std::cout << x; }), ...);

 std::list xs { "hi", "there", "world" }, ys { "lo" "thar" };

 print_seq(xs, ys);
Assign to string s the hexadecimal representation (base 16) of integer x.

E.g. 999 -> "3e7"
template <typename I>
std::string n2hexstr(I w, size_t hex_len = sizeof(I)<<1) {
    static const char* digits = "0123456789ABCDEF";
	std::string str(hex_len, '-');
    for (size_t i=0, j=(hex_len-1)*4 ; i<hex_len; ++i,j-=4)
       str[i] = digits[(w>>j) & 0x0f];
	return str;
Alternative implementation:
std::ostringstream stream;
stream << std::hex << x;
s = stream.str();
Set boolean b to true if file at path fp exists on filesystem; false otherwise.

Beware that you should never do this and then in the next instruction assume the result is still valid, this is a race condition on any multitasking OS.
bool b = std::filesystem::exists(fp);
Extract floating point value f from its string representation s
float f = std::stof(s);
Create string t from string s, keeping only ASCII characters
 copy_if(begin(src), end(src), back_inserter(dest),
         [](const auto c) { return static_cast<unsigned char>(c) <= 0x7F; });
Read a list of integer numbers from the standard input, until EOF.
vector<int> v;
for(int t;;){
	cin >> t;
Remove the last character from the string p, if this character is a forward slash /
 if('/' == s.back())
Create string s containing only the character c.
string s { 'c' };
Create the string t as the concatenation of the string s and the integer i.
auto t = s + std::to_string (i);
Delete from filesystem the file having path filepath.
auto r = std::filesystem::remove(filepath);
Initialize a constant planet with string value "Earth".
// using char*
constexpr char planet[] = "Earth";

// using string
const std::string planet{"Earth"};
Execute f32() if platform is 32-bit, or f64() if platform is 64-bit.
This can be either a compile-time condition (depending on target) or a runtime detection.
#ifdef __x86_64
#ifdef _M_AMD64
Alternative implementation:
if constexpr(sizeof(nullptr) == 8) {
} else {
Multiply all the elements of the list elements by a constant c
for (auto &it : elements)
	it *= c;
Alternative implementation:
std::transform(std::begin(elements), std::end(elements), std::begin(elements), [c](auto i){
        return i*c;
Assign to the variable x the last element of the list items.
std::vector<int> items;
int last = items.back();
Alternative implementation:
auto x = *std::crbegin(items);
Create the list ab containing all the elements of the list a, followed by all the elements of the list b.
auto ab = a;
ab.insert (ab.end (), b.begin (), b.end ());
Assign to the integer n the number of characters of the string s.
Make sure that multibyte characters are properly handled.
n can be different from the number of bytes of s.
auto utf8len(std::string_view const& s) -> size_t
	std::setlocale(LC_ALL, "en_US.utf8");

	auto n = size_t{};
	auto size = size_t{};
	auto mb = std::mbstate_t{};

	while(size < s.length())
		size += mbrlen( + size, s.length() - size, &mb);
		n += 1;

	return n;
Set n to the number of elements stored in mymap.

This is not always equal to the map capacity.
auto n = mymap.size();
Append the element x to the list s.
Alternative implementation:
Insert value v for key k in map m.
m[k] = v;
Return the center c of the rectangle with coördinates(x1,y1,x2,y2)
template <typename T> struct Point {
  T x = {};
  T y = {};
template <typename T> struct Rectangle {
  Point<T> upper_left;
  Point<T> lower_right;
template <typename T>
Point<T> get_center(const Rectangle<T>& rect) {
  return {
    .x = (rect.upper_left.x + rect.lower_right.x) / 2,
    .y = (rect.upper_left.y + rect.lower_right.y) / 2,
Create the list x containing the contents of the directory d.

x may contain files and subfolders.
No recursive subfolder listing.
auto directory_contents(auto path) {
  auto iterator = std::filesystem::directory_iterator(path);
  return std::vector<std::filesystem::path>(

auto main(int argc, char** argv) -> int {
  auto path = argc >= 2
    ? std::filesystem::path(argv[1])
    : std::filesystem::current_path();

  for (auto entry : directory_contents(path)) {
    std::cout << entry.string() << std::endl;
Produce a new list y containing the result of the function T applied to all elements e of the list x that match the predicate P.
template <typename SeqT>
auto transform_copy_if(const SeqT i, auto p, auto f)
 using namespace std;

 SeqT o;

 for_each(begin(i), end(i),
          [&](const auto& x) {

 return o;
Alternative implementation:
constexpr auto transform_matches(const auto& x, auto p, auto t) {
    auto result = x 
	| std::views::filter(p)
        | std::views::transform(t);

    std::vector<std::ranges::range_value_t<decltype(result)>> y;
    std::copy(result.begin(), result.end(), std::back_inserter(y));
    return y;
Declare an external C function with the prototype

void foo(double *a, int n);

and call it, passing an array (or a list) of size 10 to a and 10 to n.

Use only standard features of your language.
extern "C" void foo(double *a, int n);
Given a one-dimensional array a, check if any value is larger than x, and execute the procedure f if that is the case
if (std::any_of(a.begin(), a.end(), [x](auto y) { return y > x; }))
Retrieve the contents of file at path into a list of strings lines, in which each element is a line of the file.
std::ifstream file (path);
for (std::string line; std::getline(file, line); lines.push_back(line)) {}
Compute the hypotenuse h of the triangle where the sides adjacent to the square angle have lengths x and y.
auto h = std::hypot(x, y);
Read an environment variable with the name "FOO" and assign it to the string variable foo. If it does not exist or if the system does not support environment variables, assign a value of "none".
const char* tmp = std::getenv("FOO");
std::string foo = tmp ? std::string(tmp) : "none";
Declare a type t which contains a string s and an integer array n with variable size, and allocate a variable v of type t. Allocate v.s and v.n and set them to the values "Hello, world!" for s and [1,4,9,16,25], respectively. Deallocate v, automatically deallocating v.s and v.n (no memory leaks).
#include <string>
using namespace std::string_literals;
#include <vector>
#include <memory>

struct t {
    std::string s;
    std::vector<int> n;

auto v = std::make_unique<t>(
    "Hello, world!"s,
    decltype(t::n){1, 4, 9, 16, 25}

// Automatically:
void fn(){
    auto v = std::make_unique<t>(
        "Hello, world!"s,
        decltype(t::n){1, 4, 9, 16, 25}
Create the folder at path on the filesystem
namespace fs = std::filesystem;
Set the boolean b to true if path exists on the filesystem and is a directory; false otherwise.
auto b = std::filesystem::is_directory(path);
Create t consisting of 3 values having different types.

Explain if the elements of t are strongly typed or not.
std::tuple<float, std::string, bool> t(2.5f, "foo", false);
Alternative implementation:
auto t = std::make_tuple(2.5f, std::string("foo"), false);
Set i to the first index in list items at which the element x can be found, or -1 if items does not contain x.
auto it = std::find(items.begin(), items.end(), x);
i = it == items.end() ? std::distance(items.begin(), it) : -1;
Loop through list items checking a condition. Do something else if no matches are found.

A typical use case is looping through a series of containers looking for one that matches a condition. If found, an item is inserted; otherwise, a new container is created.

These are mostly used as an inner nested loop, and in a location where refactoring inner logic into a separate function reduces clarity.
bool found = false;
for (const auto item : items) {
  if (item == "baz") {
    found = true;
if (!found) {
Alternative implementation:
if (std::any_of(items.begin(), items.end(), condition))
Insert the element x at the beginning of the list items.
Alternative implementation:
Declare an optional integer argument x to procedure f, printing out "Present" and its value if it is present, "Not present" otherwise
void f(std::optional<int> x = {}) {
  std::cout << (x ? "Present" + std::to_string(x.value()) : "Not Present");
Alternative implementation:
void f(std::optional<int> x = {}) {
  if (x) {
    std::cout << "Present" << x.value();
  } else {
    std::cout << "Not present";
Remove the last element from the list items.
Assign to c the result of (a xor b)
int c = a ^ b;
Assign to string x the first word of string s consisting of exactly 3 digits, or the empty string if no such match exists.

A word containing more digits, or 3 digits as a substring fragment, must not match.
std::regex re("\\b\\d{3}\\b");
auto it = std::sregex_iterator(s.begin(), s.end(), re);
std::string x = it == std::sregex_iterator() ? "" : it->str();
Call a function f on each element e of a set x.
std::for_each(x.begin(), x.end(), f);
Alternative implementation:
for (auto e : x)
Print the contents of the list or array a on the standard output.
std::copy(a.begin(), a.end(), std::ostream_iterator<int>(std::cout, "\n"));
Print the contents of the map m to the standard output: keys and values.
for (auto entry : m) {
	std::cout << entry.first << ": " << entry.second << "\n";
Print the value of object x having custom type T, for log or debug.
std::cout << x;
Remove all the elements from list x that don't satisfy the predicate p, without allocating a new list.
Keep all the elements that do satisfy p.

For languages that don't have mutable lists, refer to idiom #57 instead.
std::list<Foo> x;
Alternative implementation:
std::erase_if(x, std::not_fn(p));
Extract integer value i from its binary string representation s (in radix 2)
E.g. "1101" -> 13
int i = std::stoi(s, nullptr, 2);
Assign to the variable x the string value "a" if calling the function condition returns true, or the value "b" otherwise.
x = condition() ? "a" : "b";
Replace all exact occurrences of "foo" with "bar" in the string list x
std::replace(x.begin(), x.end(), "foo", "bar");
Print the numbers 5, 4, ..., 0 (included), one line per number.
for (int i = 5; i >= 0; --i) {
	std::cout << i;
Declare a new list items of string elements, containing zero elements
std::vector<std::string> items;
If a variable x passed to procedure tst is of type foo, print "Same type." If it is of a type that extends foo, print "Extends type." If it is neither, print "Not related."
template<typename T>
void tst(T)
    using namespace std;

    // Strip away any references and pointer
    typedef remove_const<remove_pointer<decay<T>::type>::type>::type D;

    if(is_same<D, foo>::value)
        cout << "same type" << endl;
    else if(is_base_of<foo, D>::value)
        cout << "extends type" << endl;
        cout << "not related" << endl;
Fizz buzz is a children's counting game, and a trivial programming task used to affirm that a programmer knows the basics of a language: loops, conditions and I/O.

The typical fizz buzz game is to count from 1 to 100, saying each number in turn. When the number is divisible by 3, instead say "Fizz". When the number is divisible by 5, instead say "Buzz". When the number is divisible by both 3 and 5, say "FizzBuzz"
for(int i = 1; i <= 100; i++)
    if((i % 15) == 0) std::cout << "FizzBuzz" << std::endl;
    else if((i % 5) == 0) std::cout << "Buzz" << std::endl;
    else if((i % 3) == 0) std::cout << "Fizz" << std::endl;
    else std::cout << i << std::endl;
Alternative implementation:
for (int n=1; n<=100; n++) {
        string out="";
        if (n%3==0) {
        if (n%5==0) {
        if (out=="") {
        cout << out << "\n";
From the string s consisting of 8n binary digit characters ('0' or '1'), build the equivalent array a of n bytes.
Each chunk of 8 binary digits (2 possible values per digit) is decoded into one byte (256 possible values).
const size_t n = s.length() / 8;

vector<uint8_t> a(n);

for(size_t block = 0; block < n; block++)
    uint8_t acc = 0;
    const size_t start = block * 8;
    for(size_t offset = start; offset < start + 8; offset++)
        acc = (acc << 1) + (s[offset] - '0');

    a[block] = acc;
Remove the element e from the set x.

Explains what happens if e was already absent from x.
Read one line into the string line.

Explain what happens if EOF is reached.
int main() {
    for (std::string line; std::getline(std::cin, line);) {
        std::cout << line << std::endl;
    return 0;
Remove all the elements from the map m that don't satisfy the predicate p.
Keep all the elements that do satisfy p.

Explain if the filtering happens in-place, i.e. if m is reused or if a new map is created.
for (auto it = m.begin(); it != m.end();) {
  if (!p(it->second)) {
    it = m.erase(it);
  } else {
Create a new list a (or array, or slice) of size n, where all elements are integers initialized with the value 0.
std::vector<int> a(n, 0);
Set the boolean b to true if the set x contains the element e, false otherwise.
b = x.find(e) != x.end();
Create the string s by concatenating the strings a and b.
std::string a{"Hello"};
std::string b{"world"};
auto s = a + b;
Write a line of comments.

This line will not be compiled or executed.
// This is a comment.
Given the integer x = 8, assign to the string s the value "Our sun has 8 planets", where the number 8 was evaluated from x.
s = std::format("Our sun has {} planets", x);
Set b to true if the maps m and n have the same key/value entries, false otherwise.
b = m == n;
Set all the elements in the array x to the same value v
Set b to true if the string s is empty, false otherwise
b = s.empty();
std::exchange(x, v);