20.7 — Lambda captures

Capture clauses and capture by value

In the previous lesson (20.6 -- Introduction to lambdas (anonymous functions)), we introduced this example:

#include <algorithm>
#include <array>
#include <iostream>
#include <string_view>

int main()
{
  std::array<std::string_view, 4> arr{ "apple", "banana", "walnut", "lemon" };

  auto found{ std::find_if(arr.begin(), arr.end(),
                           [](std::string_view str)
                           {
                             return str.find("nut") != std::string_view::npos;
                           }) };

  if (found == arr.end())
  {
    std::cout << "No nuts\n";
  }
  else
  {
    std::cout << "Found " << *found << '\n';
  }

  return 0;
}

Now, let’s modify the nut example and let the user pick a substring to search for. This isn’t as intuitive as you might expect.

#include <algorithm>
#include <array>
#include <iostream>
#include <string_view>
#include <string>

int main()
{
  std::array<std::string_view, 4> arr{ "apple", "banana", "walnut", "lemon" };

  // Ask the user what to search for.
  std::cout << "search for: ";

  std::string search{};
  std::cin >> search;

  auto found{ std::find_if(arr.begin(), arr.end(), [](std::string_view str) {
    // Search for @search rather than "nut".
    return str.find(search) != std::string_view::npos; // Error: search not accessible in this scope
  }) };

  if (found == arr.end())
  {
    std::cout << "Not found\n";
  }
  else
  {
    std::cout << "Found " << *found << '\n';
  }

  return 0;
}

This code won’t compile. Unlike nested blocks, where any identifier defined in an outer block is accessible in the scope of the nested block, lambdas can only access specific kinds of identifiers: global identifiers, entities that are known at compile time, and entities with static storage duration. search fulfills none of these requirements, so the lambda can’t see it. That’s what the capture clause is there for.

The capture clause

The capture clause is used to (indirectly) give a lambda access to variables available in the surrounding scope that it normally would not have access to. All we need to do is list the entities we want to access from within the lambda as part of the capture clause. In this case, we want to give our lambda access to the value of variable search, so we add it to the capture clause:

#include <algorithm>
#include <array>
#include <iostream>
#include <string_view>
#include <string>

int main()
{
  std::array<std::string_view, 4> arr{ "apple", "banana", "walnut", "lemon" };

  std::cout << "search for: ";

  std::string search{};
  std::cin >> search;

  // Capture @search                                vvvvvv
  auto found{ std::find_if(arr.begin(), arr.end(), [search](std::string_view str) {
    return str.find(search) != std::string_view::npos;
  }) };

  if (found == arr.end())
  {
    std::cout << "Not found\n";
  }
  else
  {
    std::cout << "Found " << *found << '\n';
  }

  return 0;
}

The user can now search for an element of our array.

Output

search for: nana
Found banana

So how do captures actually work?

While it might look like our lambda in the example above is directly accessing the value of main‘s search variable, this is not the case. Lambdas might look like nested blocks, but they work slightly differently (and the distinction is important).

When a lambda definition is executed, for each variable that the lambda captures, a clone of that variable is made (with an identical name) inside the lambda. These cloned variables are initialized from the outer scope variables of the same name at this point.

Thus, in the above example, when the lambda object is created, the lambda gets its own cloned variable named search. This cloned search has the same value as main‘s search, so it behaves like we’re accessing main‘s search, but we’re not.

While these cloned variables have the same name, they don’t necessarily have the same type as the original variable. We’ll explore this in the upcoming sections of this lesson.

Captures are treated as const by default

When a lambda is called, operator() is invoked. By default, this operator() treats captures as const, meaning the lambda is not allowed to modify those captures.

In the following example, we capture the variable ammo and try to decrement it.

#include <iostream>

int main()
{
  int ammo{ 10 };

  // Define a lambda and store it in a variable called "shoot".
  auto shoot{
    [ammo]() {
      // Illegal, ammo cannot be modified.
      --ammo;

      std::cout << "Pew! " << ammo << " shot(s) left.\n";
    }
  };

  // Call the lambda
  shoot();

  std::cout << ammo << " shot(s) left\n";

  return 0;
}

The above won’t compile, because ammo is treated as const within the lambda.

Mutable captures

To allow modifications of variables that were captured, we can mark the lambda as mutable:

#include <iostream>

int main()
{
  int ammo{ 10 };

  auto shoot{
    [ammo]() mutable { // now mutable
      // We're allowed to modify ammo now
      --ammo;

      std::cout << "Pew! " << ammo << " shot(s) left.\n";
    }
  };

  shoot();
  shoot();

  std::cout << ammo << " shot(s) left\n";

  return 0;
}

Output:

Pew! 9 shot(s) left.
Pew! 8 shot(s) left.
10 shot(s) left

While this now compiles, there’s still a logic error. What happened? When the lambda was called, the lambda captured a copy of ammo. When the lambda decremented ammo from 10 to 9 to 8, it decremented its own copy, not the original ammo value in main().

Note that the value of ammo is preserved across calls to the lambda!

Capture by reference

Much like functions can change the value of arguments passed by reference, we can also capture variables by reference to allow our lambda to affect the value of the argument.

To capture a variable by reference, we prepend an ampersand (&) to the variable name in the capture. Unlike variables that are captured by value, variables that are captured by reference are non-const, unless the variable they’re capturing is const. Capture by reference should be preferred over capture by value whenever you would normally prefer passing an argument to a function by reference (e.g. for non-fundamental types).

Here’s the above code with ammo captured by reference:

#include <iostream>

int main()
{
  int ammo{ 10 };

  auto shoot{
    // We don't need mutable anymore
    [&ammo]() { // &ammo means ammo is captured by reference
      // Changes to ammo will affect main's ammo
      --ammo;

      std::cout << "Pew! " << ammo << " shot(s) left.\n";
    }
  };

  shoot();

  std::cout << ammo << " shot(s) left\n";

  return 0;
}

This produces the expected answer:

Pew! 9 shot(s) left.
9 shot(s) left

Now, let’s use a reference capture to count how many comparisons std::sort makes when it sorts an array.

#include <algorithm>
#include <array>
#include <iostream>
#include <string_view>

struct Car
{
  std::string_view make{};
  std::string_view model{};
};

int main()
{
  std::array<Car, 3> cars{ { { "Volkswagen", "Golf" },
                             { "Toyota", "Corolla" },
                             { "Honda", "Civic" } } };

  int comparisons{ 0 };

  std::sort(cars.begin(), cars.end(),
    // Capture @comparisons by reference.
    [&comparisons](const auto& a, const auto& b) {
      // We captured comparisons by reference. We can modify it without "mutable".
      ++comparisons;

      // Sort the cars by their make.
      return a.make < b.make;
  });

  std::cout << "Comparisons: " << comparisons << '\n';

  for (const auto& car : cars)
  {
    std::cout << car.make << ' ' << car.model << '\n';
  }

  return 0;
}

Possible output

Comparisons: 2
Honda Civic
Toyota Corolla
Volkswagen Golf

Capturing multiple variables

Multiple variables can be captured by separating them with a comma. This can include a mix of variables captured by value or by reference:

int health{ 33 };
int armor{ 100 };
std::vector<CEnemy> enemies{};

// Capture health and armor by value, and enemies by reference.
[health, armor, &enemies](){};

Default captures

Having to explicitly list the variables you want to capture can be burdensome. If you modify your lambda, you may forget to add or remove captured variables. Fortunately, we can enlist the compiler’s help to auto-generate a list of variables we need to capture.

A default capture (also called a capture-default) captures all variables that are mentioned in the lambda. Variables not mentioned in the lambda are not captured if a default capture is used.

To capture all used variables by value, use a capture value of =.
To capture all used variables by reference, use a capture value of &.

Here’s an example of using a default capture by value:

#include <algorithm>
#include <array>
#include <iostream>

int main()
{
  std::array areas{ 100, 25, 121, 40, 56 };

  int width{};
  int height{};

  std::cout << "Enter width and height: ";
  std::cin >> width >> height;

  auto found{ std::find_if(areas.begin(), areas.end(),
                           [=](int knownArea) { // will default capture width and height by value
                             return width * height == knownArea; // because they're mentioned here
                           }) };

  if (found == areas.end())
  {
    std::cout << "I don't know this area :(\n";
  }
  else
  {
    std::cout << "Area found :)\n";
  }

  return 0;
}

Default captures can be mixed with normal captures. We can capture some variables by value and others by reference, but each variable can only be captured once.

int health{ 33 };
int armor{ 100 };
std::vector<CEnemy> enemies{};

// Capture health and armor by value, and enemies by reference.
[health, armor, &enemies](){};

// Capture enemies by reference and everything else by value.
[=, &enemies](){};

// Capture armor by value and everything else by reference.
[&, armor](){};

// Illegal, we already said we want to capture everything by reference.
[&, &armor](){};

// Illegal, we already said we want to capture everything by value.
[=, armor](){};

// Illegal, armor appears twice.
[armor, &health, &armor](){};

// Illegal, the default capture has to be the first element in the capture group.
[armor, &](){};

Defining new variables in the lambda-capture

Sometimes we want to capture a variable with a slight modification or declare a new variable that is only visible in the scope of the lambda. We can do so by defining a variable in the lambda-capture without specifying its type.

#include <array>
#include <iostream>
#include <algorithm>

int main()
{
  std::array areas{ 100, 25, 121, 40, 56 };

  int width{};
  int height{};

  std::cout << "Enter width and height: ";
  std::cin >> width >> height;

  // We store areas, but the user entered width and height.
  // We need to calculate the area before we can search for it.
  auto found{ std::find_if(areas.begin(), areas.end(),
                           // Declare a new variable that's visible only to the lambda.
                           // The type of userArea is automatically deduced to int.
                           [userArea{ width * height }](int knownArea) {
                             return userArea == knownArea;
                           }) };

  if (found == areas.end())
  {
    std::cout << "I don't know this area :(\n";
  }
  else
  {
    std::cout << "Area found :)\n";
  }

  return 0;
}

userArea will only be calculated once when the lambda is defined. The calculated area is stored in the lambda object and is the same for every call. If a lambda is mutable and modifies a variable that was defined in the capture, the original value will be overridden.

Dangling captured variables

Variables are captured at the point where the lambda is defined. If a variable captured by reference dies before the lambda, the lambda will be left holding a dangling reference.

For example:

#include <iostream>
#include <string>

// returns a lambda
auto makeWalrus(const std::string& name)
{
  // Capture name by reference and return the lambda.
  return [&]() {
    std::cout << "I am a walrus, my name is " << name << '\n'; // Undefined behavior
  };
}

int main()
{
  // Create a new walrus whose name is Roofus.
  // sayName is the lambda returned by makeWalrus.
  auto sayName{ makeWalrus("Roofus") };

  // Call the lambda function that makeWalrus returned.
  sayName();

  return 0;
}

The call to makeWalrus creates a temporary std::string from the string literal “Roofus”. The lambda in makeWalrus captures the temporary string by reference. The temporary string dies when makeWalrus returns, but the lambda still references it. Then when we call sayName, the dangling reference is accessed, causing undefined behavior.

Note that this also happens if name is passed to makeWalrus by value. The variable name still dies at the end of makeWalrus, and the lambda is left holding a dangling reference.

If we want the captured name to be valid when the lambda is used, we need to capture it by value instead (either explicitly or using a default-capture by value).

Unintended copies of mutable lambdas

Because lambdas are objects, they can be copied. In some cases, this can cause problems. Consider the following code:

#include <iostream>

int main()
{
  int i{ 0 };

  // Create a new lambda named count
  auto count{ [i]() mutable {
    std::cout << ++i << '\n';
  } };

  count(); // invoke count

  auto otherCount{ count }; // create a copy of count

  // invoke both count and the copy
  count();
  otherCount();

  return 0;
}

Output

1
2
2

Rather than printing 1, 2, 3, the code prints 2 twice. When we created otherCount as a copy of count, we created a copy of count in its current state. count‘s i was 1, so otherCount‘s i is 1 as well. Since otherCount is a copy of count, they each have their own i.

Now let’s take a look at a slightly less obvious example:

#include <iostream>
#include <functional>

void myInvoke(const std::function<void()>& fn)
{
    fn();
}

int main()
{
    int i{ 0 };

    // Increments and prints its local copy of @i.
    auto count{ [i]() mutable {
      std::cout << ++i << '\n';
    } };

    myInvoke(count);
    myInvoke(count);
    myInvoke(count);

    return 0;
}

Output:

1
1
1

This exhibits the same problem as the prior example in a more obscure form. When myInvoke() is called, the std::function parameter is initialized with our lambda, and internally makes a copy of the lambda object. Thus, our call to fn() is actually being executed on the copy of our lambda, not the actual lambda.

If we need to pass a mutable lambda, and want to avoid the possibility of inadvertent copies being made, there are two options. One option is to use a non-capturing lambda instead -- in the above case, we could remove the capture and track our state using a static local variable instead. But static local variables can be difficult to keep track of and make our code less readable. A better option is to prevent copies of our lambda from being made in the first place. But since we can’t affect how std::function (or other standard library functions or objects) are implemented, how can we do this?

One option (h/t to reader Dck) is to put our lambda into a std::function immediately. That way, when we call myInvoke(), the reference parameter fn can bind to our std::function, and no temporary copy is made:

#include <iostream>
#include <functional>

void myInvoke(const std::function<void()>& fn)
{
    fn();
}

int main()
{
    int i{ 0 };

    // Increments and prints its local copy of @i.
    std::function count{ [i]() mutable { // lambda object stored in a std::function
      std::cout << ++i << '\n';
    } };

    myInvoke(count); // doesn't create copy when called
    myInvoke(count); // doesn't create copy when called
    myInvoke(count); // doesn't create copy when called

    return 0;
}

Our output is now as expected:

1
2
3

An alternate solution is to use a reference wrapper. C++ provides a convenient type (as part of the <functional> header) called std::reference_wrapper that allows us to pass a normal type as if it was a reference. For even more convenience, a std::reference_wrapper can be created by using the std::ref() function. By wrapping our lambda in a std::reference_wrapper, whenever anybody tries to make a copy of our lambda, they’ll make a copy of the reference_wrapper instead (avoiding making a copy of the lambda).

Here’s our updated code using std::ref:

#include <iostream>
#include <functional> // includes std::reference_wrapper and std::ref

void myInvoke(const std::function<void()>& fn)
{
    fn();
}

int main()
{
    int i{ 0 };

    // Increments and prints its local copy of @i.
    auto count{ [i]() mutable {
      std::cout << ++i << '\n';
    } };

    // std::ref(count) ensures count is treated like a reference
    // thus, anything that tries to copy count will actually copy the reference
    // ensuring that only one count exists
    myInvoke(std::ref(count));
    myInvoke(std::ref(count));
    myInvoke(std::ref(count));

    return 0;
}

Our output is now as expected:

1
2
3

What’s interesting about this method is that it works even if myInvoke takes fn by value (instead of by reference)!

Quiz time

int i{};
static int j{};

int getValue()
{
  return 0;
}

int main()
{
  int a{};
  constexpr int b{};
  static int c{};
  static constexpr int d{};
  const int e{};
  const int f{ getValue() };
  static const int g{}; 
  static const int h{ getValue() }; 

  [](){
    // Try to use the variables without explicitly capturing them.
    a;
    b;
    c;
    d;
    e;
    f;
    g;
    h;
    i;
    j;
  }();

  return 0;
}

#include <iostream>
#include <string>

int main()
{
  std::string favoriteFruit{ "grapes" };

  auto printFavoriteFruit{
    [=]() {
      std::cout << "I like " << favoriteFruit << '\n';
    }
  };

  favoriteFruit = "bananas with chocolate";

  printFavoriteFruit();

  return 0;
}

I like grapes

Start where? 4
How many? 8
I generated 8 square numbers. Do you know what each number is after multiplying it by 2?
> 32
Nice! 7 number(s) left.
> 72
Nice! 6 number(s) left.
> 50
Nice! 5 number(s) left.
> 126
126 is wrong! Try 128 next time.

Start where? 1
How many? 3
I generated 3 square numbers. Do you know what each number is after multiplying it by 4?
> 4
Nice! 2 numbers left.
> 16
Nice! 1 numbers left.
> 36
Nice! You found all numbers, good job!

Start where? 2
How many? 2
I generated 2 square numbers. Do you know what each number is after multiplying it by 4?
> 21
21 is wrong!

auto found{ std::find(/* ... */) };

// Make sure the element was found

myVector.erase(found);

int distance{ std::abs(3 - 5) }; // 2

#include <algorithm> // std::find, std::min_element
#include <cmath> // std::abs
#include <ctime>
#include <iostream>
#include <random>
#include <vector>

using Numbers = std::vector<int>;

namespace config
{
  constexpr int multiplierMin{ 2 };
  constexpr int multiplierMax{ 4 };
  constexpr int maximumWrongAnswer{ 4 };
}

int getRandomInt(int min, int max)
{
  static std::mt19937 mt{ static_cast<std::mt19937::result_type>(std::time(nullptr)) };

  return std::uniform_int_distribution{ min, max }(mt);
}

// Generates @count numbers starting at @start*@start and multiplies
// every square number by multiplier.
Numbers generateNumbers(int start, int count, int multiplier)
{
  Numbers numbers(static_cast<Numbers::size_type>(count));

  int i{ start };

  for (auto& number : numbers)
  {
    number = ((i * i) * multiplier);
    ++i;
  }

  return numbers;
}

// Asks the user to input start and amount, then calls @generateNumbers.
Numbers generateUserNumbers(int multiplier)
{
  int start{};
  int count{};

  std::cout << "Start where? ";
  std::cin >> start;

  std::cout << "How many? ";
  std::cin >> count;

  // Input validation omitted. All functions assume valid input.

  return generateNumbers(start, count, multiplier);
}

int getUserGuess()
{
  int guess{};

  std::cout << "> ";
  std::cin >> guess;

  return guess;
}

// Searches for the value @guess in @numbers and removes it.
// Returns true if the value was found. False otherwise.
bool findAndRemove(Numbers& numbers, int guess)
{
  auto found{ std::find(numbers.begin(), numbers.end(), guess) };

  if (found == numbers.end())
  {
    return false;
  }
  else
  {
    numbers.erase(found);
    return true;
  }
}

// Finds the value in @numbers that is closest to @guess.
int findClosestNumber(const Numbers& numbers, int guess)
{
  return *std::min_element(numbers.begin(), numbers.end(), [=](int a, int b) {
    return std::abs(a - guess) < std::abs(b - guess);
  });
}

void printTask(Numbers::size_type count, int multiplier)
{
  std::cout << "I generated " << count
            << " square numbers. Do you know what each number is after multiplying it by "
            << multiplier << "?\n";
}

// Called when the user guesses a number correctly.
void printSuccess(Numbers::size_type numbersLeft)
{
  std::cout << "Nice! ";

  if (numbersLeft == 0)
  {
    std::cout << "You found all numbers, good job!\n";
  }
  else
  {
    std::cout << numbersLeft << " number(s) left.\n";
  }
}

// Called when the user guesses a number that is not in the numbers.
void printFailure(const Numbers& numbers, int guess)
{
  int closest{ findClosestNumber(numbers, guess) };

  std::cout << guess << " is wrong!";

  if (std::abs(closest - guess) <= config::maximumWrongAnswer)
  {
    std::cout << " Try " << closest << " next time.\n";
  }
  else
  {
    std::cout << '\n';
  }
}

// Returns false if the game is over. True otherwise.
bool playRound(Numbers& numbers)
{
  int guess{ getUserGuess() };

  if (findAndRemove(numbers, guess))
  {
    printSuccess(numbers.size());

    return !numbers.empty();
  }
  else
  {
    printFailure(numbers, guess);
    return false;
  }
}

int main()
{
  int multiplier{ getRandomInt(config::multiplierMin, config::multiplierMax) };
  Numbers numbers{ generateUserNumbers(multiplier) };

  printTask(numbers.size(), multiplier);

  while (playRound(numbers))
    ;

  return 0;
}