Lambdas

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With the mutable keyword we can change the value of the parameter in the capture clause in the lambda function but that change is not refected in the code calling the lambda function. It's as if the parameter is local and static to the lambda expression.

File: lambdas13.cpp

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

using namespace std ;



int main()
{
  //using the mutable keyword
  //we can change the value in the lambda function
  //but that change is not refected in the code calling
  //the lambda function

  int a1 = 1;
  int b1 = 1;

  cout << "In main():" << endl;
  cout << "a1 = " << a1 << ", ";
  cout << "b1 = " << b1 << endl;
  cout << endl;

  auto add_one = [a1, &b1] () mutable {
            // modify both a1 & b1
            a1++;
            b1++;
            cout << "In add_one():" << endl;
            cout << "a1 = " << a1 << ", ";
            cout << "b1 = " << b1 << endl;
          };

  add_one();
  add_one();

  cout << endl;
  cout << "In main() After calling the lambda function:" << endl;
  cout << "a1 = " << a1 << ", ";
  cout << "b1 = " << b1 << endl;



}

Output:
$./a.exe
$ g++ lambdas13.cpp ; ./a.exe
In main():
a1 = 1, b1 = 1

In add_one():
a1 = 2, b1 = 2
In add_one():
a1 = 3, b1 = 3

In main() After calling the lambda function:
a1 = 1, b1 = 3

We are passing 2 variables in the parameters clause. The "a1" is passed by value and the "b1" is passed by reference. We are able to change the value of "a1" because of the mutable keyword. That change is kept local to the lambda

Exercise

File: mutable_ex1.cpp
1) What does the below program print ?

#include <iostream>

using namespace std ;

int main()
{
    int counter = 0;

    // Lambda capturing 'counter' by value and declared mutable
    auto increment = [counter]() mutable
    {
        // 'counter' can be modified inside the lambda because of 'mutable'
        counter++;
        cout << "Inside lambda: counter = " << counter << endl;
    };

    increment(); // Call the lambda
    increment(); // Call the lambda again
    increment(); // Call the lambda a third time

    cout << counter << endl;

    return 0;
}

Solutions

File: lambdas_ex_1s.cpp