Preparing the value using initialization captures

Another great feature of the Lambda expression coming up in C++14 is its initialization captures. The expression can capture a value of the variable and assign it to the expression's variable. Let's take a look at the following piece of code implementing the initialization captures:

    /* lambda_initialization_captures.cpp */
    #include <iostream>

    using namespace std;

    auto main() -> int
    {
      cout << "[lambda_initialization_captures.cpp]" << endl;

      // Initializing a variable
      int a = 5;
      cout << "Initial a = " << a << endl;

      // Initializing value to lambda using the variable
      auto myLambda = [&x = a]() { x += 2; };

      // Executing the Lambda
      myLambda();

      // Displaying a new value of the variable
      cout << "New a = " << a << endl;

      return 0;
     }

As we can see in the preceding code, we have an int variable named a with the value 5. The Lambda expression, myLambda, then captures the a value and executes it in the code. The result is that now the a value will be 7 since it is added by 2. The following output screenshot should appear in our console window when we run the preceding code:

From the preceding snapshot, we see that we can prepare the value to be included in the calculation inside the Lambda expression.