Lambda Expressions - Beginning Java 8 Language Features - page 165

Printer printer = msg1 -> System.out.println( msg + " " + msg1);
msg = "Hi"; // msg is changed

return printer;
}

The following snippet of code generates a compile-time error because the lambda expression accesses the msg
variable that is declared lexically after its use. In Java, forward referencing of variable names in method's scope is not
allowed. Note that the msg variable is effectively final.

public Printer test() {
// A compile-time error. The msg variable is not declared yet.
Printer printer = msg1 -> System.out.println( msg + " " + msg1);

String msg = "Hello"; // msg is effectively final

return printer;
}

Can you guess why the following snippet of code generates a compile-time error?

public Printer test() {
String msg = "Hello";
Printer printer = msg1 -> {
msg = "Hi " + msg1; // A compile-time error. Attempting to modify msg.
System.out.println( msg );
};

return printer;
}

The lambda expression accesses the local variable msg . Any local variable accessed inside a lambda expression
must be effectively final. The lambda expression attempts to modify the msg variable inside its body that causes the
compile-time error.

■ a lambda expression can access instance and class variables of a class whether they are effectively final or not.
if instance and class variables are not final, they can be modified inside the body of the lambda expressions. a lambda
expression keeps a copy of the local variables used in its body. if the local variables are reference variables, a copy of the
references is kept, not a copy of the objects.

Tip

The program in Listing 5-20 demonstrates how to access the local and instance variables inside lambda
expressions.