Вивчайте Scope and Variable Capture | Getting Started with Lambda Expressions

Understanding how lambda expressions interact with variables from their enclosing scope is essential for writing correct and predictable code. This behavior is known as variable capture. When you use a variable from the surrounding method inside a lambda expression, that variable is said to be "captured" by the lambda. However, Java enforces specific rules: only variables that are final or effectively final can be used inside a lambda. A variable is considered effectively final if its value is not changed after it is initialized, even if it is not explicitly marked as final. This restriction ensures thread safety and prevents unexpected behavior, since lambdas may be executed after the enclosing method has finished.

Main.java


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package com.example;

public class Main {
    public static void main(String[] args) {
        int baseNumber = 10; // effectively final

        Runnable printBase = () -> System.out.println("Base number is: " + baseNumber);

        printBase.run();
    }
}

Comparing lambdas to anonymous inner classes reveals important differences in variable capture. Both can access final or effectively final variables from their enclosing scope, but the way they handle the this reference is different. In a lambda, this refers to the enclosing class instance, while in an anonymous inner class, this refers to the instance of the anonymous class itself. This distinction can affect how you access class members and methods within each construct.

Main.java


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package com.example;

public class Main {
    public static void main(String[] args) {
        int number = 5;

        // Uncommenting the next line will cause a compilation error:
        // number = 6;

        Runnable r = () -> {
            // If 'number' were modified above, this would not compile
            System.out.println("Number is: " + number);
        };

        r.run();
    }
}

1. What does "effectively final" mean in the context of lambda expressions?

2. What is the difference in the meaning of `this` inside a lambda expression compared to an anonymous inner class?

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Main.java


              123456789101112
            
package com.example;

public class Main {
    public static void main(String[] args) {
        int baseNumber = 10; // effectively final

        Runnable printBase = () -> System.out.println("Base number is: " + baseNumber);

        printBase.run();
    }
}

Main.java


              123456789101112131415161718
            
package com.example;

public class Main {
    public static void main(String[] args) {
        int number = 5;

        // Uncommenting the next line will cause a compilation error:
        // number = 6;

        Runnable r = () -> {
            // If 'number' were modified above, this would not compile
            System.out.println("Number is: " + number);
        };

        r.run();
    }
}

1. What does "effectively final" mean in the context of lambda expressions?

2. What is the difference in the meaning of `this` inside a lambda expression compared to an anonymous inner class?

Все було зрозуміло?

Дякуємо за ваш відгук!

Секція 1. Розділ 4

Scope and Variable Capture

1. What does "effectively final" mean in the context of lambda expressions?

2. What is the difference in the meaning of this inside a lambda expression compared to an anonymous inner class?

Scope and Variable Capture

1. What does "effectively final" mean in the context of lambda expressions?

2. What is the difference in the meaning of this inside a lambda expression compared to an anonymous inner class?

2. What is the difference in the meaning of `this` inside a lambda expression compared to an anonymous inner class?

2. What is the difference in the meaning of `this` inside a lambda expression compared to an anonymous inner class?