Java.Java8.What variables can lambda expressions access?

Lambdas can access variables from three main places:

---

1. ✅ Parameters passed to the lambda

Just like any method, a lambda has its own parameters.

(x, y) -> x + y  // `x` and `y` are lambda parameters

2. ✅ Local variables from the enclosing scope — but only if they are effectively final

This is a key concept.

🧠 “Effectively final” means: the variable is not modified after it’s initialized, even if it’s not explicitly marked as final.

Example (valid):

int base = 10;  // effectively final
Function<Integer, Integer> addBase = x -> x + base;

Example (invalid):

int base = 10;
base = 15;  // ❌ now it's not "effectively final"

Function<Integer, Integer> addBase = x -> x + base; // Compile error

The reason: lambdas do not capture variables by reference, they capture the value of the variable at the time of lambda creation.

3. ✅ Instance and static variables

Lambdas can freely access instance variables (this.var) and static variables of the enclosing class.

Example:

class Calculator {
    int offset = 100;

    Function<Integer, Integer> addOffset() {
        return x -> x + offset;  // accessing instance variable
    }
}

🔒 Summary Table

Variable Type	Can Lambda Access?	Notes
Lambda Parameters	✅	Standard parameters
Local Variables	✅ if effectively final	Cannot be modified after declaration
Instance Variables	✅	No restrictions
Static Variables	✅	No restrictions

---

🧠 Why “effectively final”?

Java needs this restriction to:

• Ensure thread safety (lambda can be executed later in a different thread).
• Avoid mutable state confusion.
• Maintain consistency with how anonymous inner classes behave.