When a lambda expression uses an assigned local variable from its enclosing space there is an important restriction. A lambda expression may only use local variable whose value doesn't change. That restriction is referred as "variable capture" which is described as; lambda expression capture values, not variables. The local variables that a lambda expression may use are known as "effectively final". An effectively final variable is one whose value does not change after it is first assigned. There is no need to explicitly declare such a variable as final, although doing so would not be an error. Let's see it with an example, we have a local variable i which is initialized with the value 7, with in the lambda expression we are trying to change that value by assigning a new value to i. This will result in compiler error - "Local variable i defined in an enclosing scope must be final or effectively final"

@FunctionalInterface
interface IFuncInt {
    int func(int num1, int num2);
    public String toString();
}

public class LambdaVarDemo {

    public static void main(String[] args){             
        int i = 7;
        IFuncInt funcInt = (num1, num2) -> {
            i = num1 + num2;
            return i;
        };
    }   
}

．.． 从Java SE 8开始，局部类可以访问封闭块的final或有效final的局部变量和参数。在初始化后值从未改变的变量或参数实际上是final。

例如，假设变量numberLength没有声明为final，并且在PhoneNumber构造函数中添加了标记的赋值语句:

public class OutterClass {  

  int numberLength; // <== not *final*

  class PhoneNumber {

    PhoneNumber(String phoneNumber) {
        numberLength = 7;   // <== assignment to numberLength
        String currentNumber = phoneNumber.replaceAll(
            regularExpression, "");
        if (currentNumber.length() == numberLength)
            formattedPhoneNumber = currentNumber;
        else
            formattedPhoneNumber = null;
     }

  ...

  }

...

}

由于这个赋值语句，变量numberLength不再是有效的final。结果，Java编译器生成一个类似于“内部类引用的局部变量必须是final或有效的final”的错误消息，其中内部类PhoneNumber试图访问numberLength变量:

http://codeinventions.blogspot.in/2014/07/difference-between-final-and.html

http://docs.oracle.com/javase/tutorial/java/javaOO/localclasses.html

我发现解释“有效的final”最简单的方法是想象在变量声明中添加final修饰符。如果通过这个更改，程序在编译时和运行时继续以相同的方式运行，那么该变量实际上就是final。

当一个变量被初始化一次并且在其所有者类中从未发生变化时，它就是最终的或有效的最终的。我们不能在循环或内部类中初始化它。

最后:

final int number;
number = 23;

有效的最后:

int number;
number = 34;

注意:Final和Effective Final是相似的(赋值后它们的值不会改变)，但只是有效的Final变量 没有使用关键字final声明。

下面这个变量是最终的，所以我们不能改变它的值一旦初始化。如果我们尝试，我们会得到一个编译错误…

final int variable = 123;

但是如果我们创建一个这样的变量，我们可以改变它的值…

int variable = 123;
variable = 456;

但是在Java 8中，默认情况下所有变量都是final。但是代码中第二行的存在使得它不是最终的。因此，如果我们从上面的代码中删除第二行，我们的变量现在是“有效的final”…

int variable = 123;

所以. .任何赋值一次且仅一次的变量都是“有效的最终变量”。

Final是一个用关键字Final声明的变量，示例:

final double pi = 3.14 ;

它在整个程序中都是最终的，在这一行之后改变PI是不允许的。

有效的最终的:任何局部变量或参数，现在只被赋值一次(或只更新一次)。在整个项目中，它可能不是有效的最终结果。这意味着有效最终变量可能会在被分配/更新至少一次赋值后立即失去其有效最终属性。例子:

class EffectivelyFinal {
    
    public static void main(String[] args) {
        calculate(124,53);
    }
    
    public static void calculate( int operand1, int operand2){   
     int rem = 0;  //   operand1, operand2 and rem are effectively final here
     rem = operand1%2  // rem lost its effectively final property here because it gets its second assignment 
                       // operand1, operand2 are still effectively final here 
        class operators{

            void setNum(){
                operand1 =   operand2%2;  // operand1 lost its effectively final property here because it gets its second assignment
            }
            
            int add(){
                return rem + operand2;  // does not compile because rem is not effectively final
            }
            int multiply(){
                return rem * operand1;  // does not compile because both rem and operand1 are not effectively final
            }
        }   
   }    
}