Java allows for the use of the reserved word var with lambda parameters, to align its syntax with the declaration of local variables, which can now use var.

Let's modify the examples from the preceding section, adding var to the lambda parameters:

(var age) -> age > 10; 
(var age) -> age > 10? "Kid" : "Not a Kid"; 
(var age) -> {System.out.println();}; 
() -> {return Math.random() + "Number";}; 
 
(var name, var list) -> { 
                             return ( 
                                 list.stream() 
                                     .filter(e -> 
                                      e.getName().startsWith(name)) 
                                     .map(Person::getAge) 
                                     .findFirst() 
                                 ); 
                             }; 

If you are using var with lambda parameters, you must use it with all of the lambda parameters. You can't mix implicitly-typed or explicitly-typed parameters with the parameters that use var. The following code example won't compile:

(var x, y) -> x + y;                         // won't compile 
(var x, Integer y) -> x + y;                 // won't compile 

You cannot enclose the parameters of a lambda expression using round brackets (()) if you are using just one method parameter. But, you can't drop () if you are using var with your lambda parameters. Here is some sample code to illustrate this further:

(int x) -> x > 10;                         // compiles 
(x) -> x > 10;                             // compiles 
x -> x > 10;                               // compiles 
(var x) -> x > 10;                         // compiles 
var x -> x > 10;                           // Won't compile