lambda and variable

Ive difficulties understanding why the following doesn't compile when the arguments in the lambda expression are variables instead of fields.

The message given by my IDE is :

java: variable test1 is already defined in method main(java.lang.String[])

public class Java8Tester { static String test3 = "Hello"; static String test4 = "Goodbye"; public static void main(String[] hi){ String test1 = "Hello"; String test2 = "Goodbye"; betterString(test1, test2, (test1, test2) -> true); //doesn't compile betterString(test3, test4, (test3, test4) -> true);//compiles } public static String betterString(String s1, String s2, TwoStringPredicate predicate){ if(predicate.betterString(s1, s2)){ return s1; }else{ return s2; } } interface TwoStringPredicate{ boolean betterString(String s1, String s2); } }

Hi

In your code test1 and test2 variables are already declared in

String test1 = "Hello"; String test2 = "Goodbye";

You can access local variable in lambda body( if it is effectively final for Java 8 or just final for versions before Java 8),
but you can't redeclare variable (use same variable) , which was already declared in local scope.

betterString(test1, test2, (test1, test2) -> true); //doesn't compile
Here you are re-declaring test1 and test2 variables again in your lambda expression

Heyho!

betterString(test3, test4, (test3, test4) -> true);//compiles

This works because you pass the betterString method the static variables test3 and test4, and then you pass two locally created variables, that are named exactly the same as your static variables.

If you remember constructors:
class ConstructorExample { int i = 0; String k = null; ConstructorExample(int i, String k){ this.i = i; this.k = k; } }
We had to use the this keyword because the local variable names are referenced by default in case of conflict. Something similar happens in your example.
test1 and test2 are local, but test3 and test4 are class variables.

Lambdas always redeclare the parameters you give them in the parantheses. Writing (test1, test2) is just a convenience for shortly writing (String test1, String test2).

Cedric Bosch wrote:Ive difficulties understanding...

I'll leave the "functional" stuff to bods who know it better; but you can save yourself some typing by using the conditional operator instead of if at line 14, viz:
  return predicate.betterString(s1, s2) ? s1 : s2;

HIH

Winston