Java Community | Help. Code. Learn.•14mo ago

Why is this lambda illegal?

hey guys. can smb explain lambdas to me? what am i doing wrong?

public interface FunctionalInterface {
    String function(String str);
}
public class Main {
    public static void main(String[] args) {
        FunctionalInterface functionalInterface= "hello"->"hello";
    }
}

public interface FunctionalInterface {
    String function(String str);
}
public class Main {
    public static void main(String[] args) {
        FunctionalInterface functionalInterface= "hello"->"hello";
    }
}

My lambda complies to the functional interface. Its argument is of String type, and it returns String too. wt f?

11 Replies

JavaBot•14mo ago

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Unknown User•14mo ago

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0x150•14mo ago

what do you think that would do / what was your intention when testing that out

i hate SQL so much its unrealOP•14mo ago

well my lambda has to comply to the method signature of functional interface . method has to accept string, and return string. i give lambda a string, and it screams at me

0x150•14mo ago

at this point, you dont give the lambda a string you need to define the lambda giving it a string comes later defining a lambda is the same as a method but in a slightly different format and without an inherent name: (arguments) -> expression or (arguments) -> { methodBody; return expression; } since FunctionalInterface is your target lambda type, your lambda should accept one string as parameter and return another. so your lambda should look something like this: theStringArgument -> theStringArgument.doSomethingWithAString() (as an example) that makes a new lambda that takes in one string (stored under theStringArgument), calls the imaginary doSomethingWithAString() on it and returns its return value (a string) to then use that lambda, you do String result = functionalInterface.function("some string")

i hate SQL so much its unrealOP•14mo ago

@0x150 do i understand it correctly?

String greeting = "hi";
        FunctionalInterface functionalInterface= myGreeting-> "hello";// this is basically an impl of interface, not a method call
        functionalInterface.function(greeting);//actual method call

String greeting = "hi";
        FunctionalInterface functionalInterface= myGreeting-> "hello";// this is basically an impl of interface, not a method call
        functionalInterface.function(greeting);//actual method call

0x150•14mo ago

correct that is exactly what is happening the lambda is just an implementation of the interface method lambdas only work if the target interface has exactly one abstract method, for that exact reason

i hate SQL so much its unrealOP•14mo ago

okay. but then the question is why doesnt ide scream at me for myGreeting? because it isnt described anywhere

0x150•14mo ago

where the parameter? its a parameter, its not a predefined variable in fact, defining a variable with the same name causes an error because the variable already exists what you're defining there is a method parameter it's a slot an input argument gets stored in, in this case the string it applies the operation to if you already knew the value and lambdas would only be able to apply to one specific value once, they'd be a bit useless, no?

i hate SQL so much its unrealOP•14mo ago

yup yup. it would be the same as if i had a getter for an object, but it would always return hardcoded name

JavaBot•14mo ago

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