incompatible types: void cannot be converted to in

问题:

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• What does “Error: Incompatible types: void cannot be converted to …” mean? 1 answer

I'm extremely new when to comes to Java and programming in general. I am trying to create a simple program where you guess my age and if you are right it will say "correct" and if you are wrong it will say "wrong".

This is my code:

import java.util.InputMismatchException;
import java.util.Scanner; // This will import just the Scanner class.

public class GuessAge {
    public static int main(int[] args) {
       System.out.println("\nWhat is David's Age?");
       Scanner userInputScanner = new Scanner(System.in);
       int age = userInputScanner.nextLine();



        int validInput = 20;
        if (validInput == 20) {
            return System.out.println("Correct!!");
        }
        else {
            return System.out.println("Wrong....");
        }
    }
}

I get the error "incompatible types: void cannot be converted to int" but I have no void class in the code? I know my code is probably awful but if you guys could point me in the right direction that would be great. Thanks.

回答1:

Your program does not have to return an int in public static int main. Instead you can have it as void (meaning don't return anything). You should simply just print your statements and don't return them. Also the int[] should be String[] and Scanner should check for nextInt() as pointed out in comments!

import java.util.InputMismatchException;
import java.util.Scanner; // This will import just the Scanner class.

public class GuessAge {
public static void main(String[] args) {
   System.out.println("\nWhat is David's Age?");
   Scanner userInputScanner = new Scanner(System.in);
   int age = userInputScanner.nextInt();



    int validInput = 20;

    // typo in your code - compare to age
    if (validInput == age) {
        System.out.println("Correct!!");
    }
    else {
        System.out.println("Wrong....");
    }
}

}

回答2:

You are trying to return System.out.println() which is of type void. Remove the return statements from before System.out.println(), and they will still print. Note that you do not need to specify a return value in the main method.

回答3:

In your method declaration, you have public static int main(int[] args)

The word after the static keyword is a return type, and in this case your declaration is requiring the main method to return an int. To solve this, main should have a void return type, as you're only printing within main and not returning type int.