Consider the following four test methods, and what do they output?

public class Test {

    public static void main(String\[\] args) {

        System.out.println(test1());

        System.out.println(test2());

        System.out.println(test3());

        System.out.println(test4());

    }

    private static int test1() {

        int i = 1;
        try {
            return i;
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            i = 0;
        }

        return i;

    }

    private static int test2() {

        int i = 1;
        try {
            return i;
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            i = 0;
            return i;
        }

    }

    private static User test3() {

        User user = new User("u1");
        try {
            return user;
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            user = new User("u2");
        }

        return null;

    }

    private static User test4() {

        User user = new User("u1");
        try {
            return user;
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            user.setName("u2");
        }

        return null;

    }

}

public class User {

    public User(String name) {
        this.name = name;
    }

    private String name;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @Override
    public String toString() {
        return name;
    }

}

The answer is as follows:

1

0

u1

u2

conclusion

1. No matter try,finally will execute;

2. return in try, the result will be saved before finally executing. Even if there is modification in finally, the value saved in try will prevail. However, if it is a reference type, the modified property will be subject to the modified property;

3. If try/finally has return, return the return in finally directly.

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