Attempting to instantiate in loop, get unexpected type (expected variable)

Question

I'm trying to instantiate three objects in a loop so that I don't reuse the constructor text. This is because, for Uni, we get marked down if we reuse the same line of code (it's marked automatically).

My code is as follows:

EnemyShip enemy1;
EnemyShip enemy2;
EnemyShip enemy3;

public Game()
{
    for (int i = 1; i <= 3; i++) {
        getEnemyRef(i) = new EnemyShip(); //getEnemyRef unexpected type - required:variable found:value
        //enemy1 = new EnemyShip(); works normally
    }
}

The i is underlined and I am told that it is a value not variable.

Please note that I use the following code instead of a list or array because we are not allowed to use them for this task.

public EnemyShip getEnemyRef(int enemy) {
    switch (enemy) 
    {
        case 1:
        return enemy1;

        case 2:
        return enemy2;

        case 3:
        return enemy3;
    }
    return null;
}

The marking system bases how many times a line is reused from how many times it is written in source code, not how many times it is executed

you need the exact variable name when assigning values. do `EnemyShip e1 = getEnemyRef(i);` before and then `e1 = new enemyShip()` — XtremeBaumer, Jun 01 '18 at 13:42
you probably wanted something like this: `EnemyShip enemy = getEnemyRef(i);` — Lino, Jun 01 '18 at 13:43
`getEnemyRef(i)` is a method, thus returns a value. You cannot assing something to a value (thus the compiler message: `required:variable found:value`). You constructor call is not the problem, the assignment (or more precise: the left hand side of the assignment) is the problem. — Turing85, Jun 01 '18 at 13:44
It looks like you need to learn about arrays and / or Lists. — John Bollinger, Jun 01 '18 at 13:45
What are `enemy1`, `enemy2`, `enemy3`? Where do they come from? — Mikhail Batcer, Jun 01 '18 at 13:48
I have them declared at the top of the code, forgot to include them but just edited it now @Mikhail — Harry Allen, Jun 01 '18 at 13:51
I love when classes teach you bad coding practice in order to get you to do something a certain way using a certain method when in reality you would never do it that way — Zannith, Jun 01 '18 at 14:16
@HarryAllen I have updated my answer once again. You may want to have a look at it — Lino, Jun 04 '18 at 06:16

Lino · Answer 1 · 2018-06-04T06:16:26.043

0

This snippet should do the trick:

public void setEnemyRef(int enemyFlag, EnemyShip enemy){
    switch (enemyFlag){
        case 1:
            enemy1 = enemy;
            break;
        case 2:
            enemy2 = enemy;
            break;
        case 3:
            enemy3 = enemy;
            break;
    }
}

And then in the for-loop:

for (int i = 1; i <= 3; i++) {
    setEnemyRef(i, new EnemyShip());
}

edited Jun 04 '18 at 06:16

answered Jun 01 '18 at 14:03

Lino

19,604
6
47
65

This does work and meet my requirements but sadly I still lose the marks because it uses the createNew() method three times. (I lose marks for reuse (more than one usage) of equality checks, sub/add, constructor use, method calls and communication between classes) – Harry Allen Jun 01 '18 at 14:11
1

props for creativity, but you are using an array which OP said was not allowed – Zannith Jun 01 '18 at 14:19
@TravisHerbert I've updated, still not sure if that's allowed – Lino Jun 04 '18 at 06:40

score -1 · Answer 2 · answered Jun 01 '18 at 13:46

-1

You're not instantiating it correctly, try e.g.

EnemyShip variableName = getEnemryRef(i); variableName = new EnemyShip();

https://docs.oracle.com/javase/tutorial/java/javaOO/objectcreation.html

answered Jun 01 '18 at 13:46

Julian Alwandy

321
1
4
13

3

this code is equivalent to `EnemyShip variableName = new EnemyShip();` (under the assumption that `getEnemyRef(i)` is a method that never throws) and I do not think that this is what OP asks for. Please remember that [Java passes by value always](https://stackoverflow.com/questions/40480/is-java-pass-by-reference-or-pass-by-value), and it returns by value in the same manner. – Turing85 Jun 01 '18 at 13:47
Fair enough but his solution to tackle the problem does not seem optimal. – Julian Alwandy Jun 01 '18 at 13:52
@JulianAlwandy The predicament that I'm in is that I need to reduce the use of the constructor '= new EnemyShip();' so that I'm not using three times to create three obejcts. Is there a way to do that? – Harry Allen Jun 01 '18 at 13:55
@HarryAllen No. If you want three distinct objects, you need at least three constructor calls. – Turing85 Jun 01 '18 at 13:56
@Turing85 So the only possible way is to use an array/list? – Harry Allen Jun 01 '18 at 13:58
@HarryAllen I am not sure I follow your train of thoughts. Even in an array/`List<...>`, if you want to store three distring objects in them, you need to first instantiate those three objects (via three constructor calls). – Turing85 Jun 01 '18 at 13:59
@Turing85 Maybe I am using the wrong terminology. A loop such as: while (true) { EnemyShip e = new EnemyShip() list.add(e) } Could create more than one object while only using the line once? – Harry Allen Jun 01 '18 at 14:04
@HarryAllen Well... you WRITE the line only once, but the line (in your example) is executed as often as your heap memory allows it. – Turing85 Jun 01 '18 at 14:05
@Turing85 the marking system we are submitting to only counts a use as when it is written, not how many times it is used/executed. It's a bit daft – Harry Allen Jun 01 '18 at 14:13
2

@HarryAllen you may want to edit your question and add this information. It is quite crucial to answer it correctly. – Turing85 Jun 01 '18 at 14:15

Attempting to instantiate in loop, get unexpected type (expected variable)

2 Answers2