CODEWITHSUNDEEP
javanullprimitive
user502039
user502039

Reputation:

How can I check if a variable exists in Java?

I want to write to a variable only if there isn't anything already there. Here is my code so far.

if (inv[0] == null) {
    inv[0]=map.getTileId(tileX-1, tileY-1, 0);
}

It gives me this error:

java.lang.Error: Unresolved compilation problem: 
The operator == is undefined for the argument type(s) int, null

Upvotes: 5

Views: 22362

Answers (9)

Paul
Paul

Reputation: 20081

The error is because inv is an array of int, not the object wrapper Integer. Your array comes initialized for you anyway. If you wrote

int[] inv = new int[5];

you will have an array of 5 zeroes.

You should initialize your array yourself using some value that you know is invalid (e.g. if you had an array of ages, a negative value would be invalid). Check for the invalid value and if it's there, replace it.

Upvotes: 0

Hunter McMillen
Hunter McMillen

Reputation: 61520

primitive types can't be compared to null.

You can test if the number if > 0 to see if a value exists:

if(inv[0] <= 0)
{
    inv[0]=map.getTileId(tileX-1, tileY-1, 0);
}

Upvotes: -1

Bala R
Bala R

Reputation: 108967

You could try something like this 

Integer[] inv = new Integer[10];
inv[0] = 1;
inv[1] = 2;
    ....

if(inv[3] == null)
{
    inv[3] = getSomeValue();
}

Upvotes: 0

Sorceror
Sorceror

Reputation: 4843

I'm assuming from error message, that inv[] is array of int, and int in java is not an object, so it cannot have null value.. You have to compare it with 0 (default value on each index of empty int array)..

Upvotes: 2

Ondra Žižka
Ondra Žižka

Reputation: 46876

Well... int is a primitive type. That can't be null.

You can check the size of the array:

int[] arr = new int[10]; System.out.println( arr.size() );

The plain arrays are indexed from 0 to their size - 1, and no value can be missing. So in your code, you are asking whether the first member of type int is null, which can't happen - either it's a number or it will cause ArrayOutOfBoundsException.

If you want to have a "sparse array" similar to what PHP or JavaScript, you need a Map:

Map<Integer, Integer> map = new HashMap();
map.put( 1, 324 );
if( map.get( 2 ) == null ) ...

Upvotes: 0

Jon Skeet
Jon Skeet

Reputation: 1502056

I'm assuming inv is an int[].

There's no such concept as a value "existing" or not in an array. For example:

int[] x = new int[5];

has exactly the same contents as:

int[] x = new int[5];
x[3] = 0;

Now if you used an Integer[] you could use a null value to indicate "unpopulated"... is that what you want?

Arrays are always filled with the default value for the element type to start with - which is null in the case of reference types such as Integer.

Upvotes: 5

fyr
fyr

Reputation: 20869

A primitive can not be null in Java.

Upvotes: 0

T.J. Crowder
T.J. Crowder

Reputation: 1074959

I take it that inv is an int[]. You can't compare an int to null, null only applies to reference types, not primitives. You have to either assign it some kind of flag value instead (0 being popular, and the value it will have by default when you create the array), or make inv an Integer[] instead (Integer being a reference type, it is null-able).

Upvotes: 2

amit
amit

Reputation: 178491

inv is an int[], and int cannot be null, since it is a primitive and not a reference.
ints are initialized to zero in java.

Upvotes: 9

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