如何检查 Java 中是否存在变量？

Question

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

Answer 1

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

Answer 2

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 .

Answer 3

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).

Answer 4

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)..

Answer 5

A primitive can not be null in Java.

Answer 6

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 ) ...

Answer 7

You could try something like this

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

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

Answer 8

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 (eg 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.

Answer 9

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);
}