Please help me with below code , I get the same output even after changing the value
import java.util.*;
class Test {
public static void main(String[] a
I think the problem is that you think the statement ...
x = Integer.valueOf(9);
... causes that the value of '9' get 'stored' into(!) the Object on which x is referencing.
But thats wrong.
Instead the statement causes something similar as if you would call
x = new Integer(9);
If you have a look to the java source code, you will see what happens in Detail.
Here is the code of the "valueOf(int i)" method in the "Integer" class:
public static Integer valueOf(int i) {
assert IntegerCache.high >= 127;
if (i >= IntegerCache.low && i <= IntegerCache.high)
return IntegerCache.cache[i + (-IntegerCache.low)];
return new Integer(i);
}
and further, whenever the IntegerCache class is used for the first time the following script gets invoked:
static {
// high value may be configured by property
int h = 127;
String integerCacheHighPropValue =
sun.misc.VM.getSavedProperty("java.lang.Integer.IntegerCache.high");
if (integerCacheHighPropValue != null) {
int i = parseInt(integerCacheHighPropValue);
i = Math.max(i, 127);
// Maximum array size is Integer.MAX_VALUE
h = Math.min(i, Integer.MAX_VALUE - (-low) -1);
}
high = h;
cache = new Integer[(high - low) + 1];
int j = low;
for(int k = 0; k < cache.length; k++)
cache[k] = new Integer(j++);
}
You see that either a new Integer Object is created with "new Integer(i)" in the valueOf method ... ... or a reference to a Integer Object which is stored in the IntegerCache is returned.
In both cases x will reference to a new Object.
And this is why the reference to the Object in your list get lost when you call ...
x = Integer.valueOf(9);
Instead of doing so, in combination with a ListIterator use ...
i.set(Integer.valueOf(9));
... after you got the element you want to change with ...
i.next();