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Casting a Long to Integer

Posted on 2010-09-24
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Last Modified: 2012-05-10
I know that Long (int) isn't of type object. So perhaps cast is the wrong term to use.

But is there anyway to take an Object Integer and assign it a value Long?

Cheers
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Question by:directxBOB
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15 Comments
 
LVL 40

Accepted Solution

by:
gurvinder372 earned 200 total points
ID: 33753440
Long lnVar =  new Long(intValue);
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Author Comment

by:directxBOB
ID: 33753497
Sorry should have posted an example

Integer freeMemory = Runtime.getRuntime().freeMemory();
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LVL 40

Expert Comment

by:gurvinder372
ID: 33753527
Integer freeMemory = Runtime.getRuntime().freeMemory();

is not correct, since it already returns long
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LVL 40

Assisted Solution

by:gurvinder372
gurvinder372 earned 200 total points
ID: 33753534
it should be Long only

public class IntTOLong
{
      public static void main(String[] args) {
            // TODO Auto-generated method stub
            Long lnVar = Runtime.getRuntime().freeMemory();
      }
}


see this test program

let me know if i missed on something
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Author Comment

by:directxBOB
ID: 33753543
yes I'm trying to

assign Integer a Long Value
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Author Comment

by:directxBOB
ID: 33753550
The reason for using Integer is because I need to store these within a MAP and you cannot store an int or long in an MAP as it's not of type object.
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LVL 40

Expert Comment

by:gurvinder372
ID: 33753573
But it is already returning the Long only, so why int is required. Did you checked the program i shared.

Long can be put into map

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LVL 3

Assisted Solution

by:rgeers
rgeers earned 100 total points
ID: 33753601
Long freeMemory = Runtime.getRuntime().freeMemory();
int InVar = int myIntValue = unchecked((int)myLongValue);

It'll throw OverflowException in checked context if the value doesn't fit in an int
If you want to handle this you could check against int.MaxValue.
0
 
LVL 10

Assisted Solution

by:Hegemon
Hegemon earned 100 total points
ID: 33753629
int and long - (lowercase) - are primitives, that's why you can't store them in a map.
Integer and Long (uppercase) are classes wrapping these primitives.
You can put them into your map instead, i.e.

long longvalue = .... (get your long value);

myMap.put ("key", new Long (longValue));
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LVL 3

Expert Comment

by:rgeers
ID: 33753641
echange the myLongValue for freeMemory, well guess you understood this
0
 
LVL 10

Expert Comment

by:Hegemon
ID: 33753647
Sorry, the declaration should begin with

long longValue = ...
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LVL 3

Expert Comment

by:rgeers
ID: 33754618
Is your question about casting a Long to an int or about storing into a map?
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LVL 86

Expert Comment

by:CEHJ
ID: 33755693
There's no problem about Map
Map<Long, String> m = new HashMap<Long, String>();
m.put(Runtime.getRuntime().freeMemory(), "foo");


Map<String, Long> m = new HashMap<String, Long>();
m.put("foo", Runtime.getRuntime().freeMemory());

Open in new window

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LVL 92

Assisted Solution

by:objects
objects earned 100 total points
ID: 33757976
you cannout cast it, you'll need to create a new Long

Integer i = new Integer(5);
Long l = new Long(i.longValue());
0
 

Author Closing Comment

by:directxBOB
ID: 33838928
Thanks, the problem turned out that I was trying to put a primitive in the Map. So by using Integer and Long got around the problem. Thanks
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