Problems with casting and short parameters in Java

Question

I am trying to understand the following code:

public class A {
   private void doTask()  {    
      doInt(99);
   }   

   public short doInt(short s)  { 
      return 100;
   }   
}

The code gives a compiler error on line "doInt(99)".

The method doInt(short) in the type A is not applicable for the arguments (int)

Can someone explain why it's giving me the error.

Possible duplicate http://stackoverflow.com/q/5106766/269514, http://stackoverflow.com/q/2294934/269514 — Gilberto, Mar 16 '16 at 14:07
More recent duplicate question http://stackoverflow.com/q/16505959/269514 — Gilberto, Mar 16 '16 at 14:30

score 7 · Answer 1 · answered Jun 16 '12 at 13:57

7

In java, the default type of an integer number is an int. Java will automatically upcast (eg from short to int), but will issue a warning with downcasting (eg from int to short) because of possible loss of data.

The fix for your code is this:

doInt((short)99);

This explicit downcast changes the 99 (which is an int) to a short, so there's no (further) possible loss of data.

answered Jun 16 '12 at 13:57

Bohemian

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thanks very much. I am not seeing a warning. Actually it gives a compiler error. – Alan2 Jun 16 '12 at 14:00

score 4 · Answer 2 · answered Jun 16 '12 at 13:57

4

Try

public class A {
   private void doTask()  {    
      doInt((short)99);
   }   

   public short doInt(short s)  { 
      return (short)100;
   }   
}

Java thinks that all literals (such as 99 or 100) in your program is int.

If you pass big number, 12345678 for example, it will be cutted to fit to short value. Java cannot cut number without you, so you must do it yourself. That way you see that you casting int to short

answered Jun 16 '12 at 13:57

alaster

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Casting 100 to short is not necessary. When you assign a numeric literal to a short and when its value is within the legal range, the literal is assumed to be a short. – Xion345 Jun 16 '12 at 14:01
Yes, but I thought in earlier version of java you need to do this – alaster Jun 16 '12 at 14:01

score 1 · Answer 3 · answered Jun 16 '12 at 13:58

1

You need to cast 99 to a short. Please try the following

public class A {
   private void doTask()  {    
      doInt((short) 99);
   }   

   public short doInt(short s)  { 
      return 100;
   }   
}

answered Jun 16 '12 at 13:58

Chetter Hummin

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score 1 · Answer 4 · edited Jun 20 '20 at 09:12

1

int data type is generally the default choice unless there is a reason.

And hence 99 is treated as an int.

Read more details on Primitive Data Types

You need explicit casting when required.
Change doInt( 99 ) to doInt( ( short ) 99 ) and it should be working.

edited Jun 20 '20 at 09:12

Community

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answered Jun 16 '12 at 14:00

Ravinder Reddy

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Vipul · Answer 5 · 2012-06-16T14:09:19.367

1

The main cause of error is

You cannot implicitly convert nonliteral numeric types of larger storage size to short

So change your code to following.

public class A {
   private void doTask()  {    
      doInt((short)99);
   }   

   public short doInt(short s)  { 
      return (short)100;
   }   
}

edited Jun 16 '12 at 14:09

answered Jun 16 '12 at 14:00

Vipul

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Problems with casting and short parameters in Java

5 Answers5