Casting

[Tanuja Vaid]

Can you assign short to char

[Marilyn de Queiroz]

Did you try it?

[Corey McGlone]

Think about the range of a short and the range of a char. Are they the same? Are the sizes the same?
Now that you've got that, can you assign a char to a short, or vice versa?
Corey

[Tanuja Vaid]

Yes I have tried it! hence the question
I tried this code and i am getting a compiler error both on line 1 & 2
short s=10;
char c=s; //1
s=c; //2
Can somebody explain why?

[Tanuja Vaid]

Yes I have tried it! hence the question
I tried this code and i am getting a compiler error both on line 1 & 2
short s=10;
char c=s; //1
s=c; //2
Can somebody explain why?

[swapna sivaraju]

No u cant assign short to char or visa versa.
Look at their sizes....
char is 0-65535
and short is -2^17 to 2^17-1
Conversions between char and short are narrowing conversions...so cast is always needed.
so ,
short s=10;
char c=(short)s;//ok.
so this is why u get compiler error.
even conversions b/w byte and short are narrowing conversions.
Check out this topic in KM..its explained there very well.
hope this helps
Swapna

[Corey McGlone]

Well, since the compiler is giving you an error when you try to assign a short to a char, then I guess you can't do it - at least not implicitly. Why not?
What is the size of a char? 2 bytes
What is the size of a short? 2 bytes
Based on that, it would seem as if you could assign one to the other interchangably. Now, think about the other question. Are their ranges the same?
NO!
A char, in Java, is unsigned while a short is signed. Therefore, whenever you try to assign a short to a char (or vice versa) there is a chance for information loss. Due to this, the compiler views this as a narrowing conversion and refuses to implicitly perform a cast. However, if you're sure you're not going to lose data (or don't care), you can cast it explicitly, like this:

short s=10;
char c=(char)s; //1
s=(short)c; //2

I hope that helps,
Corey
By the way, please don't be upset by the answers that Marilyn and I gave earlier. From what you had posted, it looked like you had asked a question that can be looked up in any text. We were simply trying to get you to think about it - hence, really learn it. Personally, I'd rather teach you how to learn Java than teach you Java.

[Joshua Kueck]

This works though.. dont forget about compile time constants

  
final short s = 10;
char c=s; 

[Ryan Upton]

--final short s = 10;
--char c=s;
That works because the compiler can determine that the range of the final ``s" will always be in range for ``c" therefore no need to cast, right?

[Corey McGlone]

Originally posted by Ryan Upton:
--final short s = 10;
--char c=s;
That works because the compiler can determine that the range of the final ``s" will always be in range for ``c" therefore no need to cast, right?

Not quite. This works because the compiler knows that the variable s contains the value 10 (because of the attribute, final) so writing this is really no different than writing this:

final short s = 10;
char c = s;
// The above is basically the same as:
char c = 10;

That's why it works. The range of s (a short) is not always within the range of c (a char) because s can be negative while c can not.
Corey

[krussi rong]

so why the following cant work:
Long val= new Long("11");
byte b=(byte)val;

thanks
Krussi

[Corey McGlone]

Don't get confused between primitives and references. What you wrote can't work because you can't cast a reference type to a primitive type. The variable val references a Long object, while the variable b contains a byte primitive. This doesn't work for the same reason this doesn't work:

Object o = new Object();
byte b = (byte)o;

An object and a primitive are very different in Java.
However, if this is what you meant:

long var = 11;
byte b = (byte)l;

This will work. The cast of l to a byte is required because a long has a much wider range than a byte does. Therefore, if you try to assign a long to a byte, there is a chance for data loss - this is known as a narrowing conversion. But, if you apply the cast, you'll be fine.
Corey

[krussi rong]

I got it!
thanks Corey !

[Rob Ross]

Originally posted by Corey McGlone:

long var = 11;
byte b = (byte)l;

This will work. The cast of l to a byte is required because a long has a much wider range than a byte does.

I think you want your code to be this, to match your text description:
long l = 11;
byte b = (byte)l;
or
long var = 11;
byte b = (byte)var;
But not
long var = 11;
byte b = (byte)l;

[Corey McGlone]

Oops. Too many variables!