here is my doubt about Integer literal.... help

[aditya makam]

Hi everybody,

This is aditya makam. I am a new member of this group. I hope u ppl help me. Here is my doubt.

In java if you compile the following code.

class ByteTest {
public static void main ( String args [ ] ) {
ByteTest t = new ByteTest(10);
}
};

It will give CompileTime error.Because compiler is thinking that 10 is an int literal,but in Byte class only two constructors are there which take arguments as byte ,String.But if consider the following code,


class ByteTest1 {
public static void main ( String args [ ] ) {
byte b =10;
System.out.println(b);
}
};

It will get compiled and will print 10 as output.

My doubt is ,here also 10 is an integer literal and we are trying to store in byte variable.But it is not giving compile time error.Why is it so?

Plz anybody who can explain plz give me reply...to this message.


yours friendly
aditya

[Keith Lynn]

This is from the Java Language Specification.

Assignment conversion occurs when the value of an expression is assigned (�15.25) to a variable: the type of the expression must be converted to the type of the variable. Assignment contexts allow the use of an identity conversion (�5.1.1), a widening primitive conversion (�5.1.2), or a widening reference conversion (�5.1.4). In addition, a narrowing primitive conversion may be used if all of the following conditions are satisfied:

* The expression is a constant expression of type int.
* The type of the variable is byte, short, or char.
* The value of the expression (which is known at compile time, because it is a constant expression) is representable in the type of the variable.

[J Sato]

The reason the first example doesn't work is because there is no overloaded constructor that takes a byte. I've added the constructor in this example:

class ByteTest {
byte b;
 
public ByteTest(byte b) {
this.b = b;
}
 
public static void main ( String args [ ] ) {
byte myByte = 11;
ByteTest t = new ByteTest((byte) 10);
ByteTest t2 = new ByteTest(myByte);
System.out.println("t.b = " + t.b);
System.out.println("t2.b = " + t2.b);
}
}

Note that when ByteTest t is created, it's argument must be cast as a byte or it won't compile, because the compiler sees 10 as an int.

Your second example works because you declare and initialize byte b and set it to 10. Since the value 10 is a valid byte, the code complies.

Hopefully that made sense and I know what I'm talking about...

[aditya makam]

Thank you very much for your reply