Diff between two

[Sireesha Mullapudi]

Hi
Why the line2 is giving compile time error.But why not line1

int x=0x80000000;//line1
System.out.println(x);
byte y=0x80;//line2
System.out.println(y);
    

Thanks,scjp1.4

[Keith Lynn]

Because the first constant can fit into an int.

0x80 = 128 which is too large to fit into a byte.

[Sireesha Mullapudi]

Hi,
but ox80000000=2147483648,then how is it possible.Doesn't it requires explicit cast

[Praveen Babu]

Originally posted by lakshmi amulya:
Hi,
but ox80000000=2147483648,then how is it possible.Doesn't it requires explicit cast

See this,

http://java.sun.com/docs/books/jls/third_edition/html/conversions.html#5.2

[marc weber]

Originally posted by lakshmi amulya:
... but ox80000000=2147483648...

Yes, if converted directly to base 10.

But the binary representation of 0x80000000 is a one followed by 31 zeros, and that value assigned to a Java int is...

[Sanjeev Singh]

Laxmi,

but ox80000000=2147483648,then how is it possible.Doesn't it requires explicit cast

There is not precision loss as such in this case the literal is already occupying all the 32 bits designated for integar literals.

[Prashant kumar Singh]

hi runchers,
i also want to focous on

   int x=0x80000000;//line1 

In above code there is no error or fault.Actually , In java every operation is happened implicitly on 32 bit register means in int type.
So in above 0x80000000, compiler will treat it as (int)0x80000000.if there be any suffix like L or l then we have to cast it.

thanks,
Prashant Kumar Singh