int -> byte , short , char

[ankur rathi]

byte b = 23 ;

a literal ( should not be fraction ) is always int by default . But compiler automatically cast if the literal is in limit of byte . So above statement is perfect .

byte b1 = 4;
byte b2 = 6;
byte add = b1+b2;

why it doesn't work . the result will come 10 ( it is int .. i am agree ) & it is in limit of byte , do why not compiler automatically cast in this case .

help needed .
thanks .

[Mani Ram]

The compiler doesn't cast it automatically, because b1 & b2 are variables, meaning they can vary at some time. So, the compiler is bothered. It is worried that problems might occur if the values of b1 and/or b2 changes at runtime, such that the sum of them will exceed the limits of byte.

If you can convince the compiler that the values of b1 & b2 will never change, the compiler will be happy and will keep quite.
You can convince the compiler by saying

final byte b1 = 4;
final byte b2 = 6;
 
byte add = b1 + b2;

[ January 06, 2005: Message edited by: Mani Ram ]

[ankur rathi]

ultimate explanation .
thank you very much .

[ankur rathi]

float f = 2.3;

why in this case compiler doesn't cast double to float implicitly . As 2.3 is in limit of float .

thanks .