byte array initialization with Hex values

Of course you can explicitly cast the >127 values to keep the compiler's

mouth shut:byte[] array= { (byte)0xff, (byte)0x80, 0x7f, 0x00 };

kind regards,

Jos

JosAHa at 2007-7-9 5:07:39 >

> > > Jos - Kaj : 1 - 0 :-P

> >

> > Don't you mean

> >

> > 11111111 - 10000001 ;)

>

> That would still make it

>

> -1 and -127.So mathematically Jos's score is bigger :D

That was the point :)

kajbja at 2007-7-9 5:07:39 >

> My colluege got back from lunch and told me that byte alwas get

> "converted" to ints. Hmm, lots to learn in the basics.

It's a bit more subtle than that. Given a binary expression "L op R" where

L and R are both operands to the operator "op". If both L and R are

narrower than (or equal to) an int, both operand types will be cast to ints.

This casting is performed by using sign extension, i.e. the sign bit of

the originial type will be used for all the bits in the wider type.

If one of the types is wider than an int, e.g. a long, strange things can

happen if you don't pay attention. Have a look at this example:int x= 0x7fffffff;

int y= 0x7fffffff;

long z= 2;

System.out.println(x+y+z);

System.out.println(x+(y+z));

> Thanks for you help, I will spread my dukes among you :-)

Thanks for the duke thingies.

kind regards,

Jos

JosAHa at 2007-7-9 5:07:40 >