This week's book giveaway is in the OO, Patterns, UML and Refactoring forum. We're giving away four copies of Refactoring for Software Design Smells: Managing Technical Debt and have Girish Suryanarayana, Ganesh Samarthyam & Tushar Sharma on-line! See this thread for details.
Hi, can someone explain to me what happens with following: byte blue = (byte)0x98; System.out.println("blue: "+ (byte) blue); System.out.println("blue2: "+ ( ((byte) blue) & 0xff)); The print result is -104 The second + 152 Why does the 0xff makes the bytes a unsigned byte ? Can someone tell me an online reference where I can learn more about basic bit and bytes calculation ? Regards, Stefan
Just in case the reasoning behind why the result was positive isn't quite yet clear... Note that when you cast the int result to a byte, the high-order (left-most) bits were lost. The left-most bit specifies whether the integral value is positive or negative. For further example, the int value 129 in bits is 0000 0000 0000 0000 0000 0000 1000 0001 If this int value were cast to a byte, the result would be to chop of (and feed to the cat - see Cat and Mouse Games with Bits) the left-most bits 1000 0001 which is -127. I haven't yet compiled a list of easy-to-read-and-understand explanations on bit manipulations (aside from the aforementioned campfire story). In the past, I've had success finding decent explanations by searching on Google. Good Luck.