Can anyone direct me to some good sources on the Web dealing with the basics of binary numbers and how they work? I read the article in the Campfire about bit manipulation in Java and it was tremendously helpful, but I'm missing some basic things like why the heck *is* -1 represented as 11111111 in binary? Something along the lines of "Binary numbers for the village idiot" would be most helpful....thanks!

Integral numbers in Java take the form of a "signed 2's complement" value, as they do in C/C++. The high-order bit shows the sign of the value and is not part of the numeric value itself. Putting sign aside, -1 is then stored as the bitwise complement of +1. All positive integrals have 0 as the high-order bit. An integer in Java is 32 bits wide, but has only 31 bits of value, so: -1 = 11111111111111111111111111111111 1 = 00000000000000000000000000000001 It's not exactly a complement because 0 takes up a "positive" slot. This is also why you'll see the full range of an integer expressed this way: -2^31 to (2^31 - 1) There's nothing intuitive about it; just an implementation detail that maximized memory efficiency and has stuck around. This is probably more than you wanted, but I dig answering this question. Simple pleasures. You can of course search the web for "signed 2's complement" on google or whichever, but most matches are geared towards mathematics, not programming. ------------------ Michael Ernest, co-author of: [URL=http://www.amazon.com/exec/obidos/ASIN/0782128254/electricporkchop]The Complete Java [This message has been edited by Michael Ernest (edited September 27, 2001).]

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but I'm missing some basic things like why the heck *is* -1 represented as 11111111 in binary?

Hi Jody, I am sure explanations of others and the site links must have helped you, here is my attempt to simplify it. I hope you are clear with positive binary numbers and the way they are represented, simply put you can get decimal number from a binary number in this way - <table border="0" ><tr><td >0</td><td >0</td><td >0</td><td >0</td><td >0</td><td >1</td><td >0</td><td >1</td><td ></td></tr><tr><td >(128)</td><td >(64)</td><td >(32)</td><td >(16)</td><td >(8)</td><td >(4)</td><td >(2)</td><td >(1)</td><td >=> doubled on the left</td></tr><tr><td >2<sup>7</sup>)</td><td >(2<sup>6</sup>)</td><td >(2<sup>5</sup>)</td><td >(2<sup>4</sup>)</td><td >(2<sup>3</sup>)</td><td >(2<sup>2</sup>)</td><td >(2<sup>1</sup>)</td><td >(2<sup>0</sup>)</td><td ></td></tr></table> Now strike out the ones which have 0 above them and keep the ones which have 1 above them. Sum the numbers left after striking out numbers below 0. What you have is the decimal equivalent of the binary number. <table border="0" ><tr><td >0</td><td >0</td><td >0</td><td >0</td><td >0</td><td >1</td><td >0</td><td >1</td><td ></td></tr><tr><td >(128)</td><td >(64)</td><td >(32)</td><td >(16)</td><td >(8)</td><td >(4)</td><td >(2)</td><td >(1)</td><td ></td></tr></table> 4 + 1 = 5. YES!!! Now negative numbers in binary - In 2's complement system (as it is adpted by most modern computer languages to represent negative numbers), any negative number will always have its leftmost bit (MSB - Most significant Bit) set to 1. Consider a byte (8 bits) as we have seen in the above example. A byte can represent up to 256 positive numbers (2<sup>8</sup> i.e. 0 to 255). If we use it to represent signed numbers, we'll need to have leftmost bit (MSB)for sign determination. O means positive sign and 1 means negative sign. In case of negative numbers the leftmost bit (MSB) denotes the value in negative, i.e. for a signed byte having 1 in the MSB position gives it value -128 ( - 2<sup>7</sup> ). Rest of the bits denote positive value. SO we have - <table border="0" ><tr><td >1</td><td >0</td><td >0</td><td >0</td><td >0</td><td >1</td><td >0</td><td >1</td><td ></td></tr><tr><td >(-128)</td><td >(64)</td><td >(32)</td><td >(16)</td><td >(8)</td><td >(4)</td><td >(2)</td><td >(1)</td><td ></td></tr></table> -128 + 4 + 1 = -123 in decimal. Similarly, <table border="0" ><tr><td >1</td><td >1</td><td >1</td><td >1</td><td >1</td><td >1</td><td >1</td><td >1</td><td ></td></tr><tr><td >(-128)</td><td >(64)</td><td >(32)</td><td >(16)</td><td >(8)</td><td >(4)</td><td >(2)</td><td >(1)</td><td ></td></tr></table> -128 + 64 + 32 + 16 + 8 + 4 + 3 + 2 + 1 i.e -128 + 127 = -1 So 11111111 binary represents -1 decimal for a byte. Same is true for an integer as well, but an integer in java is 4 bytes (32 bits), hence 11111111 11111111 11111111 11111111 represents - 1 for a signed inter. Converting a bianry number to its 2's complement (negative number) is fairly simple. Let's say we want to find binary equivalent of -5 (minus 5). 5 is represented in binary as - 00000101

To find 2's complement, first we 'complement' the bits, converting 0 to 1 and 1 to 0. Next we add binary 1 to that number,the resulting number is the 2's complement of that number. 00000101=> 5 in binary

11111010=> complement (also called 1's complement) + 00000001=> add 1 ------------ 11111011 which is -128 + 123 = -5. Simple, isn't it You can use same method for finding 2's complement of -1 as well. (And now you also know the reason why 11111111 represents -1). I hope that now you'll be more comfortable with the binary numbers.

HTH, - Manish p.s. I don't know why I am having so many big gaps in between, I hope thet go after this edit!

[This message has been edited by Manish Hatwalne (edited September 28, 2001).]

Jody Frease
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Thank you everyone!! This is just what I needed, and light begins to dawn.... Jody