bit-shift

public class Shift { public static void main(String[] args) { for(int i = 0; i < 32; ++i){ System.out.println(-0x55555555 << i); } } } Running the above code gives the following output -1431655765 1431655766 -1431655764 1431655768 -1431655760 1431655776 -1431655744 1431655808 -1431655680 1431655936 -1431655424 1431656448 -1431654400 1431658496 -1431650304 1431666688 -1431633920 1431699456 -1431568384 1431830528 -1431306240 1432354816 -1430257664 1434451968 -1426063360 1442840576 -1409286144 1476395008 -1342177280 1610612736 -1073741824 -2147483648 While testing with an other value (64) gives a

There's a statement a co-worker of mine wrote which I don't completely understand. Unfortunately he's not available right now, so here it is (with modified names, we're working on a game in Unity). private readonly int FRUIT_LAYERS = (1 << LayerMask.NameToLayer("Apple")) | (1 << LayerMask.NameToLayer("Banana")); NameToLayer takes a string and returns an integer. I've always seen left shift operators used with the constant integer on the right side, not the left, and all the examples I'm finding via Google follow that approach. In this case, I think he's pushing Apple and Banana onto the same

Please take a look at these two pieces of pseudo-assembly code: 1) li $t0,53 sll $t1,$t0,2 srl $t2,$t0,2 sra $t3,$t0,2 print $t1 print $t2 print $t3 2) li $t0,-53 sll $t1,$t0,2 srl $t2,$t0,2 sra $t3,$t0,2 print $t1 print $t2 print $t3 in the first case the output is: 212 13 13 in the latter is: -212 107374... -14 But shouldn't : sra (-53) = - (srl 53) ? -53 = 1111111111001011 sra 2 1111111111110010(11) = -14 ^^ ^^ sign dropped extension Because the extra bits are simply dropped for both positive and negative results, the result is always rounded down if you view the shift as a division. 53 sra

May I know how can I do PHP >>> ? Such operators is not available in PHP, but is available in Javascript. I just managed to discover a function as follow: function zeroFill($a, $b) { $z = hexdec(80000000); if ($z & $a) { $a = ($a>>1); $a &= (~$z); $a |= 0x40000000; $a = ($a>>($b-1)); } else { $a = ($a>>$b); } return $a; } but unfortunately, it doesn't work perfectly. EG: -1149025787 >>> 0 Javascript returns 3145941509 PHP zeroFill() return 0 /** * The >>> javascript operator in php x86_64 * Usage: -1149025787 >>> 0 ---> rrr(-1149025787, 0) === 3145941509 * @param int $v * @param int $n *

What I need to do it implement both a bitwise left shift, and a bitwise right shift using LC-3 Assembly . Basically, every bit has to be moved over one space in the direction of the shift, and a zero fills the empty space created. Examples: Right Shift: 01001001 00100100→ Left Shift: 01001001 ←10010010 I've successfully implemented a left shift, by taking the binary string, and adding it to itself. I'm stumped on how to perform a right shift. Any thoughts would be greatly appreciated. I have AND, NOT, ADD operations, data movement operations, seven registers to store values and whole range of

How are bit shifts implemented at the hardware level when the number to shift by is unknown? I can't imagine that there would be a separate circuit for each number you can shift by (that would 64 shift circuits on a 64-bit machine), nor can I imagine that it would be a loop of shifts by one (that would take up to 64 shift cycles on a 64-bit machine). Is it some sort of compromise between the two or is there some clever trick? The circuit is called a " barrel shifter " - it's a load of multiplexers basically. It has a layer per address-bit-of-shift-required, so an 8-bit barrel shifter needs