bit-manipulation

A buddy of mine had these puzzles and this is one that is eluding me. Here is the problem, you are given a number and you want to return that number times 3 and divided by 16 rounding towards 0. Should be easy. The catch? You can only use the ! ~ & ^ | + << >> operators and of them only a combination of 12. int mult(int x){ //some code here... return y; } My attempt at it has been: int hold = x + x + x; int hold1 = 8; hold1 = hold1 & hold; hold1 = hold1 >> 3; hold = hold >> 4; hold = hold + hold1; return hold; But that doesn't seem to be working. I think I have a problem of losing bits but I

If you have the binary number 10110 how can I get it to return 5? e.g a number that tells how many bits are used? There are some likewise examples listed below: 101 should return 3 000000011 should return 2 11100 should return 5 101010101 should return 9 How can this be obtained the easiest way in Java? I have come up with the following method but can i be done faster: public static int getBitLength(int value) { if (value == 0) { return 0; } int l = 1; if (value >>> 16 > 0) { value >>= 16; l += 16; } if (value >>> 8 > 0) { value >>= 8; l += 8; } if (value >>> 4 > 0) { value >>= 4; l += 4; } if

This is part of a puzzle that I can't figure out. The function takes in three inputs. The first is an int, the second is the lower bound and the third is the upper bound. I need to test to see if that first number is within the lower and upper bound inclusive. If it is in range then return 1, else return 0. The catch is that I can only use ! ~ & ^ | + << >> operations and only a combination of 20 of them.. Also, only int variables may be used, and no if statements, loops or function calls. Range(int x, int lower, int upper){ //... some code here return retVal; } Obviously I understand the

I've a few lines of code within a project, that I can't see the value of... buffer[i] = (currentByte & 0x7F) | (currentByte & 0x80); It reads the filebuffer from a file, stored as bytes, and then transfers then to buffer[i] as shown, but I can't understand what the overall purpose is, any ideas? Thanks As the other answers already stated, (currentByte & 0x7F) | (currentByte & 0x80) is equivalent to (currentByte & 0xFF) . The JLS3 15.22.1 says this is promoted to an int : When both operands of an operator &, ^, or | are of a type that is convertible (§5.1.8) to a primitive integral type, binary

Do you think there is room for optimizations in the function haswon (see below)? I recognized that changing the argument type from __int64 to unsigned __int64 made the function faster, thus i thougt maybe there is still a chance for optimization. In more detail: I am writing a connect four game. Recently i used the Profiler Very Sleepy and recognized that the function haswon uses much of the cpu-time. The function uses a bitboard-representation of the connect-four-board for one player. The function itself i found in the sources of the fourstones benchmark. The bitboard representation is

OK, it may sound a bit complicated, but this is what I'm trying to do : Take e.g. 10101010101 And return { 0, 2, 4, 6, 8, 10 } - an array with all of the positions of bits which are set This is my code : UINT DQBitboard::firstBit(U64 bitboard) { static const int index64[64] = { 63, 0, 58, 1, 59, 47, 53, 2, 60, 39, 48, 27, 54, 33, 42, 3, 61, 51, 37, 40, 49, 18, 28, 20, 55, 30, 34, 11, 43, 14, 22, 4, 62, 57, 46, 52, 38, 26, 32, 41, 50, 36, 17, 19, 29, 10, 13, 21, 56, 45, 25, 31, 35, 16, 9, 12, 44, 24, 15, 8, 23, 7, 6, 5 }; static const U64 debruijn64 = 0x07EDD5E59A4E28C2ULL; #pragma warning

Thanks to everyone in advance - alert((~1).toString(2)); outputs: -10 But in PHP/Java it outputs 11111111111111111111111111111110 Am I missing something, why does Javascript add a "-" to the output? Thx, Sam I know Java uses two's complement to represent negative numbers, and 11111111111111111111111111111110 in binary, which is what ~1 gives, represents -2. Or, represented in binary with a negative sign, -10, which is what you got. The way you calculate the negative of 10 (in base 2) using two's complement is that you first invert all of the bits, giving you: 11111111111111111111111111111101