In the Long class, there is a handy static function numberOfTrailingZeros, which does almost what you want, except that it returns zero (instead of -1) when the input is not divisible by 2. You could handle that case in different ways. For example, extending the answer of @0x476f72616e if ((num & 0x1) == 0) return … Read more
With (val>>1)-(val & 1) you get your first result: 0 -> 0 1 -> -1 2 -> 1 The alternative result can be got with (val & 1)-(val>>1): 0 -> 0 1 -> 1 2 -> -1 0 solved Formula for Mapping three values [closed]
Let’s consider two ways to solve it that are two slow, and then merge them into one solution, that will be guaranteed to finish in milliseconds. Approach 1 (slow) Allocate an array v of size 2^16. Every time you add an element, do the following: void add(int s) { for (int i = 0; i … Read more
If you read about __builtin_clzll in http://gcc.gnu.org/onlinedocs/gcc-4.1.2/gcc/Other-Builtins.html — Built-in Function: int __builtin_clzll (unsigned long long x) Returns the number of leading 0-bits in x, starting at the most significant bit position. If x is 0, the result is undefined. From https://cs.stackexchange.com/a/29510, The maximum possible XOR of any two integers from an interval [l, r] can … Read more
this assumes 32bit int’s ((static_cast<unsigned>(-(array[0]*array[0]))&(1ul<<31))>>31) 1 solved Determine if int is not 0 with no if statements
Signed ints use Two’s complement in order to represent negative numbers. 1 represented in int32: 0000 0000 0000 0000 0000 0000 0000 0001 -1 represented in int32, using Two’s complement: 1111 1111 1111 1111 1111 1111 1111 1111 XORing the two results in: 1111 1111 1111 1111 1111 1111 1111 1110 which, for signed ints, … Read more
In C, in the result of the binary &, each bit depends on the two corresponding bits in the operands. If the bit in the same position is set (1) in both operands, it is set in the result. If it is clear (0) in either operand, it is clear in the result. For example, … Read more
The reason it works this way is because C and C++ used << for left shift and >> for right shift long before Java. Those languages have both signed and unsigned types, and for signed types the sign bit was propagated in the right-shift case. Java does not have unsigned types, so they kept the … Read more
binaryNum = binaryNum & ~ (1u << 7) here binaryNum is your binary representation of number. 7(7 because it starts from 0) is the ith position where you wants to change. solved How to change 8th position no to 0 in a 32 bit integer
The code directly translates to JS with the removal of the (int) typecast and the replacement of int with let/var/const. let tx = (texWidth * (floorX – cellX)) & (texWidth – 1); color = (color >> 1) & 8355711; // make a bit darker & is the bitwise and, >> is the bitshift right, explained … Read more
x |= -(x == 0); x |= -!x; x = x ? x : 0xFFFFFFFF; if (x == 0) x = 0xFFFFFFFF; … Benchmark and choose what’s appropriate for you 4 solved Bithack for: if x is 0, then x should be -1, else x should be untouched
You aren’t going to be able to avoid some calculation. int k = n % 10; will get you the last decimal digit, as that assignment gives k the remainder of division by 10. 3 solved How to get a right shifted number without actual calculation?
From your code, “™” is exactly what you should expect. // 0x19999999 – NOTE: you’d need an extra ‘5’ on the end to make it MAX_UNIT. eg. 0xFFFFFFFF uint32_t number = 429496729; uint8_t x1 = (number >> (8*0)) & 0xff; // 0x99 uint8_t x2 = (number >> (8*1)) & 0xff; // 0x99 uint8_t x3 = … Read more
There’s no single instruction that can do this. BMI1 blsi dst,src can isolate the lowest set bit, not the highest. i.e. x & -x. If x86 had a bit-reversed version of blsi, we could use that, but it doesn’t. But you can do much better than what you were suggesting. An all-zero input is always … Read more
I hope I understood your question right. You are looking for the XOR operator. C = A ^ B A 110101 B 101100 ——– C 011001 XOR will always be 1 if the two inputs are “different”. See: | A | B | XOR | |—+—+—–| | 0 | 0 | 0 | | 0 … Read more