That’s indeed an interesting corner case. It only occurs here because you use uint16_t for the unsigned type when you architecture use 32 bits for ìnt Here is a extract from Clause 5 Expressions from draft n4296 for C++14 (emphasize mine): 10 Many binary operators that expect operands of arithmetic or enumeration type cause conversions … Read more
The operation of “And” will be performed only if mask[i] doesn’t equal zero, else the the result of and operation will be zero. The mask should be either white or black image with single channel. you can see this link http://docs.opencv.org/2.4.13.2/modules/core/doc/operations_on_arrays.html?highlight=bitwise#bitwise-and
He meant that taking number mod 2^n is equivalent to stripping off all but the n lowest-order (right-most) bits of number. For example, if n == 2, number number mod 4 00000001 00000001 00000010 00000010 00000011 00000011 00000100 00000000 00000101 00000001 00000110 00000010 00000111 00000011 00001000 00000000 00001001 00000001 etc. So in other words, number … Read more
Let me assume that negative value is represented using two’s complement. In this case, -i can be calculated as (~i)+1 (flip bits, then add 1). For example, let me consider i = 44. Then, in binary, i = 0000 0000 0000 0000 0000 0000 0010 1100 ~i = 1111 1111 1111 1111 1111 1111 1101 … Read more
These are Bitwise Operators (reference). x & 1 produces a value that is either 1 or 0, depending on the least significant bit of x: if the last bit is 1, the result of x & 1 is 1; otherwise, it is 0. This is a bitwise AND operation. x >>= 1 means “set x … Read more
You have named functions for them. Directly from Kotlin docs Bitwise operations are represented by functions that can be called in infix form. They can be applied only to Int and Long. for example: val x = (1 shl 2) and 0x000FF000 Here is the complete list of bitwise operations: shl(bits) – signed shift left … Read more
It sets result to the (unsigned) value resulting from putting the 8 bits of value in the lowest 8 bits of result. The reason something like this is necessary is that byte is a signed type in Java. If you just wrote: int result = value; then result would end up with the value ff … Read more