c get nth byte of integer
int x = (number >> (8*n)) & 0xff; where n is 0 for the first byte, 1 for the second byte, etc.
bit-shift
Why does left shift operation invoke Undefined Behaviour when the left side operand has negative value?
The paragraph you copied is talking about unsigned types. The behavior is undefined in C++. From the last C++0x draft: The value of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are zero-filled. If E1 has an unsigned type, the value of the result is E1 × 2E2, reduced modulo one more … Read more
What is the JavaScript >>> operator and how do you use it?
It doesn’t just convert non-Numbers to Number, it converts them to Numbers that can be expressed as 32-bit unsigned ints. Although JavaScript’s Numbers are double-precision floats(*), the bitwise operators (<<, >>, &, | and ~) are defined in terms of operations on 32-bit integers. Doing a bitwise operation converts the number to a 32-bit signed … Read more
What does AND 0xFF do?
if byte1 is an 8-bit integer type then it’s pointless – if it is more than 8 bits it will essentially give you the last 8 bits of the value: 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 & 0 0 0 0 0 0 0 0 … Read more
What does a bitwise shift (left or right) do and what is it used for?
Here is an applet where you can exercise some bit-operations, including shifting. You have a collection of bits, and you move some of them beyond their bounds: 1111 1110 << 2 1111 1000 It is filled from the right with fresh zeros. 🙂 0001 1111 >> 3 0000 0011 Filled from the left. A special … Read more
warning: left shift count >= width of type
long may be a 64-bit type, but 1 is still an int. You need to make 1 a long int using the L suffix: unsigned long x = 1UL << 32; (You should also make it unsigned using the U suffix as I’ve shown, to avoid the issues of left shifting a signed integer. There’s … Read more
What is (x & 1) and (x >>= 1)?
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
Arithmetic bit-shift on a signed integer
Right shift of a negative signed number has implementation-defined behaviour. If your 8 bits are meant to represent a signed 8 bit value (as you’re talking about a “signed 32 bit integer” before switching to 8 bit examples) then you have a negative number. Shifting it right may fill “empty” bits with the original MSB … Read more
Why doesn’t left bit-shift, “
This is caused due to a combination of an undefined behaviour in C and the fact that code generated for IA-32 processors has a 5 bit mask applied on the shift count. This means that on IA-32 processors, the range of a shift count is 0-31 only. 1 From The C programming language 2 The … Read more
Java: Checking if a bit is 0 or 1 in a long
I’d use: if ((value & (1L << x)) != 0) { // The bit was set } (You may be able to get away with fewer brackets, but I never remember the precedence of bitwise operations.)