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bitshift left (<<), bitshift right (>>)
Description
From The Bitmath Tutorial in The Playground
There are two bit shift operators in C++: the left shift operator << and the right shift operator >>. These operators cause the bits in the left operand to be shifted left or right by the number of positions specified by the right operand.
More on bitwise math may be found here.
Syntax
variable << number_of_bits
variable >> number_of_bits
Parameters
variable - (byte, int, long) number_of_bits integer <= 32
Example:
int a = 5; // binary: 0000000000000101
int b = a << 3; // binary: 0000000000101000, or 40 in decimal
int c = b >> 3; // binary: 0000000000000101, or back to 5 like we started with
When you shift a value x by y bits (x << y), the leftmost y bits in x are lost, literally shifted out of existence:
int a = 5; // binary: 0000000000000101
int b = a << 14; // binary: 0100000000000000 - the first 1 in 101 was discarded
If you are certain that none of the ones in a value are being shifted into oblivion, a simple way to think of the left-shift operator is that it multiplies the left operand by 2 raised to the right operand power. For example, to generate powers of 2, the following expressions can be employed:
1 << 0 == 1
1 << 1 == 2
1 << 2 == 4
1 << 3 == 8
...
1 << 8 == 256
1 << 9 == 512
1 << 10 == 1024
...
When you shift x right by y bits (x >> y), and the highest bit in x is a 1, the behavior depends on the exact data type of x. If x is of type int, the highest bit is the sign bit, determining whether x is negative or not, as we have discussed above. In that case, the sign bit is copied into lower bits, for esoteric historical reasons:
int x = -16; // binary: 1111111111110000
int y = x >> 3; // binary: 1111111111111110
This behavior, called sign extension, is often not the behavior you want. Instead, you may wish zeros to be shifted in from the left. It turns out that the right shift rules are different for unsigned int expressions, so you can use a typecast to suppress ones being copied from the left:
int x = -16; // binary: 1111111111110000
int y = (unsigned int)x >> 3; // binary: 0001111111111110
If you are careful to avoid sign extension, you can use the right-shift operator >> as a way to divide by powers of 2. For example:
int x = 1000;
int y = x >> 3; // integer division of 1000 by 8, causing y = 125.
Reference Home
Corrections, suggestions, and new documentation should be posted to the Forum.
The text of the Arduino reference is licensed under a Creative Commons Attribution-ShareAlike 3.0 License. Code samples in the reference are released into the public domain.
взято с http://arduino.cc/en/Reference/Bitshift
| Установка бита | Сброс бита | Инверсия байта | Инверсия отдельных бит | |
| Исходное значение | 10001000 | 10001000 | 10001000 | 10001000 |
| Операция | OR (|) | AND (&) | NOT (~) | XOR (^) |
| Битовая маска | 00010000 | 01111111 | n/a | 11000000 |
| Результат | 10011000 | 00001000 | 01110111 | 01001000 |
bitshift left (<<), bitshift right (>>)
Description
From The Bitmath Tutorial in The Playground
There are two bit shift operators in C++: the left shift operator << and the right shift operator >>. These operators cause the bits in the left operand to be shifted left or right by the number of positions specified by the right operand.
More on bitwise math may be found here.
Syntax
variable << number_of_bits
variable >> number_of_bits
Parameters
variable - (byte, int, long) number_of_bits integer <= 32
Example:
int a = 5; // binary: 0000000000000101
int b = a << 3; // binary: 0000000000101000, or 40 in decimal
int c = b >> 3; // binary: 0000000000000101, or back to 5 like we started with
When you shift a value x by y bits (x << y), the leftmost y bits in x are lost, literally shifted out of existence:
int a = 5; // binary: 0000000000000101
int b = a << 14; // binary: 0100000000000000 - the first 1 in 101 was discarded
If you are certain that none of the ones in a value are being shifted into oblivion, a simple way to think of the left-shift operator is that it multiplies the left operand by 2 raised to the right operand power. For example, to generate powers of 2, the following expressions can be employed:
1 << 0 == 1
1 << 1 == 2
1 << 2 == 4
1 << 3 == 8
...
1 << 8 == 256
1 << 9 == 512
1 << 10 == 1024
...
When you shift x right by y bits (x >> y), and the highest bit in x is a 1, the behavior depends on the exact data type of x. If x is of type int, the highest bit is the sign bit, determining whether x is negative or not, as we have discussed above. In that case, the sign bit is copied into lower bits, for esoteric historical reasons:
int x = -16; // binary: 1111111111110000
int y = x >> 3; // binary: 1111111111111110
This behavior, called sign extension, is often not the behavior you want. Instead, you may wish zeros to be shifted in from the left. It turns out that the right shift rules are different for unsigned int expressions, so you can use a typecast to suppress ones being copied from the left:
int x = -16; // binary: 1111111111110000
int y = (unsigned int)x >> 3; // binary: 0001111111111110
If you are careful to avoid sign extension, you can use the right-shift operator >> as a way to divide by powers of 2. For example:
int x = 1000;
int y = x >> 3; // integer division of 1000 by 8, causing y = 125.
Reference Home
Corrections, suggestions, and new documentation should be posted to the Forum.
The text of the Arduino reference is licensed under a Creative Commons Attribution-ShareAlike 3.0 License. Code samples in the reference are released into the public domain.
взято с http://arduino.cc/en/Reference/Bitshift
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