CODEWITHSUNDEEP
c++cbit-manipulation
nimsson
nimsson

Reputation: 950

gcc rotate with carry

I want to rotate a byte (very important that it's 8 bits). I know that Windows provides a function, _rotr8 to complete this task. I want to know how I can do this in Linux because I am porting a program there. I ask this because I need to mask bits to 0. An example is this:

#define set_bit(byte,index,value) value ? \ //masking bit to 0 and one require different operators The index is used like so: 01234567 where each number is one bit
        byte |= _rotr8(0x80, index) : \ //mask bit to 1
        byte &= _rotr8(0x7F, index) //mask bit to 0

The second assignment should illustrate the importance of the 8 bit carry rotate (01111111 ror index)

Upvotes: 0

Views: 962

Answers (1)

Paul R
Paul R

Reputation: 213059

Although it's fairly simple to rotate a byte, this is a classic example of an XY problem, in that you don't actually need to rotate a byte at all in order to implement your set_bit macro. A simpler and more portable implementation would be:

#define set_bit(byte,index,value) value ? \ 
        byte |= ((uint8_t)0x80 >> index) : \  // mask bit to 1
        byte &= ~((uint8_t)0x80 >> index)     // mask bit to 0

Better yet, since this is 2014 and not 1984, make this an inline function rather than a macro:

inline uint8_t set_bit(uint8_t byte, uint8_t index, uint8_t value)
{
    const uint8_t mask = (uint8_t)0x80 >> index;
    return value ?
        byte | mask :   // mask bit to 1
        byte & ~mask;   // mask bit to 0
}

Upvotes: 6

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