Bitwise rotate right of 4-bit value - c

Question

I'm currently trying to control a stepper motor using simple full steps. This means that I'm currently outputting a sequence of values like this:
1000
0100
0010
0001
I thought an easy way to do this was just take my 4-bit value and after each step, perform a rotate right operation. "Code" obviously isn't following any kind of syntax, it's simply there to illustrate my thoughts:
step = 1000;
//Looping
Motor_Out(step)
//Rotate my step variable right by 1 bit
Rotate_Right(step, 1)
My problem is that there obviously isn't any 4-bit simple data types that I can use for this, and if I use an 8-bit unsigned int I will eventually rotate the 1 off to the MSB, which means the 4-bit value I'm actually interested in, will turn into 0000 for a few steps.
I've read that you can use structs and bit-fields to solve this, but the majority of things I read from this is telling me that it's a very bad idea.

Answer 1

With only 4 possible values you would use a table with 9 elements:
unsigned char table_right[] = { [0x1] = 0x8 , [0x2] = 0x1 , [0x4] = 0x2 , [0x8] = 0x4 };
When you need the next value you simply use the current value as the index:
unsigned char current = 0x4; //value is: 0b0100
unsigned char next = table_right[current]; //returns: 0b0010
assert( next == 0x2 );
Doing this in a loop, will loop through all four possible values.
Conveniently, passing an invalid value, will return a zero, so you can write a get function that also asserts next != 0. You should also assert value < 9 before passing the value to the array.

Answer 2

Just use an int to hold the value. When you do the rotate copy the least significant bit to bit 4 and then shift it right by 1:
int rotate(int value)
{
value |= ((value & 1) << 4); // eg 1001 becomes 11001
value >>= 1; // Now value is 1100
return value;
}

Answer 3

The arithmetic for this is simple enough that it will always be faster than the table approach:
constexpr unsigned rotate_right_4bit ( unsigned value )
{
return ( value >> 1 ) | ( ( value << 3 ) & 15 );
}
This turns into 5 lines of branch-free x86 assembly:
lea eax, [0+rdi*8]
shr edi
and eax, 15
or eax, edi
ret
Or, alternatively, if you actually like to see the indexes {3, 2, 1, 0}, then you can split them up into 2 functions, one that "increments" the index, and the other that actually computes the value:
constexpr unsigned decrement_mod4 ( unsigned index )
{
return ( index - 1 ) & 3;
}
constexpr unsigned project ( unsigned index )
{
return 1u << index;
}

Answer 4

IMO the easiest way is:
const unsigned char steps[ 4 ] = { 0x08, 0x04, 0x02, 0x01 };
int stepsIdx = 0;
...
const unsigned char step = steps[ stepsIdx++ ];
stepsIdx = stepsIdx % ( sizeof( steps ) / sizeof( steps[ 0 ] ) );

Answer 5

you can use 10001000b and mod 10000b
and you can get 01000100b 00100010b 00010001b 10001000b repeat.
for example:
char x = 0x88;
Motor_Out(x & 0xf);
Rotate_Right(step, 1);

Answer 6

if I use an 8-bit unsigned int I will eventually rotate the 1 off to the MSB
So use a shift and reinitialize the bit you want when the value goes to zero. C doesn't have a rotate operation anyway, so you'll have to do at least two shifts. (And I suppose C++ doesn't have rotates either.)
x >>= 1;
if (! x) x = 0x08;
Simple, short to write, and obvious in what it does. Yes, it'll compile into a branch (unless the processor has a conditional move operation), but until you have the profiler output to tell you it's important, you just lost more time thinking about it than those processor cycles will ever amount to.

Answer 7

Use an 8-bit data type (like e.g. uint8_t). Initialize it to zero. Set the bit you want to set in the lower four bits of the byte (e.g. value = 0x08).
For each "rotation" take the LSB (least significant bit) and save it. Shift one step right. Overwrite the fourth bit with the bit you saved.
Something like this:
#include <stdio.h>
#include <stdint.h>
uint8_t rotate_one_right(uint8_t value)
{
unsigned saved_bit = value & 1; // Save the LSB
value >>= 1; // Shift right
value |= saved_bit << 3; // Make the saved bit the nibble MSB
return value;
}
int main(void)
{
uint8_t value = 0x08; // Set the high bit in the low nibble
printf("%02hhx\n", value); // Will print 08
value = rotate_one_right(value);
printf("%02hhx\n", value); // Will print 04
value = rotate_one_right(value);
printf("%02hhx\n", value); // Will print 02
value = rotate_one_right(value);
printf("%02hhx\n", value); // Will print 01
value = rotate_one_right(value);
printf("%02hhx\n", value); // Will print 08 again
return 0;
}
Live demonstration.

Answer 8

I would make an array with the values you need and load the correct value from the array. It will take you 4 bytes, it will be fast, and solve your problems even if you start using a different motor type.
for example:
const char values[4]={1,2,4,8};
int current_value = 0;
....
if(++current_value>=4)current_value=0;
motor = values[current_value];

Answer 9

You only need to output 1, 2, 4, and 8. So you can use a counter to mark which bit to set high.
Motor_Out(8 >> i);
i = (i + 1) & 3;
If you want to drive the motor at half steps, you can use an array to store the numbers you need.
const unsigned char out[] = {0x8, 0xc, 0x4, 0x6, 0x2, 0x3, 0x1, 0x9};
Motor_out(out[i]);
i = (i + 1) & 7;
And you can rotate a 4-bit integer like this.
((i * 0x11) >> 1) & 0xf