Store 4 different values in a byte

Question

I have an assignment to do, but I have no clue where to start. I am not expecting and definitely do not want answers in code. I would like some guidance as in what to do because

Answer 1

Here is another way of doing it.

    unsigned int i = 0;

    engine_on = 1;
    gear_pos =2;
    key_pos = 2;
    brake1 = 1;
    brake2 = 1;

    i |= s1.brake2;
    i |= (s1.brake1 << 1);
    i |= (s1.key_pos << 2);
    i |= (s1.gear_pos << 4);
    i |= (s1.engine_on << 7);

Answer 2

Have you read about bit field?

struct s {
  unsigned char engine_on : 1;
  unsigned char gear_pos : 3;
  unsigned char key_pos : 2;
  unsigned char brake1 : 1;
  unsigned char brake2 : 1;
};

Answer 3

lets call a byte b, if you set b to 0, you end up with (binary) 0000 0000 (space for readability)

Now we want to pack the different parts into this byte

engine_on  0-1    1
gear_pos   0-4    3 
key_pos    0-2    2 
brake1     0-1    1
brake2     0-1    1

brake2 is simple. We can just set b to the value of brake2 and we will end up with 0000 0000, or 0000 0001 depending on if it is a 0 or a 1.

now we want to set brake 1 to b. We can do this by using a or/equal and the number itself but bitshifted to the right position. We end up with the following:

b |= (brake1 << 1) 

lets explain how I came at this:

brake1 = 0000 0001 //lets assume its a 1 not a 0)
(brake1 << 1) = 0000 0010
b = 0000 0001 //lets assume brake 2 was 1.

to 'add' the value from brake1 to b we have to set each bit if either the bit in b is 1 or if the bit in (brake1 << 1) is one. This is done by a 'bitwise-or', so we end up with:

b = b | (brake1 << 1) // which can also be written as:
b |= (brake1 << 1)

now you can also add the other parts, it also works with more bits at the same time. I hope this has been helpful