I have a char* string coming in. I need to store it accordingly.
The string can be any of those values { UK, GD, BD, ER, WR, FL}
If I want to keep them as enumerated type, which data type is the best to use. Like for 6 values three bits is enough, but how to store three bits in C?
What you want is a Bit Field:
typedef struct {
unsigned char val : 2; //use 2 bits
unsigned char : 6; // remaining 6 bits
} valContainer;
...
valContainer x;
x.val = GD;
Do note that there isn't really a way to store less than one byte, as the definition of a byte is the smallest amount of memory the computer can address. This is just a method of having names associated with different bits in a byte.
Also, of course, 2 bits is not enough for 6 values (2 bits hold 4 distinct values). So you really want at least 3 bits (8 distinct values).
Just store them as an unsigned short. Unless you're storing other things in your struct to fill out a whole word, you're WAY prematurely optimizing. The compiler will have to pad out your data anyway.
As the answer by Eric Finn suggests, you can use bit fields to store a data element of 3 bits. However, this is only good if you have something else to store in the same byte.
struct {
unsigned char value: 3;
unsigned char another: 4;
unsigned char yet_another: 5;
// 12 bits declared so far; 4 more "padding" bits are unusable
} whatever;
If you want to store an array of many such small elements, you have to do it in a different way, for example, clumping 10 elements in each 32-bit word.
int n = ...; // number of elements to store
uint32_t *data = calloc(n / 10, sizeof(*data));
for (int i = 0; i < n; i++)
{
int value = read_string_and_convert_to_int();
data[i / 10] &= ~(7 << (i % 10 * 3));
data[i / 10] |= value << (i % 10 * 3);
}
If you want to have only one element (or a few), just use enum or int.
value:2doesn't work? Also, with 6 values, two bits isn't enough. You need 3.int. Your code will have better performance if you just allow the compiler to use the platforms natural word size.