uint8_t vs unsigned char

It documents your intent - you will be storing small numbers, rather than a character.

Also it looks nicer if you're using other typedefs such as uint16_t or int32_t.

Just to be pedantic, some systems may not have an 8 bit type. According to Wikipedia:

An implementation is required to define exact-width integer types for N = 8, 16, 32, or 64 if and only if it has any type that meets the requirements. It is not required to define them for any other N, even if it supports the appropriate types.

So uint8_t isn't guaranteed to exist, though it will for all platforms where 8 bits = 1 byte. Some embedded platforms may be different, but that's getting very rare. Some systems may define char types to be 16 bits, in which case there probably won't be an 8-bit type of any kind.

Other than that (minor) issue, @Mark Ransom's answer is the best in my opinion. Use the one that most clearly shows what you're using the data for.

Also, I'm assuming you meant uint8_t (the standard typedef from C99 provided in the stdint.h header) rather than uint_8 (not part of any standard).

The whole point is to write implementation-independent code. unsigned char is not guaranteed to be an 8-bit type. uint8_t is (if available).