What is the smallest number of bytes that can store a timestamp?

Question

I want to create my own time stamp data structure in C.

DAY ( 0 - 30 ), HOUR ( 0 - 23 ), MINUTE ( 0 - 59 )

What is the smallest data structure possible?

Answer 1

Well, you could pack it all in an unsigned short (That's 2 bytes, 5 bits for Day, 5 bits for hour, 6 bits for minute)... and use some shifts and masking to get the values.

unsigned short timestamp = <some value>; // Bits: DDDDDHHHHHMMMMMM

int day = (timestamp >> 11) & 0x1F;
int hour = (timestamp >> 6) & 0x1F;
int min = (timestamp) & 0x3F;

unsigned short dup_timestamp = (short)((day << 11) | (hour << 6) | min); 

or using macros

#define DAY(x)    (((x) >> 11) & 0x1F)
#define HOUR(x)   (((x) >> 6)  & 0x1F)
#define MINUTE(x) ((x)         & 0x3F)
#define TIMESTAMP(d, h, m) ((((d) & 0x1F) << 11) | (((h) & 0x1F) << 6) | ((m) & 0x3F)

(You didn't mention month/year in your current version of the question, so I've omitted them).

[Edit: use unsigned short - not signed short.]

Answer 2

Do you mean HOUR 0-23 and MINUTE 0-59? I've heard of leap seconds but not leap minutes or hours.

(log (* 31 60 24) 2)
=> 15.446

So you can fit these values 16 bits, or 2 bytes. Whether this is a good idea or not is a completely different question.