How to remove the first 3 bytes and last one byte

Question

code :

s64 end_time;
struct timespec ts;
getrawmonotonic(&ts);
end_time = timespec_to_ns(&ts);

How to remove the first three bytes from end_time and last one byte from it?? I want to store it in a uint32. could someone tell me how to do that??

uint32 latency;
fscanf(fp, "%lu\n", latency);  //fp  is reading the end_time and storing in latency.
latency = (uint32) (latency >> 8) & 0xFFFFFFFF;

Answer 1

You can do that with bit shifting. You have to shift the value 8 bits (= 1 byte) to the right, which is done with the >> operator:

uint32_t t = (uint32_t)(end_time >> 8);
//                               ^^
//                          bit-shifting

In the following, the bytes are visualized for a better understanding. If the value end_time consisted of eight bytes with the symbolic values A B C D E F G H, what you want is D E F G:

end_time:                     A B C D E F G H
end_time >> 8:                0 A B C D E F G
(uint32_t)(end_time >> 8):            D E F G

Answer 2

How about:

u32 end_time32 = (u32) (end_time >> 24) & 0xFFFFFFFF;

Depending on your definition of first and last byte it could also be:

u32 end_time32 = (u32) (end_time >> 8) & 0xFFFFFFFF;

Example:

s64 end_time = 0x1234567890ABCDEF;
u32 end_time32 = (u32) (end_time >> 24) & 0xFFFFFFFF;

// end_time32 is now: 0x34567890

s64 end_time = 0x1234567890ABCDEF;
u32 end_time32 = (u32) (end_time >> 8) & 0xFFFFFFFF;

// end_time32 is now: 0x7890ABCD

Edit

After your updated question:

s64 latency;
fscanf(fp, "%lld", latency);  //fp  is reading the end_time and storing in latency.
u32 latency32 = (uint32) (latency >> 8) & 0xFFFFFFFF;

Answer 3

I assume by "first" and "last" you mean "most significant" and "least significant", respectively.

I.e., you have the 8 bytes:

76543210

and want to map it to the 4 bytes:

4321

This is easiest done by a shift, a mask, and a (truncating) cast:

const uint32_t time32 = (uint32_t) ((end_time >> 8) & 0xffffffff);

The mask is very likely to be optimized out by the compiler but makes it very clear what's going on.