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Home/ Questions/Q 7949941
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T02:06:56+00:00 2026-06-04T02:06:56+00:00

I have an unsigned char array whose size is 6. The content of the

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I have an unsigned char array whose size is 6. The content of the byte array is an integer (4096*number of seconds since Unix Time). I know that the byte array is big-endian.

Is there a library function in C that I can use to convert this byte array into int_64 or do I have to do it manually?

Thanks!

PS: just in case you need more information, yes, I am trying to parse an Unix timestamp. Here is the format specification of the timestamp that I dealing with.

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  1. Editorial Team
    Editorial Team
    2026-06-04T02:06:57+00:00Added an answer on June 4, 2026 at 2:06 am

    A C99 implementation may offer uint64_t (it doesn’t have to provide it if there is no native fixed-width integer that is exactly 64 bits), in which case, you could use:

    #include <stdint.h>
    
    unsigned char data[6] = { /* bytes from somewhere */ };
    uint64_t result = ((uint64_t)data[0] << 40) |
                      ((uint64_t)data[1] << 32) |
                      ((uint64_t)data[2] << 24) |
                      ((uint64_t)data[3] << 16) |
                      ((uint64_t)data[4] << 8)  |
                      ((uint64_t)data[5] << 0);
    

    If your C99 implementation doesn’t provide uint64_t you can still use unsigned long long or (I think) uint_least64_t. This will work regardless of the native endianness of the host.

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