• FizzyOrange@programming.devBanned from community
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        3 months ago

        Nw. You’re also wrong about endianness. This function would be written exactly the same irrespective of endianness:

        uint32_t u16_high_low_to_u32(uint16_t high, uint16_t low) {
          return (high << 16) | low;
        }
        

        That is endian agnostic.