What does it mean to have an "explicitly-endian" type? In the register, it's all the same thing. An integer type with an explicit representation in memory. In

What does it mean to have an "explicitly-endian" type? In the register, it's all the same thing. An integer type with an explicit representation in memory. In this case, "arithmetic works like a uint32_t, but when you're storing it to memory, do it in big-endian format". Follow the Rob Pike way [1] and regain sanity. This is the Rob Pike way, except it would involve the compiler writing the code for converting between uint32_t and uint8_t[4].