std::copy_n and reinterpret_cast

Question

uint    data1;
ushort  data2;
ushort  data3;
uchar   data4[8];

std::uint8_t buff[16];
std::uint8_t* out = buff;

out = std::copy_n(reinterpret_cast<std::uint8_t*>(&quid.data1), 4, out);
out = std::copy_n(reinterpret_cast<std::uint8_t*>(&quid.data2), 2, out);
out = std::copy_n(reinterpret_cast<std::uint8_t*>(&quid.data3), 2, out);

std::copy_n(quid.data4, 8, out);

Why will the result in out will be different if I don't use reinterpret_cast?

Answer 1

The result will be different because &x has type T *, where T is the type of x, and pointer arithmetic treats + 1 as "advancing the pointer by sizeof(T)", so that in effect you treat a pointer as a pointer into an array of elements of that type.

If you change the type of the pointer, you're going to treat the memory it's pointing to as an array of elements of a different type -- for example, treating an int as an array of chars.

Answer 2

std::copy_n(&quid.data1, 4, out);

works as if quid.data1 were declared as uint data1[4]. The result is that quid.data1 is copied to out[0] and the 3 other elements of out get a garbage.

std::copy_n(reinterpret_cast<std::uint8_t*>(&quid.data1), 4, out);

treats the contents of data1 as an array of 4 chars, that would work if sizeof(uint)==4.