Is it safe to copy the bytes of an object out to an array and back again

Question

Is it safe to copy the bytes of an object out to an array of unsigned char and then back again using memcpy, and does it leave the object unchanged?

That is, is the following safe for any t:

template <typename T>
void copy_back_and_forth(T& t) {
  unsigned char buf[sizeof(T)];
  std::memcpy(buf, &t, sizeof(T));
  std::memcpy(&t, buf, sizeof(T));
}

... and does it leave t unchanged?

Answer 1

The value of t is guaranteed to be unchanged if T is trivially copyable type, and t is not a potentially-overlapping subobject. Standard quote from latest draft:

[basic.types]

For any object (other than a potentially-overlapping subobject) of trivially copyable type T, whether or not the object holds a valid value of type T, the underlying bytes ([intro.memory]) making up the object can be copied into an array of char, unsigned char, or std​::​byte ([cstddef.syn]).³⁷ If the content of that array is copied back into the object, the object shall subsequently hold its original value.

37) By using, for example, the library functions ([headers]) std​::​memcpy or std​::​memmove.

In fact, the standard has a nearly identical example:

[ Example:

constexpr std::size_t N = sizeof(T);
char buf[N];
T obj;                          // obj initialized to its original value
std::memcpy(buf, &obj, N);      // between these two calls to std​::​memcpy, obj might be modified
std::memcpy(&obj, buf, N);      // at this point, each subobject of obj of scalar type holds its original value

— end example ]