uint64是否占用8个字节的存储空间？

Question

Official document says uint64 is an unsigned integer of 64-bits, does that mean any uint64 number should take 8 bytes storage, no matter how small or how large it is?

Edit:

Thanks for everyone's answer!

I raised the doubt when I noticed that binary.PutUvarint consumes up to 10 bytes to store a large uint64, despite that maximum uint64 should only take 8 bytes.

I then found answer to my doubt in the source code of Golang lib:

Design note:
// At most 10 bytes are needed for 64-bit values. The encoding could
// be more dense: a full 64-bit value needs an extra byte just to hold bit 63.
// Instead, the msb of the previous byte could be used to hold bit 63 since we
// know there can't be more than 64 bits. This is a trivial improvement and
// would reduce the maximum encoding length to 9 bytes. However, it breaks the
// invariant that the msb is always the "continuation bit" and thus makes the
// format incompatible with a varint encoding for larger numbers (say 128-bit).

Answer 1

According to http://golang.org/ref/spec#Size_and_alignment_guarantees:

type                                 size in bytes

byte, uint8, int8                     1
uint16, int16                         2
uint32, int32, float32                4
uint64, int64, float64, complex64     8
complex128                           16

So, yes, uint64 will always take 8 bytes.