go - uint64 是否需要 8 个字节的存储空间？

Question

官方文档说uint64是一个64位的无符号整数，是不是说任何一个uint64数都应该占用8个字节的存储空间，不管它有多大？

编辑:

谢谢大家的回答!

当我注意到 binary.PutUvarint 消耗多达 10 个字节来存储一个大的 uint64 时，我提出了疑问，尽管最大 uint64应该只占用 8 个字节。

然后我在 Golang lib 的源代码中找到了答案:

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

根据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

所以，是的，uint64总是占用 8 个字节。