官方文档说uint64是一个64位的无符号整数,这是否意味着任何uint64数字应该占用8个字节的存储空间,无论它有多小或多大？
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的源代码中找到了我的疑问：
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).
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.
