英文:
Does a uint64 take 8 bytes storage?
问题
官方文件表示uint64是一个64位的无符号整数，这是否意味着任何uint64数字都应该占用8个字节的存储空间，无论它有多小或多大？
编辑：
感谢大家的回答！
当我注意到binary.PutUvarint在存储一个大的uint64时需要最多10个字节时，我提出了这个疑问，尽管最大的uint64只需要8个字节。
然后我在Golang库的源代码中找到了对我的疑问的答案：
设计说明：
// 64位值最多需要10个字节。编码可以更紧凑：一个完整的64位值需要额外的一个字节来保存第63位。
// 相反，前一个字节的最高有效位可以用来保存第63位，因为我们知道最多只有64位。这是一个微不足道的改进，
// 可以将最大编码长度减少到9个字节。然而，它打破了最高有效位始终是“连续位”的不变性，因此使得该格式与更大的数字（比如128位）的varint编码不兼容。
英文:
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).
答案1
得分: 15
根据http://golang.org/ref/spec#Size_and_alignment_guarantees：
类型 字节大小
byte, uint8, int8 1
uint16, int16 2
uint32, int32, float32 4
uint64, int64, float64, complex64 8
complex128 16
所以，是的，uint64始终占用8个字节。
英文:
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.
答案2
得分: 1
简单来说：是的，一个64位的固定大小整数类型总是占用8个字节。除非是一种不寻常的语言，否则情况都是如此。
有些语言/平台支持可变长度的数字类型，其中内存中的存储确实取决于值，但在这种简单的情况下，你不会指定类型中的位数，因为它可能会有所不同。
英文:
Simply put: yes, a 64-bit fixed size integer type will always take 8 bytes. It would be an unusual language where that isn't the case.
There are languages/platforms which support variable-length numeric types where the storage in memory does depend on the value, but you wouldn't then specify the number of bits in the type in such a simple way, as that can vary.
答案3
得分: 1
Go编程语言规范
数值类型表示整数或浮点数值的集合。
预定义的与体系结构无关的数值类型有：
uint64 所有无符号64位整数的集合（从0到18446744073709551615）
是的，确切地说是64位或8字节。
英文:
> The Go Programming Language Specification
>
> Numeric types
>
> A numeric type represents sets of integer or floating-point values.
> The predeclared architecture-independent numeric types are:
>
> uint64 the set of all unsigned 64-bit integers (0 to 18446744073709551615)
Yes, exactly 64 bits or 8 bytes.
答案4
得分: 1
因此，64位 = 8字节
英文:
Just remember the simple rule, the variable type is usually optimized to fit certain memory space and the minimum memory space is 1 bit(s). And 8 bit(s) = 1 byte(s):
Therefore 64bit(s) = 8 byte(s)
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