doujiao1948 2013-06-25 05:38
浏览 1905


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?


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).
  • 写回答

4条回答 默认 最新

  • dongshandun4363 2013-06-25 06:19

    According to

    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.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?



  • ¥15 广告联盟的兜底广告是什么意思
  • ¥15 如何证明高斯噪声的包络公式
  • ¥150 寻找王者荣耀开发作者,合作或者解答
  • ¥15 乳腺癌数据集 相关矩阵 特征选择
  • ¥15 我的游戏账号被盗取,请问我该怎么做
  • ¥15 通关usb3.0.push文件,导致usb频繁断连
  • ¥15 有没有能解决微信公众号,只能实时拍照,没有选择相册上传功能,我不懂任何技术,,有没有给我发个软件就能搞定的方法
  • ¥15 Pythontxt文本可视化
  • ¥15 如何基于Ryu环境下使用scapy包进行数据包构造
  • ¥15 springboot国际化