base58前缀

Question

base58前缀

- 4
- 187 2019-02-01
Prefix中的
模块src/lib_crypto/base58.ml具有诸如let ed25519_public_key_hash = "\006\161\159" (* tz1(36) *).

人们如何从"\006\161\159"中获得tz1(36)?

Module Prefix in src/lib_crypto/base58.ml has lines such as let ed25519_public_key_hash = "\006\161\159" (* tz1(36) *).

How does one get "\006\161\159" from tz1(36)?

crypto base58

2 个回答

投票数

- 3
- 2019-02-01
Base58前缀将始终为设置的输出长度产生带前缀的输出.因此,输入地址为20字节+ 3字节前缀给出了一个带有tz1的36个字符长的地址.

您可以通过猜测和检查来计算这些前缀.

Base58 prefixes will always produce a prefixed output for a set length of output. So the input address is 20 bytes + the 3 byte prefix gives a 36 char long address with tz1.

You calculate these prefixes by guess and check.

Arthur B · Accepted Answer · 2019-02-01

7

2019-02-01

Base58通过添加前缀,将字节视为大字节序号并将该大字节序号写入基数58来编码字符.

因此,特定的前缀可以固定Base58中最重要的"数字".

Tezos存储库的b58_prefix.py目录中的python脚本scripts可以帮助找到这些前缀.请注意,要运行它,您将需要此版本的python lib pybitcointools

 pip install git+https://github.com/vbuterin/pybitcointools.git@aeb0a2bbb8bbfe421432d776c649650eaeb882a5#egg=master

Base58 encodes characters by appending a prefix, treating the bytes as a big endian number and writing that big endian number in base 58.

Therefore, specific prefixes can pin down the most significant "digits" in Base58.

The python script b58_prefix.py in the scripts directory of the Tezos repo can help find those prefixes. Note that to run it you will need this version of the python lib pybitcointools

pip install git+https://github.com/vbuterin/pybitcointools.git@aeb0a2bbb8bbfe421432d776c649650eaeb882a5#egg=master

如何将字节数组" \ 006 \ 161 \ 159"转换为可以传递给bitcoin.bin_to_b58check函数的magicbyte值? 我发现可以将struct.unpack与字节数组的十六进制值一起使用(生成p2sig(98))`P256_SIGNATURE=struct.unpack('> L',b'\ x36 \ xF0 \ x2C \ x34')[0],但是" \ 006 \ 161 \ 159"只有3个字节,因此无法打包为4个字节的大端整数.

How do you convert the byte array `"\006\161\159"` into a magicbyte value that can be passed to the bitcoin.bin_to_b58check function? I discovered that you can use struct.unpack with the hex values of the byte array like this (generates p2sig(98))`P256_SIGNATURE = struct.unpack('>L', b'\x36\xF0\x2C\x34')[0]`, but `"\006\161\159"` is only 3 bytes so it can't be packed into a 4-byte big-endian integer.
- 0
- 2019-02-16
- Luke Youngblood
另请参见待定的MR https://gitlab.com/tezos/tezos/-/merge_requests/1528,该文档进一步记录了该脚本.

see also the pending MR https://gitlab.com/tezos/tezos/-/merge_requests/1528 that documents this script further.
- 0
- 2020-04-08
- arvidj