python的ecdsa库有问题答案

【问题标题】：Having troubles with the ecdsa library for pythonpython的ecdsa库有问题
【发布时间】：2020-04-26 06:56:59
【问题描述】：

我想测试 Andreas Antonopolous 所著的 Mastering Bitcoin 一书中的椭圆曲线代码（从私钥派生公钥）是否可行……所以我将书中的代码复制并粘贴到我的 VSC编辑。它提示了一些错误，使我删除了参数末尾的“L”，并在打印语句后插入了必要的括号。现在还是遇到“无法导入ecdsa.util”之类的错误...

以下是我正在使用的代码，并附上它显示的错误快照：


import ecdsa
import random

from ecdsa.util import string_to_number, number_to_string

_p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
_r = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
_b = 0x0000000000000000000000000000000000000000000000000000000000000007
_a = 0x0000000000000000000000000000000000000000000000000000000000000000
_Gx = 0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798
_Gy = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
curve_secp256k1 = ecdsa.ellipticcurve.CurveFp(_p, _a, _b)
generator_secp256k1 = ecdsa.ellipticcurve.Point(curve_secp256k1, _Gx, _Gy, _r)
oid_secp256k1 = (1, 3, 132, 0, 10)
SECP256k1 = ecdsa.curves.Curve("SECP256k1", curve_secp256k1, generator_secp256k1,
oid_secp256k1)
ec_order = _r

curve = curve_secp256k1
generator = generator_secp256k1

def random_secret():
    random_char = lambda: chr(random.randint(0, 255))
    convert_to_int = lambda array: int("".join(array).encode("hex"), 16)
    byte_array = [random_char() for i in range(32)]
    return convert_to_int(byte_array)

def get_point_pubkey(point):
    if point.y() & 1:
        key = '03' + '%064x' % point.x()
    else:
        key = '02' + '%064x' % point.x()
    return key.decode('hex')

def get_point_pubkey_uncompressed(point):
    key = '04' + \
          '%064x' % point.x() + \
          '%064x' % point.y()
    return key.decode('hex')

secret = random_secret()
print ("Secret: ", secret)

point = secret * generator
print ("EC point:", point)

print ("BTC public key:", get_point_pubkey(point).encode("hex"))

point1 = ecdsa.ellipticcurve.Point(curve, point.x(), point.y(), ec_order)
assert point1 == point

【问题讨论】：

标签： python blockchain bitcoin ecdsa

【解决方案1】：

因为Mastering Bitcoin的python代码是基于python2 而python3改变了字符串编码和解码。

这是用python3正确激活的代码

import ecdsa
import os
import binascii
from ecdsa.util import string_to_number, number_to_string

_p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F
_r = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
_b = 0x0000000000000000000000000000000000000000000000000000000000000007
_a = 0x0000000000000000000000000000000000000000000000000000000000000000
_Gx = 0x79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798
_Gy = 0x483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8
curve_secp256k1 = ecdsa.ellipticcurve.CurveFp(_p, _a, _b)
generator_secp256k1 = ecdsa.ellipticcurve.Point(curve_secp256k1, _Gx, _Gy, _r)
oid_secp256k1 = (1, 3, 132, 0, 10)
SECP256k1 = ecdsa.curves.Curve(
    "SECP256k1", curve_secp256k1, generator_secp256k1, oid_secp256k1
)
ec_order = _r

curve = curve_secp256k1
generator = generator_secp256k1


def random_secret():
    convert_to_int = lambda array: int(binascii.hexlify(array), 16)
    byte_array = os.urandom(32)
    return convert_to_int(byte_array)


def get_point_pubkey(point):
    if point.y() & 1:
        key = "03" + "%064x" % point.x()
    else:
        key = "02" + "%064x" % point.x()
    return binascii.unhexlify(key)


def get_point_pubkey_uncompressed(point):
    key = "04" + "%064x" % point.x() + "%064x" % point.y()
    return binascii.unhexlify(key)


secret = random_secret()
print("Secret: ", secret)

point = secret * generator
print("EC point: ", point)

print("BTC public key: ", binascii.hexlify(get_point_pubkey(point)))

point1 = ecdsa.ellipticcurve.Point(curve, point.x(), point.y(), ec_order)
assert point1 == point

结果是这样的

Secret:  45144777120122405140812796301557992803916792041985925168483786699489296135626
EC point:  (97022596039543818601440182171533278426801313043337446429800654678890979695199,53586405348987121115764110810505844890885198959743315511392863714433787991327)
BTC public key:  b'03d680d7e3daf984534507c6ddd7d7cfa07864daf6b78d5b52bf657ecb636a9a5f'

【讨论】：