就像比特币一样,我需要使用256位的私钥用ECDSA签名256位的哈希,由于缺少python中ecdsa的文档,我陷入了绝望.

我在互联网上找到了很多代码,但是没有什么比ecdsa.sign(msg,privkey)或其他类似的工具容易,我发现的一切都是很多我不理解的数学代码,但是它们使用了ecdsa库(我不知道为什么他们不会在将用于签名的库中不添加签名功能,而是在使用该库时需要一页代码？).

这是到目前为止找到的最好的代码：

def ecdsa_sign(val, secret_exponent):
    """Return a signature for the provided hash, using the provided
    random nonce. It is absolutely vital that random_k be an unpredictable
    number in the range [1, self.public_key.point.order()-1].  If
    an attacker can guess random_k, he can compute our private key from a
    single signature. Also, if an attacker knows a few high-order
    bits (or a few low-order bits) of random_k, he can compute our private
    key from many signatures. The generation of nonces with adequate
    cryptographic strength is very difficult and far beyond the scope
    of this comment.

    May raise RuntimeError, in which case retrying with a new
    random value k is in order.
    """
    G = ecdsa.SECP256k1
    n = G.order()
    k = deterministic_generate_k(n, secret_exponent, val)
    p1 = k * G
    r = p1.x()
    if r == 0: raise RuntimeError("amazingly unlucky random number r")
    s = ( ecdsa.numbertheory.inverse_mod( k, n ) * ( val + ( secret_exponent * r ) % n ) ) % n
    if s == 0: raise RuntimeError("amazingly unlucky random number s")

    return signature_to_der(r, s)

def deterministic_generate_k(generator_order, secret_exponent, val, hash_f=hashlib.sha256):
    """
    Generate K value according to https://tools.ietf.org/html/rfc6979
    """
    n = generator_order
    order_size = (bit_length(n) + 7) // 8
    hash_size = hash_f().digest_size
    v = b'\x01' * hash_size
    k = b'\x00' * hash_size
    priv = intbytes.to_bytes(secret_exponent, length=order_size)
    shift = 8 * hash_size - bit_length(n)
    if shift > 0:
        val >>= shift
    if val > n:
        val -= n
    h1 = intbytes.to_bytes(val, length=order_size)
    k = hmac.new(k, v + b'\x00' + priv + h1, hash_f).digest()
    v = hmac.new(k, v, hash_f).digest()
    k = hmac.new(k, v + b'\x01' + priv + h1, hash_f).digest()
    v = hmac.new(k, v, hash_f).digest()

    while 1:
        t = bytearray()

        while len(t) < order_size:
            v = hmac.new(k, v, hash_f).digest()
            t.extend(v)

        k1 = intbytes.from_bytes(bytes(t))

        k1 >>= (len(t)*8 - bit_length(n))
        if k1 >= 1 and k1 < n:
            return k1

        k = hmac.new(k, v + b'\x00', hash_f).digest()
        v = hmac.new(k, v, hash_f).digest()

但是我只是不信任这样的代码,因为我不知道它的作用.另外,ecdsa_sign中的注释说,它返回给定值,秘密指数和随机数的签名.它说拥有一个随机数非常重要,但是我只是不知道那个随机数在哪里.

是否有任何简单的单线方式使用Windows上python中的任何受信任库来签名和验证ECDSA签名？

解决方法:

您可以尝试使用Python3使用python ecdsa软件包：

pip3 install ecdsa

用法：

import ecdsa

# SECP256k1 is the Bitcoin elliptic curve
sk = ecdsa.SigningKey.generate(curve=ecdsa.SECP256k1) 
vk = sk.get_verifying_key()
sig = sk.sign(b"message")
vk.verify(sig, b"message") # True

要使用公钥验证现有签名,请执行以下操作：

import ecdsa

message = b"message"
public_key = '98cedbb266d9fc38e41a169362708e0509e06b3040a5dfff6e08196f8d9e49cebfb4f4cb12aa7ac34b19f3b29a17f4e5464873f151fd699c2524e0b7843eb383'
sig = '740894121e1c7f33b174153a7349f6899d0a1d2730e9cc59f674921d8aef73532f63edb9c5dba4877074a937448a37c5c485e0d53419297967e95e9b1bef630d'

vk = ecdsa.VerifyingKey.from_string(bytes.fromhex(public_key), curve=ecdsa.SECP256k1)
vk.verify(bytes.fromhex(sig), message) # True

该软件包也与Python 2兼容