当我尝试将在 iOS 上生成的 base64 编码字符串 EC 公钥转换为 Java PublicKey 时，在 base64 字符串中没有得到正确的结果

Question

正如标题所示，我正在尝试将在 IOS 设备（Swift）上生成的 base64 编码字符串（EC 公钥）转换为 Java PublicKey，后者将用于计算两方之间的共享密钥。代码中既没有运行时也没有编译时异常/错误，它编译并成功运行并生成了一个 PublicKey 但是当我将 PublicKey (Base64.encodeToString(PublicKey.encoded), Base64.NO_WRAP) 编码回 Base64 字符串以确认是否我得到了作为参数传递的相同公钥，它们不一样。

import android.util.Base64
import org.bouncycastle.asn1.sec.SECNamedCurves
import org.bouncycastle.math.ec.ECCurve
import java.math.BigInteger
import java.nio.charset.StandardCharsets
import java.security.*
import java.security.spec.*
import javax.crypto.*
import javax.crypto.spec.IvParameterSpec
import javax.crypto.spec.SecretKeySpec
    
    
fun iosB64EncodedStrPKToPK(iOSB64EncodedPK: String): PublicKey {
    val decodedPK = Base64.decode(iOSB64EncodedPK, Base64.NO_WRAP)
    val x9ECParamSpec = SECNamedCurves.getByName("secp256r1")
    val curve = x9ECParamSpec.curve
    val point = curve.decodePoint(decodedPK)
    val xBcEC = point.affineXCoord.toBigInteger()
    val yBcEC = point.affineYCoord.toBigInteger()
    val gBcEC = x9ECParamSpec.g
    val xGBcEC = gBcEC.affineXCoord.toBigInteger()
    val yGBcEC = gBcEC.affineYCoord.toBigInteger()
    val hBcEC = x9ECParamSpec.h.toInt()
    val nBcEC = x9ECParamSpec.n
    val jPEC = ECPoint(xBcEC, yBcEC)
    val gJpEC = ECPoint(xGBcEC, yGBcEC)
    val jEllipticCurve = convertECCurveToEllipticCurve(curve, gJpEC, nBcEC, hBcEC)
    val eCParameterSpec = ECParameterSpec(jEllipticCurve, gJpEC, nBcEC, hBcEC)
    val ecPubLicKeySpec = ECPublicKeySpec(jPEC, eCParameterSpec)
    val keyFactorySpec = KeyFactory.getInstance("EC")
    return keyFactorySpec.generatePublic(ecPubLicKeySpec)
}
    
private fun convertECCurveToEllipticCurve(
    curve: ECCurve,
    ecPoint: ECPoint,
    n: BigInteger,
    h: Int
): EllipticCurve {
    val ecField = ECFieldFp(curve.field.characteristic)
    val firstCoefficient = curve.a.toBigInteger()
    val secondCoefficient = curve.b.toBigInteger()
    val ecParams = ECParameterSpec(
        EllipticCurve(ecField, firstCoefficient, secondCoefficient),
            ecPoint,
            n,
            h
        )
    return ecParams.curve
}

我传递给 iosB64EncodedStrPKToPK() 函数的公钥：BAlWWu46il/ly6Axd/qclmhEVhGth93QN5+h3JBJEKEmhKd1LfqkpCqX1cT1cQDs9nPq9Lq0/FtZitkjr7Rqd94=

我得到的输出：MFkwEwYHKoZIzj0CAQYIKoZIzj0DAQcDQgAECVZa7jqKX+XLoDF3+pyWaERWEa2H3dA3n6HckEkQoSaEp3Ut+qSkKpfVxPVxAOz2c+r0urT8W1mK2SOvtGp33g==

val pkIOS = "BAlWWu46il/ly6Axd/qclmhEVhGth93QN5+h3JBJEKEmhKd1LfqkpCqX1cT1cQDs9nPq9Lq0/FtZitkjr7Rqd94="

Log.i("SOME_TAG","PUBLIC_KEY_IOS:${Base64.encodeToString(iosB64EncodedStrPKToPK(pkIOS).encoded,Base64.NO_WRAP)}") 

我不是这方面的专家，也许有人会指引我正确的方向，并能看到我犯的错误。密码学不在我的专业领域。

我尝试过谷歌搜索、ChatGPT 和其他有见地的资源，如果您知道有关该问题的来源，我也很乐意接受。

我在 Android 环境中运行代码

我使用的BouncyCastle版本：

def bouncy_castle_version = '1.70'
implementation "org.bouncycastle:bcpkix-jdk15on:$bouncy_castle_version"
implementation "org.bouncycastle:bcprov-jdk15on:$bouncy_castle_version"