使用 x509.MarshalPKCS1PublicKey 从 go 生成 RSA 公钥

Question

我正在尝试从 go 生成一个公钥，如下所示：

reader := rand.Reader

bitSize := 2048

keypair, err := rsa.GenerateKey(reader, bitSize)

这似乎有效并产生了一些有意义的东西。接下来，我想把它的公共部分写成这样的 RSA 文件：

func PublicKeyToPemBytes(prvkey *rsa.PrivateKey) ([]byte, error) {

    var pubkey *rsa.PublicKey

    pubkey = &prvkey.PublicKey

    pubkey_bytes := x509.MarshalPKCS1PublicKey(pubkey)

    if pubkey_bytes == nil {

        return nil, errors.New("Public key could not be serialized")

    }

    pubkey_pem := pem.EncodeToMemory(

        &pem.Block{

            Type:  "RSA PUBLIC KEY",

            Bytes: pubkey_bytes,

        },

    )

    return pubkey_pem, nil

}

这会产生一些看起来或多或少像你期望的东西。这是我生成的一个虚拟密钥，只是为了显示：

-----BEGIN RSA PUBLIC KEY-----

MIIBCGKCAQEAMU6KIRUM2KACW7ISHRVRVPXG5YC7+D58Y26HV3TBHJCDNYE9Z8NE

S/XOJS58SCJL+6VLCH03RQWFLSSBZRDTAFGE4V0PTZXQ1ECUIVX6EIUWAVIKTQA9

7WEBNFU4MCHVLWFPULDAQOFP02M2WXUCI/DXCHH1R2QJCJWZKAUOERYDOP3+5YZI

CDHWX54T7GIAU6XV9M/5FH39EBLVDITK85/3RKRZIB/6SRBFSKQVWPNG69WJGIZU

YJYQNNKB8QXG5VCHRJ+OXITBWXYKFXBIKUIGE8AKUDL9OI2SR5I0HQ0AMLNCI9DA

SGHT6UQGZMVRKJC9/FVKLRQURLKMUL1AKWIDAQAB

-----END RSA PUBLIC KEY-----

但实际上并不正确：

$ grep -v -- ----- < remote.pub  | base64 -d | dumpasn1 -

Warning: Input is non-seekable, some functionality has been disabled.

  0 264: SEQUENCE {

  4 257:   [APPLICATION 2] {

       :       Error: Spurious EOC in definite-length item.

Error: Invalid data encountered at position 12: 4E 8A.

$ openssl asn1parse -in remote.pub 

    0:d=0  hl=4 l= 264 cons: SEQUENCE          

    4:d=1  hl=4 l= 257 cons: appl [ 2 ]        

    8:d=2  hl=2 l=  49 prim: EOC               

   59:d=2  hl=2 l=   8 prim: appl [ 11 ]       

   69:d=2  hl=2 l=  16 cons: appl [ 5 ]        

   71:d=3  hl=2 l=   0 prim: priv [ 19 ]       

Error in encoding

140590716953024:error:0D07209B:asn1 encoding routines:ASN1_get_object:too long:../crypto/asn1/asn1_lib.c:91:

我不知道它是如何工作的，但我认为它应该只生成一个包含两个整数的序列，其中一个是 N，另一个是 E。我不确定为什么 dumpasn1 认为它编码了一个 [应用程序 2]但无论是 dumpasn1 还是 openssl 都认为生成的内容甚至是有效的 ASN.1。

Answer 1

我试图重现你的问题，这对我有用......考虑package mainimport "fmt"import "crypto/rand"import "crypto/rsa"import "crypto/x509"import "encoding/hex"import (&nbsp; &nbsp; "encoding/pem"&nbsp; &nbsp; "log"&nbsp; &nbsp; "os")func main() {reader := rand.ReaderbitSize := 64keypair, _:= rsa.GenerateKey(reader, bitSize)fmt.Println("Public key ", &keypair.PublicKey)pubkey_bytes := x509.MarshalPKCS1PublicKey(&keypair.PublicKey)fmt.Println(hex.Dump(pubkey_bytes))block := &pem.Block{&nbsp; &nbsp; &nbsp; &nbsp; Type: "MESSAGE",&nbsp; &nbsp; &nbsp; &nbsp; Bytes: pubkey_bytes ,&nbsp; &nbsp; }&nbsp; &nbsp; if err := pem.Encode(os.Stdout, block); err != nil {&nbsp; &nbsp; &nbsp; &nbsp; log.Fatal(err)&nbsp; &nbsp; }}运行时控制台会显示Public key&nbsp; &{14927333011981288097 65537}00000000&nbsp; 30 10 02 09 00 cf 28 8a&nbsp; 49 37 1b 42 a1 02 03 01&nbsp; |0.....(.I7.B....|00000010&nbsp; 00 01&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;|..|-----BEGIN MESSAGE-----MBACCQDPKIpJNxtCoQIDAQAB-----END MESSAGE-----我使用https://asn1.io/asn1playground/将此粘贴到架构中World-Schema DEFINITIONS&nbsp; ::=&nbsp;BEGIN&nbsp;RSAPublicKey ::= SEQUENCE {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; modulus&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;INTEGER,&nbsp; -- n&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; publicExponent&nbsp; &nbsp; INTEGER&nbsp; &nbsp;-- e&nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;END点击编译将此粘贴到解码中30 10 02 09 00 cf 28 8a&nbsp; 49 37 1b 42 a1 02 03 01 00 01结果是RSAPublicKey SEQUENCE: tag = [UNIVERSAL 16] constructed; length = 16D0023E: Integer or enumerated value too long: 9; check field 'modulus' (type: INTEGER) of PDU #1 'RSAPublicKey'.&nbsp; modulus INTEGER: tag = [UNIVERSAL 2] primitive; length = 9&nbsp; &nbsp; 2147483647&nbsp; publicExponent INTEGER: tag = [UNIVERSAL 2] primitive; length = 3&nbsp; &nbsp; 65537S0012E: Decoding of PDU #1 failed with the return code '10'.我不确定他们为什么会发现错误，但 SEQUENCE 和 2 INTEGER 肯定存在（您实际上并不需要工具来查看它）&nbsp;