Golang RSA Encryption with error "failed to parse public key"

Question

I have a trouble when I just want to simply encrypt a short string.

package main

import (
    "crypto/rand"
    "crypto/rsa"
    "crypto/x509"
    "encoding/pem"
    "fmt"
    "log"
)

func main() {

    var pwd = "1234"
    var pwdB = []byte(pwd)

    fmt.Println(pwd)
    fmt.Println(pwdB)

    const pemPublicKey = `-----BEGIN PUBLIC KEY-----
MFwwDQYJKoZIhvcNAQEBBQADSwAwSAJBALE0I2XX+IzlhIfBx2KoYqcxlEU23oje
PTJzJ7GoYyK4R5gCkWV6ltyLN5G+rNkfsAnTObqIJK+sQzcqmm9up88CAwEAAQ==
-----END PUBLIC KEY-----`

    fmt.Println(pemPublicKey)

    block, _ := pem.Decode([]byte(pemPublicKey))
    if block == nil {
        panic("failed to parse PEM block containing the public key")
    }
    fmt.Println(block)

    pkey, _ := x509.ParsePKCS1PublicKey(block.Bytes)
    if pkey == nil {
        panic("failed to parse public key")
    }

    fmt.Println(pkey)

    in, err := rsa.EncryptPKCS1v15(rand.Reader, pkey, pwdB)
    if err != nil {
        log.Fatalf("encrypt: %s", err)
    }

    fmt.Println(in)
}

playground link

The error is :

panic: failed to parse public key
goroutine 1 [running]:
main.main()
    /tmp/sandbox736400947/main.go:36 +0x4a0

It seems that is the process in Parsing the public key, but I don't know what have I done wrong... What should I do next? Thanks :(

Next you should not discard the error from ParsePKCS1PublicKey. — Peter
– Peter, Commented Jul 24, 2018 at 6:46

Marc · Accepted Answer · 2018-07-24 06:19:33Z

2

You need to use x509.ParsePKIXPublicKey, check the error returned by the Parse... function, and perform a type check to make sure it is a RSA key before using it.

Your code would become:

pkey, err := x509.ParsePKIXPublicKey(block.Bytes)
if err != nil {
    panic(err)
}

rsaKey, ok := pkey.(*rsa.PublicKey)
if !ok {
    log.Fatalf("got unexpected key type: %T", pkey)
}

Please don't skip errors. In this case, ParsePKCS1PublicKey would have returned something along the lines of asn1: structure error: tags don't match ... which indicates a bad format.

answered Jul 24, 2018 at 6:19

Marc

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