'Extract ECPublicKey from a X509 certicate

I'm having trouble extracting an ECPublicKey from an X509 certifcate using Java.

The keys and certificate were created as follows

ssh-keygen -t ecdsa -f id_ecdsa
openssl pkcs8 -topk8 -in id_ecdsa -out id_ecdsa.p8
openssl req -new x509 -key id_ecdsa.p8 -out id_ecdsa.crt.der -outform der

The code used to extract the public key from the certificate is

FileInputStream fin = new FileInputStream("<path to id_ecdsa.crt.der>");
CertificateFactory cf = CertificateFactory.getInstance("X.509");
X509Certificate cert = (X509Certificate) cf.generateCertificate(fin);
PublicKey pk = cert.getPublicKey();

if (pk instanceof ECPublicKey) {
  ECPublicKey key = (ECPublicKey) pk;
  ...
} else if (pk instanceof RSAPublicKey) {
  RSAPublicKey key = (RSAPublicKey) pk;
  ...
}

For a certificate containing an RSA key all is ok. However if an ECDSA key is used the if(pk instanceof ECPublicKey) block is ignored.

A call to pk.getAlgorithm() yields "EC" which suggests the key is an ECDSA key.

Examination of pk with a debugger yields a type X509Key for ECDSA. For an RSA key the debugger yields RSAPublicKeyImpl.

N.B. java.security.* is used as the library.

Any help solving my problem would be greatly appreciated.

javapublic-keyecdsa


Solution 1:[1]

    TrustManagerFactory tmf;
    try {
        tmf = TrustManagerFactory.getInstance("X509");
        tmf.init((KeyStore) null);

        for (TrustManager trustManager : tmf.getTrustManagers()) {
            ((X509TrustManager) trustManager).checkServerTrusted(
                    chain, authType);
        }

    } catch (Exception e) {
        
    }

        
    ECPublicKey pubkey = (ECPublicKey) chain[0].getPublicKey();

Solution 2:[2]

I found that adding Bouncy Castle as a provider appears to have fixed my issue. It appears JDK is not fitted with EC support by default.

Security.addProvider(new BouncyCastleProvider());
CertificateFactory cf = CertificateFactory.getInstance("X.509", "BC");

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Yasas Weerasekara
Solution 2 John Harriott

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