JavacardOS will not accept order any more, please contact our partner Feitian online Store:
https://ftsafe.en.alibaba.com/index.html
https://ftsafe.en.alibaba.com/index.html
How to know supported encryption/sign algorithms in a Javacard
Moderator: UNKNwYSHSA
How to know supported encryption/sign algorithms in a Javacard
As the subject mentioned, how can i know all the encryption algorithms the Javacard supports?
One way.
- horse dream
- Posts: 76
- Joined: Thu May 21, 2015 11:48 pm
- Points :140
- Contact:
Re: How to know supported encryption/sign algorithms in a Javacard
The code below can check the specified algorithm your card supports or not. If the algorithm is supported, the instance creation will succeed. Otherwise,CryptoException.NO_SUCH_ALGORITHM is thrown. e.g. ALG_DES_CBC_NOPAD algorithm:
Code: Select all
try {
m_cipher = Cipher.getInstance(ALG_DES_CBC_NOPAD, false);
supported = true;
}
catch (CryptoException e) {
if (e.getReason() == CryptoException.NO_SUCH_ALGORITHM))
{
supported = false;
}
else {
// other errors
}
}
Re: How to know supported encryption/sign algorithms in a Javacard
That's so nice of you. But,it takes too long to test all the possible algorithms successively . Is there any method to know all the algorithms supported by the given card at a time ?
One way.
- horse dream
- Posts: 76
- Joined: Thu May 21, 2015 11:48 pm
- Points :140
- Contact:
Re: How to know supported encryption/sign algorithms in a Javacard
Of course,you can see JavaCard Algorithms Support Test Tool, which can meet your need. Hope to help you.
- UNKNwYSHSA
- Posts: 630
- Joined: Thu May 21, 2015 4:05 am
- Points :3053
- Contact:
Re: How to know supported encryption/sign algorithms in a Javacard
So simple, ask your card provider. If you did not know the card provider, you can only test yourself.
sense and simplicity
Re: How to know supported encryption/sign algorithms in a Javacard
UNKNwYSHSA wrote:So simple, ask your card provider. If you did not know the card provider, you can only test yourself.
Hi,UNKNwYSHSA
Thanks for your answer anyway.
One way.
Who is online
Users browsing this forum: No registered users and 42 guests