How do I generate random RSA public and private keys (RSAParameters) using RSACryptoServiceProvider class? Each time I create a new instance of RSACryptoServiceProvider, I end up exporting the same keys.
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- 1Show your code, don't make us guess.President James K. Polk– President James K. Polk2010-11-20 17:09:01 +00:00Commented Nov 20, 2010 at 17:09
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2 Answers
I did some test on the following code, and the exported parameters are always different:
var rsaAlgo1 = new RSACryptoServiceProvider(); var rsaAlgo2 = new RSACryptoServiceProvider(); var xml1 = rsaAlgo1.ToXmlString(true); var xml2 = rsaAlgo2.ToXmlString(true); if (xml1 != xml2) { // it always goes here... } 
3 Comments
Rasmus Faber
This code works, but it is a very dangerous style. The
Create()-method you are calling is actually a static method on the abstract RSA-class (the superclass to RSACryptoServiceProvider) and creates an instance of the default implementation of the RSA-class. On most systems this will be RSACryptoServiceProvider - but not on all. A better style would be either var rsa = RSA.Create() or var rsa = new RSACryptoServiceProvider.
CodesInChaos
Note that for practical use you should specify a key size (e.g. 2048 bits) instead of using the default constructor.
Nayef
Randomness in Cryptography can't be checked by a simple comparison. It has to meet four requirements. Check them out cs.ucsb.edu/~koc/cren/docs/w06/rng.pdf
Using the following code you should never get all the same keys out
var rsa = new RSACryptoServiceProvider(); var rsaParams = rsa.ExportParameters(true); However you should note that the Exponent key can be the same and if often is 65537(0x010001)
"Choose an integer e such that 1 < e < φ(n) and gcd(e, φ(n)) = 1; i.e., e and φ(n) are coprime. e is released as the public key exponent. e having a short bit-length and small Hamming weight results in more efficient encryption – most commonly 216 + 1 = 65,537. However, much smaller values of e (such as 3) have been shown to be less secure in some settings." RSA wiki
