Even with a null terminator appended to the char I'm seeing the same
problems.
On 17 October 2012 09:58, Ben Laurie b...@links.org wrote:
On Wed, Oct 17, 2012 at 9:52 AM, Brent Evans brentevan...@gmail.com
wrote:
Hi,
I'm currently trying to use the openSSL library to perform DES3
encryption
on a string. The result from this encryption then has a base64 operation
performed on it, before this is passed to a Java application to decode
the
base64 and unencrypt it.
In the implementation below PKCS#5 padding is applied to the string
before
it is encrypted.
The strange thing at the moment as that the DES3 encryption is returning
extra, 'unexpected' data. This results in the base64 returning an
incorrect
result, however if I only base64 up until the first line termination (\n)
then the base64 operation will return the correct string, with the Java
application then successfully unbasing and decoding it.
Any ideas?
Yeah. You pass outtext to the string constructor. Where do you think
the NUL terminator will be?
Also - why are you using ECB mode? Not generally a good idea.
std::string encrypt(const std::string plainText)
{
int dif = 8 - (plainText.length() % 8);
int length = (plainText.length() + dif);
char *outtext = new char[length];
// pad the string so that it conforms to PKCS#5
char padChar = static_castunsigned char(dif);
std::string padStr;
padStr.assign(dif, padChar);
std::string unencryptedStr = plainText;
unencryptedStr += padStr;
DES_key_schedule ksched1;
DES_key_schedule ksched2;
DES_key_schedule ksched3;
DES_set_key((DES_cblock *)abcdefgh, ksched1);
DES_set_key((DES_cblock *)ijklmnop, ksched2);
DES_set_key((DES_cblock *)qrstuvwx, ksched3);
for (int i = 0; i length; i += 8)
{
DES_ecb3_encrypt((DES_cblock *)(unencryptedStr.c_str() + i),
(DES_cblock *)(outtext +
i),
ksched1, ksched2,
ksched3, DES_ENCRYPT);
}
std::string result = std::string(outtext);
return result;
}
Thanks,
Brent
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