> char* hex = "0000"; > BN_hex2bn(&bignum, hex); > buffer = (char*)malloc(2); > BN_bn2bin(bignum, buffer); > > > wont' do the same thing as > > buffer = (char*) malloc(2); > memset(buffer, 0, 2);
No, how would it? How does BN_bn2bin() know that buffer is 2 bytes? There's no way for the function to know how many bytes your buffer is. Infact, since BN_num_bytes() returns 0, the buffer you pass it won't ever even be touched, so no data will be set. -Brad ______________________________________________________________________ OpenSSL Project http://www.openssl.org Development Mailing List [email protected] Automated List Manager [EMAIL PROTECTED]
