Hi,
the link you posted below are very useful!
However, after many trials and errors, I created a little program to
derive a public key from an x-only coordinate but, in the last step,
it fails, namely in the function
'EC_POINT_set_compressed_coordinates_GFp' called in the function
'loadKey_xOnly'.
Here the whole code:
#include <iostream>
#include <string>
#include <cstring>
#include <openssl/ec.h>
#include <openssl/ecdsa.h>
#include <openssl/obj_mac.h>
#include <openssl/err.h>
char *ossl_err_as_string (void)
{ BIO *bio = BIO_new (BIO_s_mem ());
ERR_print_errors (bio);
char *buf = NULL;
size_t len = BIO_get_mem_data (bio, &buf);
char *ret = (char *) calloc (1, 1 + len);
if (ret)
memcpy (ret, buf, len);
BIO_free (bio);
return ret;
}
void printAndExit (std::string s) {
std::cerr << "ERROR!\n" << s << std::endl << ossl_err_as_string () <<
std::endl;
exit(EXIT_FAILURE);
}
EC_KEY* loadKey_xOnly (std::string& xOnly) {
EC_GROUP* group = EC_GROUP_new_by_curve_name (NID_secp256k1);
if (group == NULL)
printAndExit ("group null");
EC_GROUP_set_point_conversion_form (group, POINT_CONVERSION_COMPRESSED);
EC_POINT* pubPoint = EC_POINT_new (group);
if (pubPoint == NULL)
printAndExit ("pibpoint == null");
BN_CTX* bnCtx = BN_CTX_new ();
BIGNUM* x = BN_new ();
if (x == NULL)
printAndExit ("X == NULL");
if (BN_hex2bn (&x, (const char*) xOnly.c_str()) == 0)
printAndExit ("hex2bn == 0");
std::cout << "x_BN:\t" << BN_bn2hex (x) << std::endl;
std::cout << "x_BA:\t" << xOnly << std::endl;
if (EC_POINT_set_compressed_coordinates_GFp (group, pubPoint, x, 0,
bnCtx) == 0)
printAndExit ("set_compr_coord == 0");
}
int main () {
std::string xOnlyStr =
"c16b4ce0532f5dc9d09114fe121d3956ae84f9eb677a0d4bdac1d3af7a91950c";
EC_KEY* key = loadKey_xOnly (xOnlyStr);
return 0;
}
The string 'xOnlyStr' contains a real x-only coordinate of a verification key.
Could you help me?
Thank you,
Luca Di Mauro
Luca Di Mauro <luca.dimauro@xxxxxxx> ha scritto:
Thank you very much for the reply!
Yes, I have also the additional information about on which of two
solutions I should take.
I'll check the guides you linked below.
Luca Di Mauro
Nicola Tuveri <nic.tuv@xxxxxxxxx> ha scritto:
Hi,
with traditional EC from the x coordinate alone you can't really do that,
because there are always 2 possible solutions for y (in R the curve is
symmetrical on the x axis).
The standards define a "compressed point" format in which you can send the
coordinate x and an additional bit to select which of the 2 y solutions to
pick.
You can read more about it in EC_GROUP_set_point_conversion_form at
https://www.openssl.org/docs/man1.1.1/man3/EC_GROUP_copy.html
and in EC_POINT_set_compressed_coordinates at
https://www.openssl.org/docs/man1.1.1/man3/EC_POINT_new.html
Hope this helps,
Nicola Tuveri
On Fri, Oct 18, 2019, 11:31 Luca Di Mauro <luca.dimauro@xxxxxxx> wrote:
Hello all,
I don't know if it is the correct mailing list to ask this, so I'm
sorry if it is the wrong palce.
I'm using openssl v1.1, and I'm trying to compute both the X and Y
coordinates of an elliptic curve point starting from a single
coordinate (X or Y).
How can i perform that in C/C++ language using libssl? I searched on
google for a couple of days but i haven't found a solution.
Luca Di Mauro
