Modular Inverse for ECDSA in C++

I've seen several variations on "modular inverse" but I can't seem to identify the proper way of utilizing it for C++. I'm not sure how this python code translates into c++:

def inverse(x, p):

"""
Calculate the modular inverse of x ( mod p )
the modular inverse is a number such that:
(inverse(x, p) * x) % p == 1
you could think of this as: 1/x
"""
inv1 = 1
inv2 = 0
while p != 1 and p!=0:
inv1, inv2 = inv2, inv1 - inv2 * (x / p)
x, p = p, x % p

return inv2

My current coding is here: http://coliru.stacked-crooked.com/a/74648b16c2692525

But it only shows like the first public key properly, after that it messes up.

Mine

Posted 2014-06-24T06:13:38.543

Reputation: 1 142

Answers

#include <openssl/bn.h>

[...]

BN_mod_inverse ( a, b, c, ctx );

user15732

Posted 2014-06-24T06:13:38.543

Reputation: 86

I downloaded that library, but I don't know what to do with your inverse. I'm only sending two variables to my inverse to be worked with. I think it works out to the number to be modding and the number being modded by. – Mine – 2014-06-24T07:29:01.090

I'll take a shot, I don't code in C++ often. Mostly do Java, and I am completely unfamiliar with python. I have tested this code.

int inverse(int x,int p)
{
  int orig = p;
  int inv1 = 1;
  int inv2 = 0;
  while(p != 1 && p != 0)
  {
    int temp = inv2;
    inv2 = inv1 - inv2 * (x/p);
    inv1 = temp;
    temp = p;
    p = x % p;
    x = temp;
  }
  while(inv2 < 0)
    inv2 += orig;
  return inv2;
}

It turns out that python does % differently than C++. The while(inv2 < 0) statement corrects the sign error caused by the difference. There is probably a more elegant way of doing this, but this was quick and easy.

Bogoth

Posted 2014-06-24T06:13:38.543

Reputation: 1