int i, k;
m0 = m[0];
+ assert(m0 >> (BIGNUM_INT_BITS-1) == 1);
if (mlen > 1)
m1 = m[1];
else
len = mod[0];
r = bn_power_2(BIGNUM_INT_BITS * len);
inv = modinv(mod, r);
+ assert(inv); /* cannot fail, since mod is odd and r is a power of 2 */
/*
* Multiply the base by r mod n, to get it into Montgomery
int pqlen, mlen, rlen, i, j;
Bignum result;
+ /*
+ * The most significant word of mod needs to be non-zero. It
+ * should already be, but let's make sure.
+ */
+ assert(mod[mod[0]] != 0);
+
/* Allocate m of size mlen, copy mod to m */
/* We use big endian internally */
mlen = mod[0];
int mshift;
int plen, mlen, i, j;
+ /*
+ * The most significant word of mod needs to be non-zero. It
+ * should already be, but let's make sure.
+ */
+ assert(mod[mod[0]] != 0);
+
/* Allocate m of size mlen, copy mod to m */
/* We use big endian internally */
mlen = mod[0];
Bignum x = copybn(One);
int sign = +1;
+ assert(number[number[0]] != 0);
+ assert(modulus[modulus[0]] != 0);
+
while (bignum_cmp(b, One) != 0) {
- Bignum t = newbn(b[0]);
- Bignum q = newbn(a[0]);
+ Bignum t, q;
+
+ if (bignum_cmp(b, Zero) == 0) {
+ /*
+ * Found a common factor between the inputs, so we cannot
+ * return a modular inverse at all.
+ */
+ freebn(b);
+ freebn(a);
+ freebn(xp);
+ freebn(x);
+ return NULL;
+ }
+
+ t = newbn(b[0]);
+ q = newbn(a[0]);
bigdivmod(a, b, t, q);
while (t[0] > 1 && t[t[0]] == 0)
t[0]--;