- { if (!BN_mul(a,x,y)) goto err; }
- if (!BN_rshift(d,a,nb)) goto err;
- if (!BN_mul(b,d,i)) goto err;
- if (!BN_rshift(c,b,nb)) goto err;
- if (!BN_mul(b,m,c)) goto err;
- if (!BN_sub(r,a,b)) goto err;
+ ca=x; /* Just do the mod */
+
+ ret = BN_div_recp(NULL,r,ca,recp,ctx);
+err:
+ BN_CTX_end(ctx);
+ bn_check_top(r);
+ return(ret);
+ }
+
+int BN_div_recp(BIGNUM *dv, BIGNUM *rem, const BIGNUM *m,
+ BN_RECP_CTX *recp, BN_CTX *ctx)
+ {
+ int i,j,ret=0;
+ BIGNUM *a,*b,*d,*r;
+
+ BN_CTX_start(ctx);
+ a=BN_CTX_get(ctx);
+ b=BN_CTX_get(ctx);
+ if (dv != NULL)
+ d=dv;
+ else
+ d=BN_CTX_get(ctx);
+ if (rem != NULL)
+ r=rem;
+ else
+ r=BN_CTX_get(ctx);
+ if (a == NULL || b == NULL || d == NULL || r == NULL) goto err;
+
+ if (BN_ucmp(m,&(recp->N)) < 0)
+ {
+ BN_zero(d);
+ if (!BN_copy(r,m)) return 0;
+ BN_CTX_end(ctx);
+ return(1);
+ }
+
+ /* We want the remainder
+ * Given input of ABCDEF / ab
+ * we need multiply ABCDEF by 3 digests of the reciprocal of ab
+ *
+ */
+
+ /* i := max(BN_num_bits(m), 2*BN_num_bits(N)) */
+ i=BN_num_bits(m);
+ j=recp->num_bits<<1;
+ if (j>i) i=j;
+
+ /* Nr := round(2^i / N) */
+ if (i != recp->shift)
+ recp->shift=BN_reciprocal(&(recp->Nr),&(recp->N),
+ i,ctx); /* BN_reciprocal returns i, or -1 for an error */
+ if (recp->shift == -1) goto err;
+
+ /* d := |round(round(m / 2^BN_num_bits(N)) * recp->Nr / 2^(i - BN_num_bits(N)))|
+ * = |round(round(m / 2^BN_num_bits(N)) * round(2^i / N) / 2^(i - BN_num_bits(N)))|
+ * <= |(m / 2^BN_num_bits(N)) * (2^i / N) * (2^BN_num_bits(N) / 2^i)|
+ * = |m/N|
+ */
+ if (!BN_rshift(a,m,recp->num_bits)) goto err;
+ if (!BN_mul(b,a,&(recp->Nr),ctx)) goto err;
+ if (!BN_rshift(d,b,i-recp->num_bits)) goto err;
+ d->neg=0;
+
+ if (!BN_mul(b,&(recp->N),d,ctx)) goto err;
+ if (!BN_usub(r,m,b)) goto err;
+ r->neg=0;
+
+#if 1