Run util/openssl-format-source -v -c .

Reviewed-by: Tim Hudson <tjh@openssl.org>

1 files changed, 124 insertions, 123 deletions

diff --git a/crypto/bn/bn_kron.c b/crypto/bn/bn_kron.c
index 6c0cd08210..71808321d5 100644
--- a/crypto/bn/bn_kron.c
+++ b/crypto/bn/bn_kron.c
@@ -7,7 +7,7 @@
  * are met:
  *
  * 1. Redistributions of source code must retain the above copyright
- *    notice, this list of conditions and the following disclaimer. 
+ *    notice, this list of conditions and the following disclaimer.
  *
  * 2. Redistributions in binary form must reproduce the above copyright
  *    notice, this list of conditions and the following disclaimer in
@@ -61,125 +61,126 @@
 
 /* Returns -2 for errors because both -1 and 0 are valid results. */
 int BN_kronecker(const BIGNUM *a, const BIGNUM *b, BN_CTX *ctx)
-	{
-	int i;
-	int ret = -2; /* avoid 'uninitialized' warning */
-	int err = 0;
-	BIGNUM *A, *B, *tmp;
-	/*-
-	 * In 'tab', only odd-indexed entries are relevant:
-	 * For any odd BIGNUM n,
-	 *     tab[BN_lsw(n) & 7]
-	 * is $(-1)^{(n^2-1)/8}$ (using TeX notation).
-	 * Note that the sign of n does not matter.
-	 */
-	static const int tab[8] = {0, 1, 0, -1, 0, -1, 0, 1};
-
-	bn_check_top(a);
-	bn_check_top(b);
-
-	BN_CTX_start(ctx);
-	A = BN_CTX_get(ctx);
-	B = BN_CTX_get(ctx);
-	if (B == NULL) goto end;
-	
-	err = !BN_copy(A, a);
-	if (err) goto end;
-	err = !BN_copy(B, b);
-	if (err) goto end;
-
-	/*
-	 * Kronecker symbol, imlemented according to Henri Cohen,
-	 * "A Course in Computational Algebraic Number Theory"
-	 * (algorithm 1.4.10).
-	 */
-
-	/* Cohen's step 1: */
-
-	if (BN_is_zero(B))
-		{
-		ret = BN_abs_is_word(A, 1);
-		goto end;
- 		}
-	
-	/* Cohen's step 2: */
-
-	if (!BN_is_odd(A) && !BN_is_odd(B))
-		{
-		ret = 0;
-		goto end;
-		}
-
-	/* now  B  is non-zero */
-	i = 0;
-	while (!BN_is_bit_set(B, i))
-		i++;
-	err = !BN_rshift(B, B, i);
-	if (err) goto end;
-	if (i & 1)
-		{
-		/* i is odd */
-		/* (thus  B  was even, thus  A  must be odd!)  */
-
-		/* set 'ret' to $(-1)^{(A^2-1)/8}$ */
-		ret = tab[BN_lsw(A) & 7];
-		}
-	else
-		{
-		/* i is even */
-		ret = 1;
-		}
-	
-	if (B->neg)
-		{
-		B->neg = 0;
-		if (A->neg)
-			ret = -ret;
-		}
-
-	/* now  B  is positive and odd, so what remains to be done is
-	 * to compute the Jacobi symbol  (A/B)  and multiply it by 'ret' */
-
-	while (1)
-		{
-		/* Cohen's step 3: */
-
-		/*  B  is positive and odd */
-
-		if (BN_is_zero(A))
-			{
-			ret = BN_is_one(B) ? ret : 0;
-			goto end;
-			}
-
-		/* now  A  is non-zero */
-		i = 0;
-		while (!BN_is_bit_set(A, i))
-			i++;
-		err = !BN_rshift(A, A, i);
-		if (err) goto end;
-		if (i & 1)
-			{
-			/* i is odd */
-			/* multiply 'ret' by  $(-1)^{(B^2-1)/8}$ */
-			ret = ret * tab[BN_lsw(B) & 7];
-			}
-	
-		/* Cohen's step 4: */
-		/* multiply 'ret' by  $(-1)^{(A-1)(B-1)/4}$ */
-		if ((A->neg ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)
-			ret = -ret;
-		
-		/* (A, B) := (B mod |A|, |A|) */
-		err = !BN_nnmod(B, B, A, ctx);
-		if (err) goto end;
-		tmp = A; A = B; B = tmp;
-		tmp->neg = 0;
-		}
-end:
-	BN_CTX_end(ctx);
-	if (err)
-		return -2;
-	else
-		return ret;
-	}
+{
+    int i;
+    int ret = -2;               /* avoid 'uninitialized' warning */
+    int err = 0;
+    BIGNUM *A, *B, *tmp;
+        /*-
+         * In 'tab', only odd-indexed entries are relevant:
+         * For any odd BIGNUM n,
+         *     tab[BN_lsw(n) & 7]
+         * is $(-1)^{(n^2-1)/8}$ (using TeX notation).
+         * Note that the sign of n does not matter.
+         */
+    static const int tab[8] = { 0, 1, 0, -1, 0, -1, 0, 1 };
+
+    bn_check_top(a);
+    bn_check_top(b);
+
+    BN_CTX_start(ctx);
+    A = BN_CTX_get(ctx);
+    B = BN_CTX_get(ctx);
+    if (B == NULL)
+        goto end;
+
+    err = !BN_copy(A, a);
+    if (err)
+        goto end;
+    err = !BN_copy(B, b);
+    if (err)
+        goto end;
+
+    /*
+     * Kronecker symbol, imlemented according to Henri Cohen,
+     * "A Course in Computational Algebraic Number Theory"
+     * (algorithm 1.4.10).
+     */
+
+    /* Cohen's step 1: */
+
+    if (BN_is_zero(B)) {
+        ret = BN_abs_is_word(A, 1);
+        goto end;
+    }
+
+    /* Cohen's step 2: */
+
+    if (!BN_is_odd(A) && !BN_is_odd(B)) {
+        ret = 0;
+        goto end;
+    }
+
+    /* now  B  is non-zero */
+    i = 0;
+    while (!BN_is_bit_set(B, i))
+        i++;
+    err = !BN_rshift(B, B, i);
+    if (err)
+        goto end;
+    if (i & 1) {
+        /* i is odd */
+        /* (thus  B  was even, thus  A  must be odd!)  */
+
+        /* set 'ret' to $(-1)^{(A^2-1)/8}$ */
+        ret = tab[BN_lsw(A) & 7];
+    } else {
+        /* i is even */
+        ret = 1;
+    }
+
+    if (B->neg) {
+        B->neg = 0;
+        if (A->neg)
+            ret = -ret;
+    }
+
+    /*
+     * now B is positive and odd, so what remains to be done is to compute
+     * the Jacobi symbol (A/B) and multiply it by 'ret'
+     */
+
+    while (1) {
+        /* Cohen's step 3: */
+
+        /*  B  is positive and odd */
+
+        if (BN_is_zero(A)) {
+            ret = BN_is_one(B) ? ret : 0;
+            goto end;
+        }
+
+        /* now  A  is non-zero */
+        i = 0;
+        while (!BN_is_bit_set(A, i))
+            i++;
+        err = !BN_rshift(A, A, i);
+        if (err)
+            goto end;
+        if (i & 1) {
+            /* i is odd */
+            /* multiply 'ret' by  $(-1)^{(B^2-1)/8}$ */
+            ret = ret * tab[BN_lsw(B) & 7];
+        }
+
+        /* Cohen's step 4: */
+        /* multiply 'ret' by  $(-1)^{(A-1)(B-1)/4}$ */
+        if ((A->neg ? ~BN_lsw(A) : BN_lsw(A)) & BN_lsw(B) & 2)
+            ret = -ret;
+
+        /* (A, B) := (B mod |A|, |A|) */
+        err = !BN_nnmod(B, B, A, ctx);
+        if (err)
+            goto end;
+        tmp = A;
+        A = B;
+        B = tmp;
+        tmp->neg = 0;
+    }
+ end:
+    BN_CTX_end(ctx);
+    if (err)
+        return -2;
+    else
+        return ret;
+}