block-sha1: make the 'ntohl()' part of the first SHA1 loop

This helps a teeny bit.  But what I -really- want to do is to avoid the
whole 80-array loop, and do the xor updates as I go along..

Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
Signed-off-by: Junio C Hamano <gitster@pobox.com>

diff --git a/block-sha1/sha1.c b/block-sha1/sha1.c
index 698e435..13da511 100644 (file)
--- a/block-sha1/sha1.c
+++ b/block-sha1/sha1.c
@@ -100,27 +100,31 @@ static void blk_SHA1Block(blk_SHA_CTX *ctx, const unsigned int *data)
        unsigned int A,B,C,D,E,TEMP;
        unsigned int W[80];
 
-       for (t = 0; t < 16; t++)
-               W[t] = htonl(data[t]);
-
-       /* Unroll it? */
-       for (t = 16; t <= 79; t++)
-               W[t] = SHA_ROL(W[t-3] ^ W[t-8] ^ W[t-14] ^ W[t-16], 1);
-
        A = ctx->H[0];
        B = ctx->H[1];
        C = ctx->H[2];
        D = ctx->H[3];
        E = ctx->H[4];
 
-#define T_0_19(t) \
+#define T_0_15(t) \
+       TEMP = htonl(data[t]); W[t] = TEMP; \
+       TEMP += SHA_ROL(A,5) + (((C^D)&B)^D)     + E + 0x5a827999; \
+       E = D; D = C; C = SHA_ROR(B, 2); B = A; A = TEMP; \
+
+       T_0_15( 0); T_0_15( 1); T_0_15( 2); T_0_15( 3); T_0_15( 4);
+       T_0_15( 5); T_0_15( 6); T_0_15( 7); T_0_15( 8); T_0_15( 9);
+       T_0_15(10); T_0_15(11); T_0_15(12); T_0_15(13); T_0_15(14);
+       T_0_15(15);
+
+       /* Unroll it? */
+       for (t = 16; t <= 79; t++)
+               W[t] = SHA_ROL(W[t-3] ^ W[t-8] ^ W[t-14] ^ W[t-16], 1);
+
+#define T_16_19(t) \
        TEMP = SHA_ROL(A,5) + (((C^D)&B)^D)     + E + W[t] + 0x5a827999; \
        E = D; D = C; C = SHA_ROR(B, 2); B = A; A = TEMP;
 
-       T_0_19( 0); T_0_19( 1); T_0_19( 2); T_0_19( 3); T_0_19( 4);
-       T_0_19( 5); T_0_19( 6); T_0_19( 7); T_0_19( 8); T_0_19( 9);
-       T_0_19(10); T_0_19(11); T_0_19(12); T_0_19(13); T_0_19(14);
-       T_0_19(15); T_0_19(16); T_0_19(17); T_0_19(18); T_0_19(19);
+       T_16_19(16); T_16_19(17); T_16_19(18); T_16_19(19);
 
 #define T_20_39(t) \
        TEMP = SHA_ROL(A,5) + (B^C^D)           + E + W[t] + 0x6ed9eba1; \