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[CRYPTO] sha1: Avoid useless memcpy()
The current code unconditionally copy the first block for every call to sha1_update(). This can be avoided if there is no pending partial block. This is always the case on the first call to sha1_update() (if the length is >= 64 of course. Furthermore, temp does need to be called if sha_transform is never invoked. Also consolidate the sha_transform calls into one to reduce code size. Signed-off-by: Nicolas Pitre <nico@cam.org> Signed-off-by: Herbert Xu <herbert@gondor.apana.org.au>
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1 changed files with 17 additions and 8 deletions
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@ -50,22 +50,31 @@ static void sha1_update(void *ctx, const u8 *data, unsigned int len)
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{
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struct sha1_ctx *sctx = ctx;
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unsigned int i, j;
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u32 temp[SHA_WORKSPACE_WORDS];
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const u8 *src;
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j = (sctx->count >> 3) & 0x3f;
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sctx->count += len << 3;
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i = 0;
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src = data;
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if ((j + len) > 63) {
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memcpy(&sctx->buffer[j], data, (i = 64-j));
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sha_transform(sctx->state, sctx->buffer, temp);
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for ( ; i + 63 < len; i += 64) {
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sha_transform(sctx->state, &data[i], temp);
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u32 temp[SHA_WORKSPACE_WORDS];
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if (j) {
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memcpy(&sctx->buffer[j], data, (i = 64-j));
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src = sctx->buffer;
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}
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do {
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sha_transform(sctx->state, src, temp);
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i += 64;
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src = &data[i];
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} while (i + 63 < len);
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memset(temp, 0, sizeof(temp));
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j = 0;
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}
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else i = 0;
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memset(temp, 0, sizeof(temp));
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memcpy(&sctx->buffer[j], &data[i], len - i);
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memcpy(&sctx->buffer[j], src, len - i);
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}
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