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hash_md5_sha: use common finalization routine for MD5 and sha1/256. -15 bytes

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Signed-off-by: Denys Vlasenko <dvlasenk@redhat.com>
This commit is contained in:
Denys Vlasenko 2010-10-18 14:48:30 +02:00
parent b5aa1d95a1
commit c48a5c607d
2 changed files with 110 additions and 103 deletions
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46
include/libbb.h
View File

@ -1515,35 +1515,14 @@ enum {
};
void FAST_FUNC read_base64(FILE *src_stream, FILE *dst_stream, int flags);
typedef struct sha1_ctx_t {
uint32_t hash[8]; /* 5, +3 elements for sha256 */
uint64_t total64; /* must be directly after hash[] */
uint8_t wbuffer[64]; /* NB: always correctly aligned for uint64_t */
void (*process_block)(struct sha1_ctx_t*) FAST_FUNC;
} sha1_ctx_t;
void sha1_begin(sha1_ctx_t *ctx) FAST_FUNC;
void sha1_hash(sha1_ctx_t *ctx, const void *data, size_t length) FAST_FUNC;
void sha1_end(sha1_ctx_t *ctx, void *resbuf) FAST_FUNC;
typedef struct sha1_ctx_t sha256_ctx_t;
void sha256_begin(sha256_ctx_t *ctx) FAST_FUNC;
#define sha256_hash sha1_hash
#define sha256_end sha1_end
typedef struct sha512_ctx_t {
uint64_t hash[8];
uint64_t total64[2]; /* must be directly after hash[] */
uint8_t wbuffer[128]; /* NB: always correctly aligned for uint64_t */
} sha512_ctx_t;
void sha512_begin(sha512_ctx_t *ctx) FAST_FUNC;
void sha512_hash(sha512_ctx_t *ctx, const void *buffer, size_t len) FAST_FUNC;
void sha512_end(sha512_ctx_t *ctx, void *resbuf) FAST_FUNC;
#if 1
typedef struct md5_ctx_t {
char wbuffer[64]; /* NB: always correctly aligned for uint64_t */
uint64_t total64;
uint32_t A;
uint32_t B;
uint32_t C;
uint32_t D;
uint64_t total64;
char wbuffer[64]; /* NB: always correctly aligned for uint64_t */
} md5_ctx_t;
#else
/* libbb/md5prime.c uses a bit different one: */
@ -1556,6 +1535,27 @@ typedef struct md5_ctx_t {
void md5_begin(md5_ctx_t *ctx) FAST_FUNC;
void md5_hash(md5_ctx_t *ctx, const void *data, size_t length) FAST_FUNC;
void md5_end(md5_ctx_t *ctx, void *resbuf) FAST_FUNC;
typedef struct sha1_ctx_t {
uint8_t wbuffer[64]; /* NB: always correctly aligned for uint64_t */
uint64_t total64; /* must be directly before hash[] */
uint32_t hash[8]; /* 5, +3 elements for sha256 */
void (*process_block)(struct sha1_ctx_t*) FAST_FUNC;
} sha1_ctx_t;
void sha1_begin(sha1_ctx_t *ctx) FAST_FUNC;
void sha1_hash(sha1_ctx_t *ctx, const void *data, size_t length) FAST_FUNC;
void sha1_end(sha1_ctx_t *ctx, void *resbuf) FAST_FUNC;
typedef struct sha1_ctx_t sha256_ctx_t;
void sha256_begin(sha256_ctx_t *ctx) FAST_FUNC;
#define sha256_hash sha1_hash
#define sha256_end sha1_end
typedef struct sha512_ctx_t {
uint64_t total64[2]; /* must be directly before hash[] */
uint64_t hash[8];
uint8_t wbuffer[128]; /* NB: always correctly aligned for uint64_t */
} sha512_ctx_t;
void sha512_begin(sha512_ctx_t *ctx) FAST_FUNC;
void sha512_hash(sha512_ctx_t *ctx, const void *buffer, size_t len) FAST_FUNC;
void sha512_end(sha512_ctx_t *ctx, void *resbuf) FAST_FUNC;
uint32_t *crc32_filltable(uint32_t *tbl256, int endian) FAST_FUNC;

167
libbb/hash_md5_sha.c
View File

@ -1,4 +1,72 @@
/* vi: set sw=4 ts=4: */
/*
* Utility routines.
*
* Copyright (C) 2010 Denys Vlasenko
*
* Licensed under GPLv2 or later, see file LICENSE in this source tree.
*/
#include "libbb.h"
/* gcc 4.2.1 optimizes rotr64 better with inline than with macro
* (for rotX32, there is no difference). Why? My guess is that
* macro requires clever common subexpression elimination heuristics
* in gcc, while inline basically forces it to happen.
*/
//#define rotl32(x,n) (((x) << (n)) | ((x) >> (32 - (n))))
static ALWAYS_INLINE uint32_t rotl32(uint32_t x, unsigned n)
{
return (x << n) | (x >> (32 - n));
}
//#define rotr32(x,n) (((x) >> (n)) | ((x) << (32 - (n))))
static ALWAYS_INLINE uint32_t rotr32(uint32_t x, unsigned n)
{
return (x >> n) | (x << (32 - n));
}
/* rotr64 in needed for sha512 only: */
//#define rotr64(x,n) (((x) >> (n)) | ((x) << (64 - (n))))
static ALWAYS_INLINE uint64_t rotr64(uint64_t x, unsigned n)
{
return (x >> n) | (x << (64 - n));
}
typedef struct common64_ctx_t {
char wbuffer[64]; /* NB: always correctly aligned for uint64_t */
uint64_t total64;
} common64_ctx_t;
typedef void FAST_FUNC process_block64_func(void*);
static void FAST_FUNC common64_end(void *vctx, process_block64_func process_block64, int swap_needed)
{
common64_ctx_t *ctx = vctx;
unsigned bufpos = ctx->total64 & 63;
/* Pad the buffer to the next 64-byte boundary with 0x80,0,0,0... */
ctx->wbuffer[bufpos++] = 0x80;
/* This loop iterates either once or twice, no more, no less */
while (1) {
unsigned remaining = 64 - bufpos;
memset(ctx->wbuffer + bufpos, 0, remaining);
/* Do we have enough space for the length count? */
if (remaining >= 8) {
/* Store the 64-bit counter of bits in the buffer */
uint64_t t = ctx->total64 << 3;
if (swap_needed)
t = bb_bswap_64(t);
/* wbuffer is suitably aligned for this */
*(uint64_t *) (&ctx->wbuffer[64 - 8]) = t;
}
process_block64(ctx);
if (remaining >= 8)
break;
bufpos = 0;
}
}
/*
* Based on shasum from http://www.netsw.org/crypto/hash/
* Majorly hacked up to use Dr Brian Gladman's sha1 code
@ -28,31 +96,6 @@
* then rebuild and compare "shaNNNsum bigfile" results.
*/
#include "libbb.h"
/* gcc 4.2.1 optimizes rotr64 better with inline than with macro
* (for rotX32, there is no difference). Why? My guess is that
* macro requires clever common subexpression elimination heuristics
* in gcc, while inline basically forces it to happen.
*/
//#define rotl32(x,n) (((x) << (n)) | ((x) >> (32 - (n))))
static ALWAYS_INLINE uint32_t rotl32(uint32_t x, unsigned n)
{
return (x << n) | (x >> (32 - n));
}
//#define rotr32(x,n) (((x) >> (n)) | ((x) << (32 - (n))))
static ALWAYS_INLINE uint32_t rotr32(uint32_t x, unsigned n)
{
return (x >> n) | (x << (32 - n));
}
/* rotr64 in needed for sha512 only: */
//#define rotr64(x,n) (((x) >> (n)) | ((x) << (64 - (n))))
static ALWAYS_INLINE uint64_t rotr64(uint64_t x, unsigned n)
{
return (x >> n) | (x << (64 - n));
}
static void FAST_FUNC sha1_process_block64(sha1_ctx_t *ctx)
{
unsigned t;
@ -308,6 +351,8 @@ void FAST_FUNC sha1_begin(sha1_ctx_t *ctx)
}
static const uint32_t init256[] = {
0,
0,
0x6a09e667,
0xbb67ae85,
0x3c6ef372,
@ -316,10 +361,10 @@ static const uint32_t init256[] = {
0x9b05688c,
0x1f83d9ab,
0x5be0cd19,
0,
0,
};
static const uint32_t init512_lo[] = {
0,
0,
0xf3bcc908,
0x84caa73b,
0xfe94f82b,
@ -328,16 +373,14 @@ static const uint32_t init512_lo[] = {
0x2b3e6c1f,
0xfb41bd6b,
0x137e2179,
0,
0,
};
/* Initialize structure containing state of computation.
(FIPS 180-2:5.3.2) */
void FAST_FUNC sha256_begin(sha256_ctx_t *ctx)
{
memcpy(ctx->hash, init256, sizeof(init256));
/*ctx->total64 = 0; - done by extending init256 with two 32-bit zeros */
memcpy(&ctx->total64, init256, sizeof(init256));
/*ctx->total64 = 0; - done by prepending two 32-bit zeros to init256 */
ctx->process_block = sha256_process_block64;
}
@ -346,9 +389,10 @@ void FAST_FUNC sha256_begin(sha256_ctx_t *ctx)
void FAST_FUNC sha512_begin(sha512_ctx_t *ctx)
{
int i;
/* Two extra iterations zero out ctx->total64[] */
for (i = 0; i < 8+2; i++)
ctx->hash[i] = ((uint64_t)(init256[i]) << 32) + init512_lo[i];
/* Two extra iterations zero out ctx->total64[2] */
uint64_t *tp = ctx->total64;
for (i = 0; i < 2+8; i++)
tp[i] = ((uint64_t)(init256[i]) << 32) + init512_lo[i];
/*ctx->total64[0] = ctx->total64[1] = 0; - already done */
}
@ -448,37 +492,19 @@ void FAST_FUNC sha512_hash(sha512_ctx_t *ctx, const void *buffer, size_t len)
/* Used also for sha256 */
void FAST_FUNC sha1_end(sha1_ctx_t *ctx, void *resbuf)
{
unsigned bufpos = ctx->total64 & 63;
unsigned hash_size;
/* Pad the buffer to the next 64-byte boundary with 0x80,0,0,0... */
ctx->wbuffer[bufpos++] = 0x80;
/* SHA stores total in BE, need to swap on LE arches: */
common64_end(ctx, (process_block64_func*) ctx->process_block, /*swap_needed:*/ BB_LITTLE_ENDIAN);
/* This loop iterates either once or twice, no more, no less */
while (1) {
unsigned remaining = 64 - bufpos;
memset(ctx->wbuffer + bufpos, 0, remaining);
/* Do we have enough space for the length count? */
if (remaining >= 8) {
/* Store the 64-bit counter of bits in the buffer in BE format */
uint64_t t = ctx->total64 << 3;
t = SWAP_BE64(t);
/* wbuffer is suitably aligned for this */
*(uint64_t *) (&ctx->wbuffer[64 - 8]) = t;
}
ctx->process_block(ctx);
if (remaining >= 8)
break;
bufpos = 0;
}
bufpos = (ctx->process_block == sha1_process_block64) ? 5 : 8;
hash_size = (ctx->process_block == sha1_process_block64) ? 5 : 8;
/* This way we do not impose alignment constraints on resbuf: */
if (BB_LITTLE_ENDIAN) {
unsigned i;
for (i = 0; i < bufpos; ++i)
for (i = 0; i < hash_size; ++i)
ctx->hash[i] = SWAP_BE32(ctx->hash[i]);
}
memcpy(resbuf, ctx->hash, sizeof(ctx->hash[0]) * bufpos);
memcpy(resbuf, ctx->hash, sizeof(ctx->hash[0]) * hash_size);
}
void FAST_FUNC sha512_end(sha512_ctx_t *ctx, void *resbuf)
@ -566,7 +592,7 @@ void FAST_FUNC md5_begin(md5_ctx_t *ctx)
#define FI(b, c, d) (c ^ (b | ~d))
/* Hash a single block, 64 bytes long and 4-byte aligned */
static void md5_process_block64(md5_ctx_t *ctx)
static void FAST_FUNC md5_process_block64(md5_ctx_t *ctx)
{
#if MD5_SIZE_VS_SPEED > 0
/* Before we start, one word to the strange constants.
@ -927,27 +953,8 @@ void FAST_FUNC md5_hash(md5_ctx_t *ctx, const void *buffer, size_t len)
*/
void FAST_FUNC md5_end(md5_ctx_t *ctx, void *resbuf)
{
unsigned bufpos = ctx->total64 & 63;
/* Pad the buffer to the next 64-byte boundary with 0x80,0,0,0... */
ctx->wbuffer[bufpos++] = 0x80;
/* This loop iterates either once or twice, no more, no less */
while (1) {
unsigned remaining = 64 - bufpos;
memset(ctx->wbuffer + bufpos, 0, remaining);
/* Do we have enough space for the length count? */
if (remaining >= 8) {
/* Store the 64-bit counter of bits in the buffer in LE format */
uint64_t t = ctx->total64 << 3;
t = SWAP_LE64(t);
/* wbuffer is suitably aligned for this */
*(uint64_t *) (&ctx->wbuffer[64 - 8]) = t;
}
md5_process_block64(ctx);
if (remaining >= 8)
break;
bufpos = 0;
}
/* MD5 stores total in LE, need to swap on BE arches: */
common64_end(ctx, (process_block64_func*) md5_process_block64, /*swap_needed:*/ BB_BIG_ENDIAN);
/* The MD5 result is in little endian byte order.
* We (ab)use the fact that A-D are consecutive in memory.