zfs/module/icp/algs/blake3/blake3.c

733 lines
26 KiB
C

/*
* CDDL HEADER START
*
* The contents of this file are subject to the terms of the
* Common Development and Distribution License (the "License").
* You may not use this file except in compliance with the License.
*
* You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE
* or https://opensource.org/licenses/CDDL-1.0.
* See the License for the specific language governing permissions
* and limitations under the License.
*
* When distributing Covered Code, include this CDDL HEADER in each
* file and include the License file at usr/src/OPENSOLARIS.LICENSE.
* If applicable, add the following below this CDDL HEADER, with the
* fields enclosed by brackets "[]" replaced with your own identifying
* information: Portions Copyright [yyyy] [name of copyright owner]
*
* CDDL HEADER END
*/
/*
* Based on BLAKE3 v1.3.1, https://github.com/BLAKE3-team/BLAKE3
* Copyright (c) 2019-2020 Samuel Neves and Jack O'Connor
* Copyright (c) 2021-2022 Tino Reichardt <milky-zfs@mcmilk.de>
*/
#include <sys/zfs_context.h>
#include <sys/blake3.h>
#include "blake3_impl.h"
/*
* We need 1056 byte stack for blake3_compress_subtree_wide()
* - we define this pragma to make gcc happy
*/
#if defined(__GNUC__)
#pragma GCC diagnostic ignored "-Wframe-larger-than="
#endif
/* internal used */
typedef struct {
uint32_t input_cv[8];
uint64_t counter;
uint8_t block[BLAKE3_BLOCK_LEN];
uint8_t block_len;
uint8_t flags;
} output_t;
/* internal flags */
enum blake3_flags {
CHUNK_START = 1 << 0,
CHUNK_END = 1 << 1,
PARENT = 1 << 2,
ROOT = 1 << 3,
KEYED_HASH = 1 << 4,
DERIVE_KEY_CONTEXT = 1 << 5,
DERIVE_KEY_MATERIAL = 1 << 6,
};
/* internal start */
static void chunk_state_init(blake3_chunk_state_t *ctx,
const uint32_t key[8], uint8_t flags)
{
memcpy(ctx->cv, key, BLAKE3_KEY_LEN);
ctx->chunk_counter = 0;
memset(ctx->buf, 0, BLAKE3_BLOCK_LEN);
ctx->buf_len = 0;
ctx->blocks_compressed = 0;
ctx->flags = flags;
}
static void chunk_state_reset(blake3_chunk_state_t *ctx,
const uint32_t key[8], uint64_t chunk_counter)
{
memcpy(ctx->cv, key, BLAKE3_KEY_LEN);
ctx->chunk_counter = chunk_counter;
ctx->blocks_compressed = 0;
memset(ctx->buf, 0, BLAKE3_BLOCK_LEN);
ctx->buf_len = 0;
}
static size_t chunk_state_len(const blake3_chunk_state_t *ctx)
{
return (BLAKE3_BLOCK_LEN * (size_t)ctx->blocks_compressed) +
((size_t)ctx->buf_len);
}
static size_t chunk_state_fill_buf(blake3_chunk_state_t *ctx,
const uint8_t *input, size_t input_len)
{
size_t take = BLAKE3_BLOCK_LEN - ((size_t)ctx->buf_len);
if (take > input_len) {
take = input_len;
}
uint8_t *dest = ctx->buf + ((size_t)ctx->buf_len);
memcpy(dest, input, take);
ctx->buf_len += (uint8_t)take;
return (take);
}
static uint8_t chunk_state_maybe_start_flag(const blake3_chunk_state_t *ctx)
{
if (ctx->blocks_compressed == 0) {
return (CHUNK_START);
} else {
return (0);
}
}
static output_t make_output(const uint32_t input_cv[8],
const uint8_t *block, uint8_t block_len,
uint64_t counter, uint8_t flags)
{
output_t ret;
memcpy(ret.input_cv, input_cv, 32);
memcpy(ret.block, block, BLAKE3_BLOCK_LEN);
ret.block_len = block_len;
ret.counter = counter;
ret.flags = flags;
return (ret);
}
/*
* Chaining values within a given chunk (specifically the compress_in_place
* interface) are represented as words. This avoids unnecessary bytes<->words
* conversion overhead in the portable implementation. However, the hash_many
* interface handles both user input and parent node blocks, so it accepts
* bytes. For that reason, chaining values in the CV stack are represented as
* bytes.
*/
static void output_chaining_value(const blake3_ops_t *ops,
const output_t *ctx, uint8_t cv[32])
{
uint32_t cv_words[8];
memcpy(cv_words, ctx->input_cv, 32);
ops->compress_in_place(cv_words, ctx->block, ctx->block_len,
ctx->counter, ctx->flags);
store_cv_words(cv, cv_words);
}
static void output_root_bytes(const blake3_ops_t *ops, const output_t *ctx,
uint64_t seek, uint8_t *out, size_t out_len)
{
uint64_t output_block_counter = seek / 64;
size_t offset_within_block = seek % 64;
uint8_t wide_buf[64];
while (out_len > 0) {
ops->compress_xof(ctx->input_cv, ctx->block, ctx->block_len,
output_block_counter, ctx->flags | ROOT, wide_buf);
size_t available_bytes = 64 - offset_within_block;
size_t memcpy_len;
if (out_len > available_bytes) {
memcpy_len = available_bytes;
} else {
memcpy_len = out_len;
}
memcpy(out, wide_buf + offset_within_block, memcpy_len);
out += memcpy_len;
out_len -= memcpy_len;
output_block_counter += 1;
offset_within_block = 0;
}
}
static void chunk_state_update(const blake3_ops_t *ops,
blake3_chunk_state_t *ctx, const uint8_t *input, size_t input_len)
{
if (ctx->buf_len > 0) {
size_t take = chunk_state_fill_buf(ctx, input, input_len);
input += take;
input_len -= take;
if (input_len > 0) {
ops->compress_in_place(ctx->cv, ctx->buf,
BLAKE3_BLOCK_LEN, ctx->chunk_counter,
ctx->flags|chunk_state_maybe_start_flag(ctx));
ctx->blocks_compressed += 1;
ctx->buf_len = 0;
memset(ctx->buf, 0, BLAKE3_BLOCK_LEN);
}
}
while (input_len > BLAKE3_BLOCK_LEN) {
ops->compress_in_place(ctx->cv, input, BLAKE3_BLOCK_LEN,
ctx->chunk_counter,
ctx->flags|chunk_state_maybe_start_flag(ctx));
ctx->blocks_compressed += 1;
input += BLAKE3_BLOCK_LEN;
input_len -= BLAKE3_BLOCK_LEN;
}
size_t take = chunk_state_fill_buf(ctx, input, input_len);
input += take;
input_len -= take;
}
static output_t chunk_state_output(const blake3_chunk_state_t *ctx)
{
uint8_t block_flags =
ctx->flags | chunk_state_maybe_start_flag(ctx) | CHUNK_END;
return (make_output(ctx->cv, ctx->buf, ctx->buf_len, ctx->chunk_counter,
block_flags));
}
static output_t parent_output(const uint8_t block[BLAKE3_BLOCK_LEN],
const uint32_t key[8], uint8_t flags)
{
return (make_output(key, block, BLAKE3_BLOCK_LEN, 0, flags | PARENT));
}
/*
* Given some input larger than one chunk, return the number of bytes that
* should go in the left subtree. This is the largest power-of-2 number of
* chunks that leaves at least 1 byte for the right subtree.
*/
static size_t left_len(size_t content_len)
{
/*
* Subtract 1 to reserve at least one byte for the right side.
* content_len
* should always be greater than BLAKE3_CHUNK_LEN.
*/
size_t full_chunks = (content_len - 1) / BLAKE3_CHUNK_LEN;
return (round_down_to_power_of_2(full_chunks) * BLAKE3_CHUNK_LEN);
}
/*
* Use SIMD parallelism to hash up to MAX_SIMD_DEGREE chunks at the same time
* on a single thread. Write out the chunk chaining values and return the
* number of chunks hashed. These chunks are never the root and never empty;
* those cases use a different codepath.
*/
static size_t compress_chunks_parallel(const blake3_ops_t *ops,
const uint8_t *input, size_t input_len, const uint32_t key[8],
uint64_t chunk_counter, uint8_t flags, uint8_t *out)
{
const uint8_t *chunks_array[MAX_SIMD_DEGREE];
size_t input_position = 0;
size_t chunks_array_len = 0;
while (input_len - input_position >= BLAKE3_CHUNK_LEN) {
chunks_array[chunks_array_len] = &input[input_position];
input_position += BLAKE3_CHUNK_LEN;
chunks_array_len += 1;
}
ops->hash_many(chunks_array, chunks_array_len, BLAKE3_CHUNK_LEN /
BLAKE3_BLOCK_LEN, key, chunk_counter, B_TRUE, flags, CHUNK_START,
CHUNK_END, out);
/*
* Hash the remaining partial chunk, if there is one. Note that the
* empty chunk (meaning the empty message) is a different codepath.
*/
if (input_len > input_position) {
uint64_t counter = chunk_counter + (uint64_t)chunks_array_len;
blake3_chunk_state_t chunk_state;
chunk_state_init(&chunk_state, key, flags);
chunk_state.chunk_counter = counter;
chunk_state_update(ops, &chunk_state, &input[input_position],
input_len - input_position);
output_t output = chunk_state_output(&chunk_state);
output_chaining_value(ops, &output, &out[chunks_array_len *
BLAKE3_OUT_LEN]);
return (chunks_array_len + 1);
} else {
return (chunks_array_len);
}
}
/*
* Use SIMD parallelism to hash up to MAX_SIMD_DEGREE parents at the same time
* on a single thread. Write out the parent chaining values and return the
* number of parents hashed. (If there's an odd input chaining value left over,
* return it as an additional output.) These parents are never the root and
* never empty; those cases use a different codepath.
*/
static size_t compress_parents_parallel(const blake3_ops_t *ops,
const uint8_t *child_chaining_values, size_t num_chaining_values,
const uint32_t key[8], uint8_t flags, uint8_t *out)
{
const uint8_t *parents_array[MAX_SIMD_DEGREE_OR_2];
size_t parents_array_len = 0;
while (num_chaining_values - (2 * parents_array_len) >= 2) {
parents_array[parents_array_len] = &child_chaining_values[2 *
parents_array_len * BLAKE3_OUT_LEN];
parents_array_len += 1;
}
ops->hash_many(parents_array, parents_array_len, 1, key, 0, B_FALSE,
flags | PARENT, 0, 0, out);
/* If there's an odd child left over, it becomes an output. */
if (num_chaining_values > 2 * parents_array_len) {
memcpy(&out[parents_array_len * BLAKE3_OUT_LEN],
&child_chaining_values[2 * parents_array_len *
BLAKE3_OUT_LEN], BLAKE3_OUT_LEN);
return (parents_array_len + 1);
} else {
return (parents_array_len);
}
}
/*
* The wide helper function returns (writes out) an array of chaining values
* and returns the length of that array. The number of chaining values returned
* is the dyanmically detected SIMD degree, at most MAX_SIMD_DEGREE. Or fewer,
* if the input is shorter than that many chunks. The reason for maintaining a
* wide array of chaining values going back up the tree, is to allow the
* implementation to hash as many parents in parallel as possible.
*
* As a special case when the SIMD degree is 1, this function will still return
* at least 2 outputs. This guarantees that this function doesn't perform the
* root compression. (If it did, it would use the wrong flags, and also we
* wouldn't be able to implement exendable ouput.) Note that this function is
* not used when the whole input is only 1 chunk long; that's a different
* codepath.
*
* Why not just have the caller split the input on the first update(), instead
* of implementing this special rule? Because we don't want to limit SIMD or
* multi-threading parallelism for that update().
*/
static size_t blake3_compress_subtree_wide(const blake3_ops_t *ops,
const uint8_t *input, size_t input_len, const uint32_t key[8],
uint64_t chunk_counter, uint8_t flags, uint8_t *out)
{
/*
* Note that the single chunk case does *not* bump the SIMD degree up
* to 2 when it is 1. If this implementation adds multi-threading in
* the future, this gives us the option of multi-threading even the
* 2-chunk case, which can help performance on smaller platforms.
*/
if (input_len <= (size_t)(ops->degree * BLAKE3_CHUNK_LEN)) {
return (compress_chunks_parallel(ops, input, input_len, key,
chunk_counter, flags, out));
}
/*
* With more than simd_degree chunks, we need to recurse. Start by
* dividing the input into left and right subtrees. (Note that this is
* only optimal as long as the SIMD degree is a power of 2. If we ever
* get a SIMD degree of 3 or something, we'll need a more complicated
* strategy.)
*/
size_t left_input_len = left_len(input_len);
size_t right_input_len = input_len - left_input_len;
const uint8_t *right_input = &input[left_input_len];
uint64_t right_chunk_counter = chunk_counter +
(uint64_t)(left_input_len / BLAKE3_CHUNK_LEN);
/*
* Make space for the child outputs. Here we use MAX_SIMD_DEGREE_OR_2
* to account for the special case of returning 2 outputs when the
* SIMD degree is 1.
*/
uint8_t cv_array[2 * MAX_SIMD_DEGREE_OR_2 * BLAKE3_OUT_LEN];
size_t degree = ops->degree;
if (left_input_len > BLAKE3_CHUNK_LEN && degree == 1) {
/*
* The special case: We always use a degree of at least two,
* to make sure there are two outputs. Except, as noted above,
* at the chunk level, where we allow degree=1. (Note that the
* 1-chunk-input case is a different codepath.)
*/
degree = 2;
}
uint8_t *right_cvs = &cv_array[degree * BLAKE3_OUT_LEN];
/*
* Recurse! If this implementation adds multi-threading support in the
* future, this is where it will go.
*/
size_t left_n = blake3_compress_subtree_wide(ops, input, left_input_len,
key, chunk_counter, flags, cv_array);
size_t right_n = blake3_compress_subtree_wide(ops, right_input,
right_input_len, key, right_chunk_counter, flags, right_cvs);
/*
* The special case again. If simd_degree=1, then we'll have left_n=1
* and right_n=1. Rather than compressing them into a single output,
* return them directly, to make sure we always have at least two
* outputs.
*/
if (left_n == 1) {
memcpy(out, cv_array, 2 * BLAKE3_OUT_LEN);
return (2);
}
/* Otherwise, do one layer of parent node compression. */
size_t num_chaining_values = left_n + right_n;
return compress_parents_parallel(ops, cv_array,
num_chaining_values, key, flags, out);
}
/*
* Hash a subtree with compress_subtree_wide(), and then condense the resulting
* list of chaining values down to a single parent node. Don't compress that
* last parent node, however. Instead, return its message bytes (the
* concatenated chaining values of its children). This is necessary when the
* first call to update() supplies a complete subtree, because the topmost
* parent node of that subtree could end up being the root. It's also necessary
* for extended output in the general case.
*
* As with compress_subtree_wide(), this function is not used on inputs of 1
* chunk or less. That's a different codepath.
*/
static void compress_subtree_to_parent_node(const blake3_ops_t *ops,
const uint8_t *input, size_t input_len, const uint32_t key[8],
uint64_t chunk_counter, uint8_t flags, uint8_t out[2 * BLAKE3_OUT_LEN])
{
uint8_t cv_array[MAX_SIMD_DEGREE_OR_2 * BLAKE3_OUT_LEN];
size_t num_cvs = blake3_compress_subtree_wide(ops, input, input_len,
key, chunk_counter, flags, cv_array);
/*
* If MAX_SIMD_DEGREE is greater than 2 and there's enough input,
* compress_subtree_wide() returns more than 2 chaining values. Condense
* them into 2 by forming parent nodes repeatedly.
*/
uint8_t out_array[MAX_SIMD_DEGREE_OR_2 * BLAKE3_OUT_LEN / 2];
while (num_cvs > 2) {
num_cvs = compress_parents_parallel(ops, cv_array, num_cvs, key,
flags, out_array);
memcpy(cv_array, out_array, num_cvs * BLAKE3_OUT_LEN);
}
memcpy(out, cv_array, 2 * BLAKE3_OUT_LEN);
}
static void hasher_init_base(BLAKE3_CTX *ctx, const uint32_t key[8],
uint8_t flags)
{
memcpy(ctx->key, key, BLAKE3_KEY_LEN);
chunk_state_init(&ctx->chunk, key, flags);
ctx->cv_stack_len = 0;
ctx->ops = blake3_impl_get_ops();
}
/*
* As described in hasher_push_cv() below, we do "lazy merging", delaying
* merges until right before the next CV is about to be added. This is
* different from the reference implementation. Another difference is that we
* aren't always merging 1 chunk at a time. Instead, each CV might represent
* any power-of-two number of chunks, as long as the smaller-above-larger
* stack order is maintained. Instead of the "count the trailing 0-bits"
* algorithm described in the spec, we use a "count the total number of
* 1-bits" variant that doesn't require us to retain the subtree size of the
* CV on top of the stack. The principle is the same: each CV that should
* remain in the stack is represented by a 1-bit in the total number of chunks
* (or bytes) so far.
*/
static void hasher_merge_cv_stack(BLAKE3_CTX *ctx, uint64_t total_len)
{
size_t post_merge_stack_len = (size_t)popcnt(total_len);
while (ctx->cv_stack_len > post_merge_stack_len) {
uint8_t *parent_node =
&ctx->cv_stack[(ctx->cv_stack_len - 2) * BLAKE3_OUT_LEN];
output_t output =
parent_output(parent_node, ctx->key, ctx->chunk.flags);
output_chaining_value(ctx->ops, &output, parent_node);
ctx->cv_stack_len -= 1;
}
}
/*
* In reference_impl.rs, we merge the new CV with existing CVs from the stack
* before pushing it. We can do that because we know more input is coming, so
* we know none of the merges are root.
*
* This setting is different. We want to feed as much input as possible to
* compress_subtree_wide(), without setting aside anything for the chunk_state.
* If the user gives us 64 KiB, we want to parallelize over all 64 KiB at once
* as a single subtree, if at all possible.
*
* This leads to two problems:
* 1) This 64 KiB input might be the only call that ever gets made to update.
* In this case, the root node of the 64 KiB subtree would be the root node
* of the whole tree, and it would need to be ROOT finalized. We can't
* compress it until we know.
* 2) This 64 KiB input might complete a larger tree, whose root node is
* similarly going to be the the root of the whole tree. For example, maybe
* we have 196 KiB (that is, 128 + 64) hashed so far. We can't compress the
* node at the root of the 256 KiB subtree until we know how to finalize it.
*
* The second problem is solved with "lazy merging". That is, when we're about
* to add a CV to the stack, we don't merge it with anything first, as the
* reference impl does. Instead we do merges using the *previous* CV that was
* added, which is sitting on top of the stack, and we put the new CV
* (unmerged) on top of the stack afterwards. This guarantees that we never
* merge the root node until finalize().
*
* Solving the first problem requires an additional tool,
* compress_subtree_to_parent_node(). That function always returns the top
* *two* chaining values of the subtree it's compressing. We then do lazy
* merging with each of them separately, so that the second CV will always
* remain unmerged. (That also helps us support extendable output when we're
* hashing an input all-at-once.)
*/
static void hasher_push_cv(BLAKE3_CTX *ctx, uint8_t new_cv[BLAKE3_OUT_LEN],
uint64_t chunk_counter)
{
hasher_merge_cv_stack(ctx, chunk_counter);
memcpy(&ctx->cv_stack[ctx->cv_stack_len * BLAKE3_OUT_LEN], new_cv,
BLAKE3_OUT_LEN);
ctx->cv_stack_len += 1;
}
void
Blake3_Init(BLAKE3_CTX *ctx)
{
hasher_init_base(ctx, BLAKE3_IV, 0);
}
void
Blake3_InitKeyed(BLAKE3_CTX *ctx, const uint8_t key[BLAKE3_KEY_LEN])
{
uint32_t key_words[8];
load_key_words(key, key_words);
hasher_init_base(ctx, key_words, KEYED_HASH);
}
static void
Blake3_Update2(BLAKE3_CTX *ctx, const void *input, size_t input_len)
{
/*
* Explicitly checking for zero avoids causing UB by passing a null
* pointer to memcpy. This comes up in practice with things like:
* std::vector<uint8_t> v;
* blake3_hasher_update(&hasher, v.data(), v.size());
*/
if (input_len == 0) {
return;
}
const uint8_t *input_bytes = (const uint8_t *)input;
/*
* If we have some partial chunk bytes in the internal chunk_state, we
* need to finish that chunk first.
*/
if (chunk_state_len(&ctx->chunk) > 0) {
size_t take = BLAKE3_CHUNK_LEN - chunk_state_len(&ctx->chunk);
if (take > input_len) {
take = input_len;
}
chunk_state_update(ctx->ops, &ctx->chunk, input_bytes, take);
input_bytes += take;
input_len -= take;
/*
* If we've filled the current chunk and there's more coming,
* finalize this chunk and proceed. In this case we know it's
* not the root.
*/
if (input_len > 0) {
output_t output = chunk_state_output(&ctx->chunk);
uint8_t chunk_cv[32];
output_chaining_value(ctx->ops, &output, chunk_cv);
hasher_push_cv(ctx, chunk_cv, ctx->chunk.chunk_counter);
chunk_state_reset(&ctx->chunk, ctx->key,
ctx->chunk.chunk_counter + 1);
} else {
return;
}
}
/*
* Now the chunk_state is clear, and we have more input. If there's
* more than a single chunk (so, definitely not the root chunk), hash
* the largest whole subtree we can, with the full benefits of SIMD
* (and maybe in the future, multi-threading) parallelism. Two
* restrictions:
* - The subtree has to be a power-of-2 number of chunks. Only
* subtrees along the right edge can be incomplete, and we don't know
* where the right edge is going to be until we get to finalize().
* - The subtree must evenly divide the total number of chunks up
* until this point (if total is not 0). If the current incomplete
* subtree is only waiting for 1 more chunk, we can't hash a subtree
* of 4 chunks. We have to complete the current subtree first.
* Because we might need to break up the input to form powers of 2, or
* to evenly divide what we already have, this part runs in a loop.
*/
while (input_len > BLAKE3_CHUNK_LEN) {
size_t subtree_len = round_down_to_power_of_2(input_len);
uint64_t count_so_far =
ctx->chunk.chunk_counter * BLAKE3_CHUNK_LEN;
/*
* Shrink the subtree_len until it evenly divides the count so
* far. We know that subtree_len itself is a power of 2, so we
* can use a bitmasking trick instead of an actual remainder
* operation. (Note that if the caller consistently passes
* power-of-2 inputs of the same size, as is hopefully
* typical, this loop condition will always fail, and
* subtree_len will always be the full length of the input.)
*
* An aside: We don't have to shrink subtree_len quite this
* much. For example, if count_so_far is 1, we could pass 2
* chunks to compress_subtree_to_parent_node. Since we'll get
* 2 CVs back, we'll still get the right answer in the end,
* and we might get to use 2-way SIMD parallelism. The problem
* with this optimization, is that it gets us stuck always
* hashing 2 chunks. The total number of chunks will remain
* odd, and we'll never graduate to higher degrees of
* parallelism. See
* https://github.com/BLAKE3-team/BLAKE3/issues/69.
*/
while ((((uint64_t)(subtree_len - 1)) & count_so_far) != 0) {
subtree_len /= 2;
}
/*
* The shrunken subtree_len might now be 1 chunk long. If so,
* hash that one chunk by itself. Otherwise, compress the
* subtree into a pair of CVs.
*/
uint64_t subtree_chunks = subtree_len / BLAKE3_CHUNK_LEN;
if (subtree_len <= BLAKE3_CHUNK_LEN) {
blake3_chunk_state_t chunk_state;
chunk_state_init(&chunk_state, ctx->key,
ctx->chunk.flags);
chunk_state.chunk_counter = ctx->chunk.chunk_counter;
chunk_state_update(ctx->ops, &chunk_state, input_bytes,
subtree_len);
output_t output = chunk_state_output(&chunk_state);
uint8_t cv[BLAKE3_OUT_LEN];
output_chaining_value(ctx->ops, &output, cv);
hasher_push_cv(ctx, cv, chunk_state.chunk_counter);
} else {
/*
* This is the high-performance happy path, though
* getting here depends on the caller giving us a long
* enough input.
*/
uint8_t cv_pair[2 * BLAKE3_OUT_LEN];
compress_subtree_to_parent_node(ctx->ops, input_bytes,
subtree_len, ctx->key, ctx-> chunk.chunk_counter,
ctx->chunk.flags, cv_pair);
hasher_push_cv(ctx, cv_pair, ctx->chunk.chunk_counter);
hasher_push_cv(ctx, &cv_pair[BLAKE3_OUT_LEN],
ctx->chunk.chunk_counter + (subtree_chunks / 2));
}
ctx->chunk.chunk_counter += subtree_chunks;
input_bytes += subtree_len;
input_len -= subtree_len;
}
/*
* If there's any remaining input less than a full chunk, add it to
* the chunk state. In that case, also do a final merge loop to make
* sure the subtree stack doesn't contain any unmerged pairs. The
* remaining input means we know these merges are non-root. This merge
* loop isn't strictly necessary here, because hasher_push_chunk_cv
* already does its own merge loop, but it simplifies
* blake3_hasher_finalize below.
*/
if (input_len > 0) {
chunk_state_update(ctx->ops, &ctx->chunk, input_bytes,
input_len);
hasher_merge_cv_stack(ctx, ctx->chunk.chunk_counter);
}
}
void
Blake3_Update(BLAKE3_CTX *ctx, const void *input, size_t todo)
{
size_t done = 0;
const uint8_t *data = input;
const size_t block_max = 1024 * 64;
/* max feed buffer to leave the stack size small */
while (todo != 0) {
size_t block = (todo >= block_max) ? block_max : todo;
Blake3_Update2(ctx, data + done, block);
done += block;
todo -= block;
}
}
void
Blake3_Final(const BLAKE3_CTX *ctx, uint8_t *out)
{
Blake3_FinalSeek(ctx, 0, out, BLAKE3_OUT_LEN);
}
void
Blake3_FinalSeek(const BLAKE3_CTX *ctx, uint64_t seek, uint8_t *out,
size_t out_len)
{
/*
* Explicitly checking for zero avoids causing UB by passing a null
* pointer to memcpy. This comes up in practice with things like:
* std::vector<uint8_t> v;
* blake3_hasher_finalize(&hasher, v.data(), v.size());
*/
if (out_len == 0) {
return;
}
/* If the subtree stack is empty, then the current chunk is the root. */
if (ctx->cv_stack_len == 0) {
output_t output = chunk_state_output(&ctx->chunk);
output_root_bytes(ctx->ops, &output, seek, out, out_len);
return;
}
/*
* If there are any bytes in the chunk state, finalize that chunk and
* do a roll-up merge between that chunk hash and every subtree in the
* stack. In this case, the extra merge loop at the end of
* blake3_hasher_update guarantees that none of the subtrees in the
* stack need to be merged with each other first. Otherwise, if there
* are no bytes in the chunk state, then the top of the stack is a
* chunk hash, and we start the merge from that.
*/
output_t output;
size_t cvs_remaining;
if (chunk_state_len(&ctx->chunk) > 0) {
cvs_remaining = ctx->cv_stack_len;
output = chunk_state_output(&ctx->chunk);
} else {
/* There are always at least 2 CVs in the stack in this case. */
cvs_remaining = ctx->cv_stack_len - 2;
output = parent_output(&ctx->cv_stack[cvs_remaining * 32],
ctx->key, ctx->chunk.flags);
}
while (cvs_remaining > 0) {
cvs_remaining -= 1;
uint8_t parent_block[BLAKE3_BLOCK_LEN];
memcpy(parent_block, &ctx->cv_stack[cvs_remaining * 32], 32);
output_chaining_value(ctx->ops, &output, &parent_block[32]);
output = parent_output(parent_block, ctx->key,
ctx->chunk.flags);
}
output_root_bytes(ctx->ops, &output, seek, out, out_len);
}