1 /*
2  * CDDL HEADER START
3  *
4  * The contents of this file are subject to the terms of the
5  * Common Development and Distribution License (the "License").
6  * You may not use this file except in compliance with the License.
7  *
8  * You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE
9  * or https://opensource.org/licenses/CDDL-1.0.
10  * See the License for the specific language governing permissions
11  * and limitations under the License.
12  *
13  * When distributing Covered Code, include this CDDL HEADER in each
14  * file and include the License file at usr/src/OPENSOLARIS.LICENSE.
15  * If applicable, add the following below this CDDL HEADER, with the
16  * fields enclosed by brackets "[]" replaced with your own identifying
17  * information: Portions Copyright [yyyy] [name of copyright owner]
18  *
19  * CDDL HEADER END
20  */
21 
22 /*
23  * Based on BLAKE3 v1.3.1, https://github.com/BLAKE3-team/BLAKE3
24  * Copyright (c) 2019-2020 Samuel Neves and Jack O'Connor
25  * Copyright (c) 2021-2022 Tino Reichardt <milky-zfs@mcmilk.de>
26  */
27 
28 #include <sys/simd.h>
29 #include <sys/zfs_context.h>
30 #include <sys/blake3.h>
31 
32 #include "blake3_impl.h"
33 
34 /*
35  * We need 1056 byte stack for blake3_compress_subtree_wide()
36  * - we define this pragma to make gcc happy
37  */
38 #if defined(__GNUC__)
39 #pragma GCC diagnostic ignored "-Wframe-larger-than="
40 #endif
41 
42 /* internal used */
43 typedef struct {
44 	uint32_t input_cv[8];
45 	uint64_t counter;
46 	uint8_t block[BLAKE3_BLOCK_LEN];
47 	uint8_t block_len;
48 	uint8_t flags;
49 } output_t;
50 
51 /* internal flags */
52 enum blake3_flags {
53 	CHUNK_START		= 1 << 0,
54 	CHUNK_END		= 1 << 1,
55 	PARENT			= 1 << 2,
56 	ROOT			= 1 << 3,
57 	KEYED_HASH		= 1 << 4,
58 	DERIVE_KEY_CONTEXT	= 1 << 5,
59 	DERIVE_KEY_MATERIAL	= 1 << 6,
60 };
61 
62 /* internal start */
chunk_state_init(blake3_chunk_state_t * ctx,const uint32_t key[8],uint8_t flags)63 static void chunk_state_init(blake3_chunk_state_t *ctx,
64     const uint32_t key[8], uint8_t flags)
65 {
66 	memcpy(ctx->cv, key, BLAKE3_KEY_LEN);
67 	ctx->chunk_counter = 0;
68 	memset(ctx->buf, 0, BLAKE3_BLOCK_LEN);
69 	ctx->buf_len = 0;
70 	ctx->blocks_compressed = 0;
71 	ctx->flags = flags;
72 }
73 
chunk_state_reset(blake3_chunk_state_t * ctx,const uint32_t key[8],uint64_t chunk_counter)74 static void chunk_state_reset(blake3_chunk_state_t *ctx,
75     const uint32_t key[8], uint64_t chunk_counter)
76 {
77 	memcpy(ctx->cv, key, BLAKE3_KEY_LEN);
78 	ctx->chunk_counter = chunk_counter;
79 	ctx->blocks_compressed = 0;
80 	memset(ctx->buf, 0, BLAKE3_BLOCK_LEN);
81 	ctx->buf_len = 0;
82 }
83 
chunk_state_len(const blake3_chunk_state_t * ctx)84 static size_t chunk_state_len(const blake3_chunk_state_t *ctx)
85 {
86 	return (BLAKE3_BLOCK_LEN * (size_t)ctx->blocks_compressed) +
87 	    ((size_t)ctx->buf_len);
88 }
89 
chunk_state_fill_buf(blake3_chunk_state_t * ctx,const uint8_t * input,size_t input_len)90 static size_t chunk_state_fill_buf(blake3_chunk_state_t *ctx,
91     const uint8_t *input, size_t input_len)
92 {
93 	size_t take = BLAKE3_BLOCK_LEN - ((size_t)ctx->buf_len);
94 	if (take > input_len) {
95 		take = input_len;
96 	}
97 	uint8_t *dest = ctx->buf + ((size_t)ctx->buf_len);
98 	memcpy(dest, input, take);
99 	ctx->buf_len += (uint8_t)take;
100 	return (take);
101 }
102 
chunk_state_maybe_start_flag(const blake3_chunk_state_t * ctx)103 static uint8_t chunk_state_maybe_start_flag(const blake3_chunk_state_t *ctx)
104 {
105 	if (ctx->blocks_compressed == 0) {
106 		return (CHUNK_START);
107 	} else {
108 		return (0);
109 	}
110 }
111 
make_output(const uint32_t input_cv[8],const uint8_t * block,uint8_t block_len,uint64_t counter,uint8_t flags)112 static output_t make_output(const uint32_t input_cv[8],
113     const uint8_t *block, uint8_t block_len,
114     uint64_t counter, uint8_t flags)
115 {
116 	output_t ret;
117 	memcpy(ret.input_cv, input_cv, 32);
118 	memcpy(ret.block, block, BLAKE3_BLOCK_LEN);
119 	ret.block_len = block_len;
120 	ret.counter = counter;
121 	ret.flags = flags;
122 	return (ret);
123 }
124 
125 /*
126  * Chaining values within a given chunk (specifically the compress_in_place
127  * interface) are represented as words. This avoids unnecessary bytes<->words
128  * conversion overhead in the portable implementation. However, the hash_many
129  * interface handles both user input and parent node blocks, so it accepts
130  * bytes. For that reason, chaining values in the CV stack are represented as
131  * bytes.
132  */
output_chaining_value(const blake3_ops_t * ops,const output_t * ctx,uint8_t cv[32])133 static void output_chaining_value(const blake3_ops_t *ops,
134     const output_t *ctx, uint8_t cv[32])
135 {
136 	uint32_t cv_words[8];
137 	memcpy(cv_words, ctx->input_cv, 32);
138 	ops->compress_in_place(cv_words, ctx->block, ctx->block_len,
139 	    ctx->counter, ctx->flags);
140 	store_cv_words(cv, cv_words);
141 }
142 
output_root_bytes(const blake3_ops_t * ops,const output_t * ctx,uint64_t seek,uint8_t * out,size_t out_len)143 static void output_root_bytes(const blake3_ops_t *ops, const output_t *ctx,
144     uint64_t seek, uint8_t *out, size_t out_len)
145 {
146 	uint64_t output_block_counter = seek / 64;
147 	size_t offset_within_block = seek % 64;
148 	uint8_t wide_buf[64];
149 	while (out_len > 0) {
150 		ops->compress_xof(ctx->input_cv, ctx->block, ctx->block_len,
151 		    output_block_counter, ctx->flags | ROOT, wide_buf);
152 		size_t available_bytes = 64 - offset_within_block;
153 		size_t memcpy_len;
154 		if (out_len > available_bytes) {
155 			memcpy_len = available_bytes;
156 		} else {
157 			memcpy_len = out_len;
158 		}
159 		memcpy(out, wide_buf + offset_within_block, memcpy_len);
160 		out += memcpy_len;
161 		out_len -= memcpy_len;
162 		output_block_counter += 1;
163 		offset_within_block = 0;
164 	}
165 }
166 
chunk_state_update(const blake3_ops_t * ops,blake3_chunk_state_t * ctx,const uint8_t * input,size_t input_len)167 static void chunk_state_update(const blake3_ops_t *ops,
168     blake3_chunk_state_t *ctx, const uint8_t *input, size_t input_len)
169 {
170 	if (ctx->buf_len > 0) {
171 		size_t take = chunk_state_fill_buf(ctx, input, input_len);
172 		input += take;
173 		input_len -= take;
174 		if (input_len > 0) {
175 			ops->compress_in_place(ctx->cv, ctx->buf,
176 			    BLAKE3_BLOCK_LEN, ctx->chunk_counter,
177 			    ctx->flags|chunk_state_maybe_start_flag(ctx));
178 			ctx->blocks_compressed += 1;
179 			ctx->buf_len = 0;
180 			memset(ctx->buf, 0, BLAKE3_BLOCK_LEN);
181 		}
182 	}
183 
184 	while (input_len > BLAKE3_BLOCK_LEN) {
185 		ops->compress_in_place(ctx->cv, input, BLAKE3_BLOCK_LEN,
186 		    ctx->chunk_counter,
187 		    ctx->flags|chunk_state_maybe_start_flag(ctx));
188 		ctx->blocks_compressed += 1;
189 		input += BLAKE3_BLOCK_LEN;
190 		input_len -= BLAKE3_BLOCK_LEN;
191 	}
192 
193 	chunk_state_fill_buf(ctx, input, input_len);
194 }
195 
chunk_state_output(const blake3_chunk_state_t * ctx)196 static output_t chunk_state_output(const blake3_chunk_state_t *ctx)
197 {
198 	uint8_t block_flags =
199 	    ctx->flags | chunk_state_maybe_start_flag(ctx) | CHUNK_END;
200 	return (make_output(ctx->cv, ctx->buf, ctx->buf_len, ctx->chunk_counter,
201 	    block_flags));
202 }
203 
parent_output(const uint8_t block[BLAKE3_BLOCK_LEN],const uint32_t key[8],uint8_t flags)204 static output_t parent_output(const uint8_t block[BLAKE3_BLOCK_LEN],
205     const uint32_t key[8], uint8_t flags)
206 {
207 	return (make_output(key, block, BLAKE3_BLOCK_LEN, 0, flags | PARENT));
208 }
209 
210 /*
211  * Given some input larger than one chunk, return the number of bytes that
212  * should go in the left subtree. This is the largest power-of-2 number of
213  * chunks that leaves at least 1 byte for the right subtree.
214  */
left_len(size_t content_len)215 static size_t left_len(size_t content_len)
216 {
217 	/*
218 	 * Subtract 1 to reserve at least one byte for the right side.
219 	 * content_len
220 	 * should always be greater than BLAKE3_CHUNK_LEN.
221 	 */
222 	size_t full_chunks = (content_len - 1) / BLAKE3_CHUNK_LEN;
223 	return (round_down_to_power_of_2(full_chunks) * BLAKE3_CHUNK_LEN);
224 }
225 
226 /*
227  * Use SIMD parallelism to hash up to MAX_SIMD_DEGREE chunks at the same time
228  * on a single thread. Write out the chunk chaining values and return the
229  * number of chunks hashed. These chunks are never the root and never empty;
230  * those cases use a different codepath.
231  */
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)232 static size_t compress_chunks_parallel(const blake3_ops_t *ops,
233     const uint8_t *input, size_t input_len, const uint32_t key[8],
234     uint64_t chunk_counter, uint8_t flags, uint8_t *out)
235 {
236 	const uint8_t *chunks_array[MAX_SIMD_DEGREE];
237 	size_t input_position = 0;
238 	size_t chunks_array_len = 0;
239 	while (input_len - input_position >= BLAKE3_CHUNK_LEN) {
240 		chunks_array[chunks_array_len] = &input[input_position];
241 		input_position += BLAKE3_CHUNK_LEN;
242 		chunks_array_len += 1;
243 	}
244 
245 	ops->hash_many(chunks_array, chunks_array_len, BLAKE3_CHUNK_LEN /
246 	    BLAKE3_BLOCK_LEN, key, chunk_counter, B_TRUE, flags, CHUNK_START,
247 	    CHUNK_END, out);
248 
249 	/*
250 	 * Hash the remaining partial chunk, if there is one. Note that the
251 	 * empty chunk (meaning the empty message) is a different codepath.
252 	 */
253 	if (input_len > input_position) {
254 		uint64_t counter = chunk_counter + (uint64_t)chunks_array_len;
255 		blake3_chunk_state_t chunk_state;
256 		chunk_state_init(&chunk_state, key, flags);
257 		chunk_state.chunk_counter = counter;
258 		chunk_state_update(ops, &chunk_state, &input[input_position],
259 		    input_len - input_position);
260 		output_t output = chunk_state_output(&chunk_state);
261 		output_chaining_value(ops, &output, &out[chunks_array_len *
262 		    BLAKE3_OUT_LEN]);
263 		return (chunks_array_len + 1);
264 	} else {
265 		return (chunks_array_len);
266 	}
267 }
268 
269 /*
270  * Use SIMD parallelism to hash up to MAX_SIMD_DEGREE parents at the same time
271  * on a single thread. Write out the parent chaining values and return the
272  * number of parents hashed. (If there's an odd input chaining value left over,
273  * return it as an additional output.) These parents are never the root and
274  * never empty; those cases use a different codepath.
275  */
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)276 static size_t compress_parents_parallel(const blake3_ops_t *ops,
277     const uint8_t *child_chaining_values, size_t num_chaining_values,
278     const uint32_t key[8], uint8_t flags, uint8_t *out)
279 {
280 	const uint8_t *parents_array[MAX_SIMD_DEGREE_OR_2] = {0};
281 	size_t parents_array_len = 0;
282 
283 	while (num_chaining_values - (2 * parents_array_len) >= 2) {
284 		parents_array[parents_array_len] = &child_chaining_values[2 *
285 		    parents_array_len * BLAKE3_OUT_LEN];
286 		parents_array_len += 1;
287 	}
288 
289 	ops->hash_many(parents_array, parents_array_len, 1, key, 0, B_FALSE,
290 	    flags | PARENT, 0, 0, out);
291 
292 	/* If there's an odd child left over, it becomes an output. */
293 	if (num_chaining_values > 2 * parents_array_len) {
294 		memcpy(&out[parents_array_len * BLAKE3_OUT_LEN],
295 		    &child_chaining_values[2 * parents_array_len *
296 		    BLAKE3_OUT_LEN], BLAKE3_OUT_LEN);
297 		return (parents_array_len + 1);
298 	} else {
299 		return (parents_array_len);
300 	}
301 }
302 
303 /*
304  * The wide helper function returns (writes out) an array of chaining values
305  * and returns the length of that array. The number of chaining values returned
306  * is the dyanmically detected SIMD degree, at most MAX_SIMD_DEGREE. Or fewer,
307  * if the input is shorter than that many chunks. The reason for maintaining a
308  * wide array of chaining values going back up the tree, is to allow the
309  * implementation to hash as many parents in parallel as possible.
310  *
311  * As a special case when the SIMD degree is 1, this function will still return
312  * at least 2 outputs. This guarantees that this function doesn't perform the
313  * root compression. (If it did, it would use the wrong flags, and also we
314  * wouldn't be able to implement exendable ouput.) Note that this function is
315  * not used when the whole input is only 1 chunk long; that's a different
316  * codepath.
317  *
318  * Why not just have the caller split the input on the first update(), instead
319  * of implementing this special rule? Because we don't want to limit SIMD or
320  * multi-threading parallelism for that update().
321  */
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)322 static size_t blake3_compress_subtree_wide(const blake3_ops_t *ops,
323     const uint8_t *input, size_t input_len, const uint32_t key[8],
324     uint64_t chunk_counter, uint8_t flags, uint8_t *out)
325 {
326 	/*
327 	 * Note that the single chunk case does *not* bump the SIMD degree up
328 	 * to 2 when it is 1. If this implementation adds multi-threading in
329 	 * the future, this gives us the option of multi-threading even the
330 	 * 2-chunk case, which can help performance on smaller platforms.
331 	 */
332 	if (input_len <= (size_t)(ops->degree * BLAKE3_CHUNK_LEN)) {
333 		return (compress_chunks_parallel(ops, input, input_len, key,
334 		    chunk_counter, flags, out));
335 	}
336 
337 
338 	/*
339 	 * With more than simd_degree chunks, we need to recurse. Start by
340 	 * dividing the input into left and right subtrees. (Note that this is
341 	 * only optimal as long as the SIMD degree is a power of 2. If we ever
342 	 * get a SIMD degree of 3 or something, we'll need a more complicated
343 	 * strategy.)
344 	 */
345 	size_t left_input_len = left_len(input_len);
346 	size_t right_input_len = input_len - left_input_len;
347 	const uint8_t *right_input = &input[left_input_len];
348 	uint64_t right_chunk_counter = chunk_counter +
349 	    (uint64_t)(left_input_len / BLAKE3_CHUNK_LEN);
350 
351 	/*
352 	 * Make space for the child outputs. Here we use MAX_SIMD_DEGREE_OR_2
353 	 * to account for the special case of returning 2 outputs when the
354 	 * SIMD degree is 1.
355 	 */
356 	uint8_t cv_array[2 * MAX_SIMD_DEGREE_OR_2 * BLAKE3_OUT_LEN];
357 	size_t degree = ops->degree;
358 	if (left_input_len > BLAKE3_CHUNK_LEN && degree == 1) {
359 
360 		/*
361 		 * The special case: We always use a degree of at least two,
362 		 * to make sure there are two outputs. Except, as noted above,
363 		 * at the chunk level, where we allow degree=1. (Note that the
364 		 * 1-chunk-input case is a different codepath.)
365 		 */
366 		degree = 2;
367 	}
368 	uint8_t *right_cvs = &cv_array[degree * BLAKE3_OUT_LEN];
369 
370 	/*
371 	 * Recurse! If this implementation adds multi-threading support in the
372 	 * future, this is where it will go.
373 	 */
374 	size_t left_n = blake3_compress_subtree_wide(ops, input, left_input_len,
375 	    key, chunk_counter, flags, cv_array);
376 	size_t right_n = blake3_compress_subtree_wide(ops, right_input,
377 	    right_input_len, key, right_chunk_counter, flags, right_cvs);
378 
379 	/*
380 	 * The special case again. If simd_degree=1, then we'll have left_n=1
381 	 * and right_n=1. Rather than compressing them into a single output,
382 	 * return them directly, to make sure we always have at least two
383 	 * outputs.
384 	 */
385 	if (left_n == 1) {
386 		memcpy(out, cv_array, 2 * BLAKE3_OUT_LEN);
387 		return (2);
388 	}
389 
390 	/* Otherwise, do one layer of parent node compression. */
391 	size_t num_chaining_values = left_n + right_n;
392 	return compress_parents_parallel(ops, cv_array,
393 	    num_chaining_values, key, flags, out);
394 }
395 
396 /*
397  * Hash a subtree with compress_subtree_wide(), and then condense the resulting
398  * list of chaining values down to a single parent node. Don't compress that
399  * last parent node, however. Instead, return its message bytes (the
400  * concatenated chaining values of its children). This is necessary when the
401  * first call to update() supplies a complete subtree, because the topmost
402  * parent node of that subtree could end up being the root. It's also necessary
403  * for extended output in the general case.
404  *
405  * As with compress_subtree_wide(), this function is not used on inputs of 1
406  * chunk or less. That's a different codepath.
407  */
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])408 static void compress_subtree_to_parent_node(const blake3_ops_t *ops,
409     const uint8_t *input, size_t input_len, const uint32_t key[8],
410     uint64_t chunk_counter, uint8_t flags, uint8_t out[2 * BLAKE3_OUT_LEN])
411 {
412 	uint8_t cv_array[MAX_SIMD_DEGREE_OR_2 * BLAKE3_OUT_LEN];
413 	size_t num_cvs = blake3_compress_subtree_wide(ops, input, input_len,
414 	    key, chunk_counter, flags, cv_array);
415 
416 	/*
417 	 * If MAX_SIMD_DEGREE is greater than 2 and there's enough input,
418 	 * compress_subtree_wide() returns more than 2 chaining values. Condense
419 	 * them into 2 by forming parent nodes repeatedly.
420 	 */
421 	uint8_t out_array[MAX_SIMD_DEGREE_OR_2 * BLAKE3_OUT_LEN / 2];
422 	while (num_cvs > 2) {
423 		num_cvs = compress_parents_parallel(ops, cv_array, num_cvs, key,
424 		    flags, out_array);
425 		memcpy(cv_array, out_array, num_cvs * BLAKE3_OUT_LEN);
426 	}
427 	memcpy(out, cv_array, 2 * BLAKE3_OUT_LEN);
428 }
429 
hasher_init_base(BLAKE3_CTX * ctx,const uint32_t key[8],uint8_t flags)430 static void hasher_init_base(BLAKE3_CTX *ctx, const uint32_t key[8],
431     uint8_t flags)
432 {
433 	memcpy(ctx->key, key, BLAKE3_KEY_LEN);
434 	chunk_state_init(&ctx->chunk, key, flags);
435 	ctx->cv_stack_len = 0;
436 	ctx->ops = blake3_get_ops();
437 }
438 
439 /*
440  * As described in hasher_push_cv() below, we do "lazy merging", delaying
441  * merges until right before the next CV is about to be added. This is
442  * different from the reference implementation. Another difference is that we
443  * aren't always merging 1 chunk at a time. Instead, each CV might represent
444  * any power-of-two number of chunks, as long as the smaller-above-larger
445  * stack order is maintained. Instead of the "count the trailing 0-bits"
446  * algorithm described in the spec, we use a "count the total number of
447  * 1-bits" variant that doesn't require us to retain the subtree size of the
448  * CV on top of the stack. The principle is the same: each CV that should
449  * remain in the stack is represented by a 1-bit in the total number of chunks
450  * (or bytes) so far.
451  */
hasher_merge_cv_stack(BLAKE3_CTX * ctx,uint64_t total_len)452 static void hasher_merge_cv_stack(BLAKE3_CTX *ctx, uint64_t total_len)
453 {
454 	size_t post_merge_stack_len = (size_t)popcnt(total_len);
455 	while (ctx->cv_stack_len > post_merge_stack_len) {
456 		uint8_t *parent_node =
457 		    &ctx->cv_stack[(ctx->cv_stack_len - 2) * BLAKE3_OUT_LEN];
458 		output_t output =
459 		    parent_output(parent_node, ctx->key, ctx->chunk.flags);
460 		output_chaining_value(ctx->ops, &output, parent_node);
461 		ctx->cv_stack_len -= 1;
462 	}
463 }
464 
465 /*
466  * In reference_impl.rs, we merge the new CV with existing CVs from the stack
467  * before pushing it. We can do that because we know more input is coming, so
468  * we know none of the merges are root.
469  *
470  * This setting is different. We want to feed as much input as possible to
471  * compress_subtree_wide(), without setting aside anything for the chunk_state.
472  * If the user gives us 64 KiB, we want to parallelize over all 64 KiB at once
473  * as a single subtree, if at all possible.
474  *
475  * This leads to two problems:
476  * 1) This 64 KiB input might be the only call that ever gets made to update.
477  *    In this case, the root node of the 64 KiB subtree would be the root node
478  *    of the whole tree, and it would need to be ROOT finalized. We can't
479  *    compress it until we know.
480  * 2) This 64 KiB input might complete a larger tree, whose root node is
481  *    similarly going to be the the root of the whole tree. For example, maybe
482  *    we have 196 KiB (that is, 128 + 64) hashed so far. We can't compress the
483  *    node at the root of the 256 KiB subtree until we know how to finalize it.
484  *
485  * The second problem is solved with "lazy merging". That is, when we're about
486  * to add a CV to the stack, we don't merge it with anything first, as the
487  * reference impl does. Instead we do merges using the *previous* CV that was
488  * added, which is sitting on top of the stack, and we put the new CV
489  * (unmerged) on top of the stack afterwards. This guarantees that we never
490  * merge the root node until finalize().
491  *
492  * Solving the first problem requires an additional tool,
493  * compress_subtree_to_parent_node(). That function always returns the top
494  * *two* chaining values of the subtree it's compressing. We then do lazy
495  * merging with each of them separately, so that the second CV will always
496  * remain unmerged. (That also helps us support extendable output when we're
497  * hashing an input all-at-once.)
498  */
hasher_push_cv(BLAKE3_CTX * ctx,uint8_t new_cv[BLAKE3_OUT_LEN],uint64_t chunk_counter)499 static void hasher_push_cv(BLAKE3_CTX *ctx, uint8_t new_cv[BLAKE3_OUT_LEN],
500     uint64_t chunk_counter)
501 {
502 	hasher_merge_cv_stack(ctx, chunk_counter);
503 	memcpy(&ctx->cv_stack[ctx->cv_stack_len * BLAKE3_OUT_LEN], new_cv,
504 	    BLAKE3_OUT_LEN);
505 	ctx->cv_stack_len += 1;
506 }
507 
508 void
Blake3_Init(BLAKE3_CTX * ctx)509 Blake3_Init(BLAKE3_CTX *ctx)
510 {
511 	hasher_init_base(ctx, BLAKE3_IV, 0);
512 }
513 
514 void
Blake3_InitKeyed(BLAKE3_CTX * ctx,const uint8_t key[BLAKE3_KEY_LEN])515 Blake3_InitKeyed(BLAKE3_CTX *ctx, const uint8_t key[BLAKE3_KEY_LEN])
516 {
517 	uint32_t key_words[8];
518 	load_key_words(key, key_words);
519 	hasher_init_base(ctx, key_words, KEYED_HASH);
520 }
521 
522 static void
Blake3_Update2(BLAKE3_CTX * ctx,const void * input,size_t input_len)523 Blake3_Update2(BLAKE3_CTX *ctx, const void *input, size_t input_len)
524 {
525 	/*
526 	 * Explicitly checking for zero avoids causing UB by passing a null
527 	 * pointer to memcpy. This comes up in practice with things like:
528 	 *   std::vector<uint8_t> v;
529 	 *   blake3_hasher_update(&hasher, v.data(), v.size());
530 	 */
531 	if (input_len == 0) {
532 		return;
533 	}
534 
535 	const uint8_t *input_bytes = (const uint8_t *)input;
536 
537 	/*
538 	 * If we have some partial chunk bytes in the internal chunk_state, we
539 	 * need to finish that chunk first.
540 	 */
541 	if (chunk_state_len(&ctx->chunk) > 0) {
542 		size_t take = BLAKE3_CHUNK_LEN - chunk_state_len(&ctx->chunk);
543 		if (take > input_len) {
544 			take = input_len;
545 		}
546 		chunk_state_update(ctx->ops, &ctx->chunk, input_bytes, take);
547 		input_bytes += take;
548 		input_len -= take;
549 		/*
550 		 * If we've filled the current chunk and there's more coming,
551 		 * finalize this chunk and proceed. In this case we know it's
552 		 * not the root.
553 		 */
554 		if (input_len > 0) {
555 			output_t output = chunk_state_output(&ctx->chunk);
556 			uint8_t chunk_cv[32];
557 			output_chaining_value(ctx->ops, &output, chunk_cv);
558 			hasher_push_cv(ctx, chunk_cv, ctx->chunk.chunk_counter);
559 			chunk_state_reset(&ctx->chunk, ctx->key,
560 			    ctx->chunk.chunk_counter + 1);
561 		} else {
562 			return;
563 		}
564 	}
565 
566 	/*
567 	 * Now the chunk_state is clear, and we have more input. If there's
568 	 * more than a single chunk (so, definitely not the root chunk), hash
569 	 * the largest whole subtree we can, with the full benefits of SIMD
570 	 * (and maybe in the future, multi-threading) parallelism. Two
571 	 * restrictions:
572 	 * - The subtree has to be a power-of-2 number of chunks. Only
573 	 *   subtrees along the right edge can be incomplete, and we don't know
574 	 *   where the right edge is going to be until we get to finalize().
575 	 * - The subtree must evenly divide the total number of chunks up
576 	 *   until this point (if total is not 0). If the current incomplete
577 	 *   subtree is only waiting for 1 more chunk, we can't hash a subtree
578 	 *   of 4 chunks. We have to complete the current subtree first.
579 	 * Because we might need to break up the input to form powers of 2, or
580 	 * to evenly divide what we already have, this part runs in a loop.
581 	 */
582 	while (input_len > BLAKE3_CHUNK_LEN) {
583 		size_t subtree_len = round_down_to_power_of_2(input_len);
584 		uint64_t count_so_far =
585 		    ctx->chunk.chunk_counter * BLAKE3_CHUNK_LEN;
586 		/*
587 		 * Shrink the subtree_len until it evenly divides the count so
588 		 * far. We know that subtree_len itself is a power of 2, so we
589 		 * can use a bitmasking trick instead of an actual remainder
590 		 * operation. (Note that if the caller consistently passes
591 		 * power-of-2 inputs of the same size, as is hopefully
592 		 * typical, this loop condition will always fail, and
593 		 * subtree_len will always be the full length of the input.)
594 		 *
595 		 * An aside: We don't have to shrink subtree_len quite this
596 		 * much. For example, if count_so_far is 1, we could pass 2
597 		 * chunks to compress_subtree_to_parent_node. Since we'll get
598 		 * 2 CVs back, we'll still get the right answer in the end,
599 		 * and we might get to use 2-way SIMD parallelism. The problem
600 		 * with this optimization, is that it gets us stuck always
601 		 * hashing 2 chunks. The total number of chunks will remain
602 		 * odd, and we'll never graduate to higher degrees of
603 		 * parallelism. See
604 		 * https://github.com/BLAKE3-team/BLAKE3/issues/69.
605 		 */
606 		while ((((uint64_t)(subtree_len - 1)) & count_so_far) != 0) {
607 			subtree_len /= 2;
608 		}
609 		/*
610 		 * The shrunken subtree_len might now be 1 chunk long. If so,
611 		 * hash that one chunk by itself. Otherwise, compress the
612 		 * subtree into a pair of CVs.
613 		 */
614 		uint64_t subtree_chunks = subtree_len / BLAKE3_CHUNK_LEN;
615 		if (subtree_len <= BLAKE3_CHUNK_LEN) {
616 			blake3_chunk_state_t chunk_state;
617 			chunk_state_init(&chunk_state, ctx->key,
618 			    ctx->chunk.flags);
619 			chunk_state.chunk_counter = ctx->chunk.chunk_counter;
620 			chunk_state_update(ctx->ops, &chunk_state, input_bytes,
621 			    subtree_len);
622 			output_t output = chunk_state_output(&chunk_state);
623 			uint8_t cv[BLAKE3_OUT_LEN];
624 			output_chaining_value(ctx->ops, &output, cv);
625 			hasher_push_cv(ctx, cv, chunk_state.chunk_counter);
626 		} else {
627 			/*
628 			 * This is the high-performance happy path, though
629 			 * getting here depends on the caller giving us a long
630 			 * enough input.
631 			 */
632 			uint8_t cv_pair[2 * BLAKE3_OUT_LEN];
633 			compress_subtree_to_parent_node(ctx->ops, input_bytes,
634 			    subtree_len, ctx->key, ctx-> chunk.chunk_counter,
635 			    ctx->chunk.flags, cv_pair);
636 			hasher_push_cv(ctx, cv_pair, ctx->chunk.chunk_counter);
637 			hasher_push_cv(ctx, &cv_pair[BLAKE3_OUT_LEN],
638 			    ctx->chunk.chunk_counter + (subtree_chunks / 2));
639 		}
640 		ctx->chunk.chunk_counter += subtree_chunks;
641 		input_bytes += subtree_len;
642 		input_len -= subtree_len;
643 	}
644 
645 	/*
646 	 * If there's any remaining input less than a full chunk, add it to
647 	 * the chunk state. In that case, also do a final merge loop to make
648 	 * sure the subtree stack doesn't contain any unmerged pairs. The
649 	 * remaining input means we know these merges are non-root. This merge
650 	 * loop isn't strictly necessary here, because hasher_push_chunk_cv
651 	 * already does its own merge loop, but it simplifies
652 	 * blake3_hasher_finalize below.
653 	 */
654 	if (input_len > 0) {
655 		chunk_state_update(ctx->ops, &ctx->chunk, input_bytes,
656 		    input_len);
657 		hasher_merge_cv_stack(ctx, ctx->chunk.chunk_counter);
658 	}
659 }
660 
661 void
Blake3_Update(BLAKE3_CTX * ctx,const void * input,size_t todo)662 Blake3_Update(BLAKE3_CTX *ctx, const void *input, size_t todo)
663 {
664 	size_t done = 0;
665 	const uint8_t *data = input;
666 	const size_t block_max = 1024 * 64;
667 
668 	/* max feed buffer to leave the stack size small */
669 	while (todo != 0) {
670 		size_t block = (todo >= block_max) ? block_max : todo;
671 		Blake3_Update2(ctx, data + done, block);
672 		done += block;
673 		todo -= block;
674 	}
675 }
676 
677 void
Blake3_Final(const BLAKE3_CTX * ctx,uint8_t * out)678 Blake3_Final(const BLAKE3_CTX *ctx, uint8_t *out)
679 {
680 	Blake3_FinalSeek(ctx, 0, out, BLAKE3_OUT_LEN);
681 }
682 
683 void
Blake3_FinalSeek(const BLAKE3_CTX * ctx,uint64_t seek,uint8_t * out,size_t out_len)684 Blake3_FinalSeek(const BLAKE3_CTX *ctx, uint64_t seek, uint8_t *out,
685     size_t out_len)
686 {
687 	/*
688 	 * Explicitly checking for zero avoids causing UB by passing a null
689 	 * pointer to memcpy. This comes up in practice with things like:
690 	 *   std::vector<uint8_t> v;
691 	 *   blake3_hasher_finalize(&hasher, v.data(), v.size());
692 	 */
693 	if (out_len == 0) {
694 		return;
695 	}
696 	/* If the subtree stack is empty, then the current chunk is the root. */
697 	if (ctx->cv_stack_len == 0) {
698 		output_t output = chunk_state_output(&ctx->chunk);
699 		output_root_bytes(ctx->ops, &output, seek, out, out_len);
700 		return;
701 	}
702 	/*
703 	 * If there are any bytes in the chunk state, finalize that chunk and
704 	 * do a roll-up merge between that chunk hash and every subtree in the
705 	 * stack. In this case, the extra merge loop at the end of
706 	 * blake3_hasher_update guarantees that none of the subtrees in the
707 	 * stack need to be merged with each other first. Otherwise, if there
708 	 * are no bytes in the chunk state, then the top of the stack is a
709 	 * chunk hash, and we start the merge from that.
710 	 */
711 	output_t output;
712 	size_t cvs_remaining;
713 	if (chunk_state_len(&ctx->chunk) > 0) {
714 		cvs_remaining = ctx->cv_stack_len;
715 		output = chunk_state_output(&ctx->chunk);
716 	} else {
717 		/* There are always at least 2 CVs in the stack in this case. */
718 		cvs_remaining = ctx->cv_stack_len - 2;
719 		output = parent_output(&ctx->cv_stack[cvs_remaining * 32],
720 		    ctx->key, ctx->chunk.flags);
721 	}
722 	while (cvs_remaining > 0) {
723 		cvs_remaining -= 1;
724 		uint8_t parent_block[BLAKE3_BLOCK_LEN];
725 		memcpy(parent_block, &ctx->cv_stack[cvs_remaining * 32], 32);
726 		output_chaining_value(ctx->ops, &output, &parent_block[32]);
727 		output = parent_output(parent_block, ctx->key,
728 		    ctx->chunk.flags);
729 	}
730 	output_root_bytes(ctx->ops, &output, seek, out, out_len);
731 }
732