xref: /freebsd/sys/contrib/openzfs/module/icp/algs/blake3/blake3.c (revision 3078531de10dcae44b253a35125c949ff4235284)
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/zfs_context.h>
29 #include <sys/blake3.h>
30 
31 #include "blake3_impl.h"
32 
33 /*
34  * We need 1056 byte stack for blake3_compress_subtree_wide()
35  * - we define this pragma to make gcc happy
36  */
37 #if defined(__GNUC__)
38 #pragma GCC diagnostic ignored "-Wframe-larger-than="
39 #endif
40 
41 /* internal used */
42 typedef struct {
43 	uint32_t input_cv[8];
44 	uint64_t counter;
45 	uint8_t block[BLAKE3_BLOCK_LEN];
46 	uint8_t block_len;
47 	uint8_t flags;
48 } output_t;
49 
50 /* internal flags */
51 enum blake3_flags {
52 	CHUNK_START		= 1 << 0,
53 	CHUNK_END		= 1 << 1,
54 	PARENT			= 1 << 2,
55 	ROOT			= 1 << 3,
56 	KEYED_HASH		= 1 << 4,
57 	DERIVE_KEY_CONTEXT	= 1 << 5,
58 	DERIVE_KEY_MATERIAL	= 1 << 6,
59 };
60 
61 /* internal start */
62 static void chunk_state_init(blake3_chunk_state_t *ctx,
63     const uint32_t key[8], uint8_t flags)
64 {
65 	memcpy(ctx->cv, key, BLAKE3_KEY_LEN);
66 	ctx->chunk_counter = 0;
67 	memset(ctx->buf, 0, BLAKE3_BLOCK_LEN);
68 	ctx->buf_len = 0;
69 	ctx->blocks_compressed = 0;
70 	ctx->flags = flags;
71 }
72 
73 static void chunk_state_reset(blake3_chunk_state_t *ctx,
74     const uint32_t key[8], uint64_t chunk_counter)
75 {
76 	memcpy(ctx->cv, key, BLAKE3_KEY_LEN);
77 	ctx->chunk_counter = chunk_counter;
78 	ctx->blocks_compressed = 0;
79 	memset(ctx->buf, 0, BLAKE3_BLOCK_LEN);
80 	ctx->buf_len = 0;
81 }
82 
83 static size_t chunk_state_len(const blake3_chunk_state_t *ctx)
84 {
85 	return (BLAKE3_BLOCK_LEN * (size_t)ctx->blocks_compressed) +
86 	    ((size_t)ctx->buf_len);
87 }
88 
89 static size_t chunk_state_fill_buf(blake3_chunk_state_t *ctx,
90     const uint8_t *input, size_t input_len)
91 {
92 	size_t take = BLAKE3_BLOCK_LEN - ((size_t)ctx->buf_len);
93 	if (take > input_len) {
94 		take = input_len;
95 	}
96 	uint8_t *dest = ctx->buf + ((size_t)ctx->buf_len);
97 	memcpy(dest, input, take);
98 	ctx->buf_len += (uint8_t)take;
99 	return (take);
100 }
101 
102 static uint8_t chunk_state_maybe_start_flag(const blake3_chunk_state_t *ctx)
103 {
104 	if (ctx->blocks_compressed == 0) {
105 		return (CHUNK_START);
106 	} else {
107 		return (0);
108 	}
109 }
110 
111 static output_t make_output(const uint32_t input_cv[8],
112     const uint8_t *block, uint8_t block_len,
113     uint64_t counter, uint8_t flags)
114 {
115 	output_t ret;
116 	memcpy(ret.input_cv, input_cv, 32);
117 	memcpy(ret.block, block, BLAKE3_BLOCK_LEN);
118 	ret.block_len = block_len;
119 	ret.counter = counter;
120 	ret.flags = flags;
121 	return (ret);
122 }
123 
124 /*
125  * Chaining values within a given chunk (specifically the compress_in_place
126  * interface) are represented as words. This avoids unnecessary bytes<->words
127  * conversion overhead in the portable implementation. However, the hash_many
128  * interface handles both user input and parent node blocks, so it accepts
129  * bytes. For that reason, chaining values in the CV stack are represented as
130  * bytes.
131  */
132 static void output_chaining_value(const blake3_impl_ops_t *ops,
133     const output_t *ctx, uint8_t cv[32])
134 {
135 	uint32_t cv_words[8];
136 	memcpy(cv_words, ctx->input_cv, 32);
137 	ops->compress_in_place(cv_words, ctx->block, ctx->block_len,
138 	    ctx->counter, ctx->flags);
139 	store_cv_words(cv, cv_words);
140 }
141 
142 static void output_root_bytes(const blake3_impl_ops_t *ops, const output_t *ctx,
143     uint64_t seek, uint8_t *out, size_t out_len)
144 {
145 	uint64_t output_block_counter = seek / 64;
146 	size_t offset_within_block = seek % 64;
147 	uint8_t wide_buf[64];
148 	while (out_len > 0) {
149 		ops->compress_xof(ctx->input_cv, ctx->block, ctx->block_len,
150 		    output_block_counter, ctx->flags | ROOT, wide_buf);
151 		size_t available_bytes = 64 - offset_within_block;
152 		size_t memcpy_len;
153 		if (out_len > available_bytes) {
154 			memcpy_len = available_bytes;
155 		} else {
156 			memcpy_len = out_len;
157 		}
158 		memcpy(out, wide_buf + offset_within_block, memcpy_len);
159 		out += memcpy_len;
160 		out_len -= memcpy_len;
161 		output_block_counter += 1;
162 		offset_within_block = 0;
163 	}
164 }
165 
166 static void chunk_state_update(const blake3_impl_ops_t *ops,
167     blake3_chunk_state_t *ctx, const uint8_t *input, size_t input_len)
168 {
169 	if (ctx->buf_len > 0) {
170 		size_t take = chunk_state_fill_buf(ctx, input, input_len);
171 		input += take;
172 		input_len -= take;
173 		if (input_len > 0) {
174 			ops->compress_in_place(ctx->cv, ctx->buf,
175 			    BLAKE3_BLOCK_LEN, ctx->chunk_counter,
176 			    ctx->flags|chunk_state_maybe_start_flag(ctx));
177 			ctx->blocks_compressed += 1;
178 			ctx->buf_len = 0;
179 			memset(ctx->buf, 0, BLAKE3_BLOCK_LEN);
180 		}
181 	}
182 
183 	while (input_len > BLAKE3_BLOCK_LEN) {
184 		ops->compress_in_place(ctx->cv, input, BLAKE3_BLOCK_LEN,
185 		    ctx->chunk_counter,
186 		    ctx->flags|chunk_state_maybe_start_flag(ctx));
187 		ctx->blocks_compressed += 1;
188 		input += BLAKE3_BLOCK_LEN;
189 		input_len -= BLAKE3_BLOCK_LEN;
190 	}
191 
192 	size_t take = chunk_state_fill_buf(ctx, input, input_len);
193 	input += take;
194 	input_len -= take;
195 }
196 
197 static output_t chunk_state_output(const blake3_chunk_state_t *ctx)
198 {
199 	uint8_t block_flags =
200 	    ctx->flags | chunk_state_maybe_start_flag(ctx) | CHUNK_END;
201 	return (make_output(ctx->cv, ctx->buf, ctx->buf_len, ctx->chunk_counter,
202 	    block_flags));
203 }
204 
205 static output_t parent_output(const uint8_t block[BLAKE3_BLOCK_LEN],
206     const uint32_t key[8], uint8_t flags)
207 {
208 	return (make_output(key, block, BLAKE3_BLOCK_LEN, 0, flags | PARENT));
209 }
210 
211 /*
212  * Given some input larger than one chunk, return the number of bytes that
213  * should go in the left subtree. This is the largest power-of-2 number of
214  * chunks that leaves at least 1 byte for the right subtree.
215  */
216 static size_t left_len(size_t content_len)
217 {
218 	/*
219 	 * Subtract 1 to reserve at least one byte for the right side.
220 	 * content_len
221 	 * should always be greater than BLAKE3_CHUNK_LEN.
222 	 */
223 	size_t full_chunks = (content_len - 1) / BLAKE3_CHUNK_LEN;
224 	return (round_down_to_power_of_2(full_chunks) * BLAKE3_CHUNK_LEN);
225 }
226 
227 /*
228  * Use SIMD parallelism to hash up to MAX_SIMD_DEGREE chunks at the same time
229  * on a single thread. Write out the chunk chaining values and return the
230  * number of chunks hashed. These chunks are never the root and never empty;
231  * those cases use a different codepath.
232  */
233 static size_t compress_chunks_parallel(const blake3_impl_ops_t *ops,
234     const uint8_t *input, size_t input_len, const uint32_t key[8],
235     uint64_t chunk_counter, uint8_t flags, uint8_t *out)
236 {
237 	const uint8_t *chunks_array[MAX_SIMD_DEGREE];
238 	size_t input_position = 0;
239 	size_t chunks_array_len = 0;
240 	while (input_len - input_position >= BLAKE3_CHUNK_LEN) {
241 		chunks_array[chunks_array_len] = &input[input_position];
242 		input_position += BLAKE3_CHUNK_LEN;
243 		chunks_array_len += 1;
244 	}
245 
246 	ops->hash_many(chunks_array, chunks_array_len, BLAKE3_CHUNK_LEN /
247 	    BLAKE3_BLOCK_LEN, key, chunk_counter, B_TRUE, flags, CHUNK_START,
248 	    CHUNK_END, out);
249 
250 	/*
251 	 * Hash the remaining partial chunk, if there is one. Note that the
252 	 * empty chunk (meaning the empty message) is a different codepath.
253 	 */
254 	if (input_len > input_position) {
255 		uint64_t counter = chunk_counter + (uint64_t)chunks_array_len;
256 		blake3_chunk_state_t chunk_state;
257 		chunk_state_init(&chunk_state, key, flags);
258 		chunk_state.chunk_counter = counter;
259 		chunk_state_update(ops, &chunk_state, &input[input_position],
260 		    input_len - input_position);
261 		output_t output = chunk_state_output(&chunk_state);
262 		output_chaining_value(ops, &output, &out[chunks_array_len *
263 		    BLAKE3_OUT_LEN]);
264 		return (chunks_array_len + 1);
265 	} else {
266 		return (chunks_array_len);
267 	}
268 }
269 
270 /*
271  * Use SIMD parallelism to hash up to MAX_SIMD_DEGREE parents at the same time
272  * on a single thread. Write out the parent chaining values and return the
273  * number of parents hashed. (If there's an odd input chaining value left over,
274  * return it as an additional output.) These parents are never the root and
275  * never empty; those cases use a different codepath.
276  */
277 static size_t compress_parents_parallel(const blake3_impl_ops_t *ops,
278     const uint8_t *child_chaining_values, size_t num_chaining_values,
279     const uint32_t key[8], uint8_t flags, uint8_t *out)
280 {
281 	const uint8_t *parents_array[MAX_SIMD_DEGREE_OR_2];
282 	size_t parents_array_len = 0;
283 
284 	while (num_chaining_values - (2 * parents_array_len) >= 2) {
285 		parents_array[parents_array_len] = &child_chaining_values[2 *
286 		    parents_array_len * BLAKE3_OUT_LEN];
287 		parents_array_len += 1;
288 	}
289 
290 	ops->hash_many(parents_array, parents_array_len, 1, key, 0, B_FALSE,
291 	    flags | PARENT, 0, 0, out);
292 
293 	/* If there's an odd child left over, it becomes an output. */
294 	if (num_chaining_values > 2 * parents_array_len) {
295 		memcpy(&out[parents_array_len * BLAKE3_OUT_LEN],
296 		    &child_chaining_values[2 * parents_array_len *
297 		    BLAKE3_OUT_LEN], BLAKE3_OUT_LEN);
298 		return (parents_array_len + 1);
299 	} else {
300 		return (parents_array_len);
301 	}
302 }
303 
304 /*
305  * The wide helper function returns (writes out) an array of chaining values
306  * and returns the length of that array. The number of chaining values returned
307  * is the dyanmically detected SIMD degree, at most MAX_SIMD_DEGREE. Or fewer,
308  * if the input is shorter than that many chunks. The reason for maintaining a
309  * wide array of chaining values going back up the tree, is to allow the
310  * implementation to hash as many parents in parallel as possible.
311  *
312  * As a special case when the SIMD degree is 1, this function will still return
313  * at least 2 outputs. This guarantees that this function doesn't perform the
314  * root compression. (If it did, it would use the wrong flags, and also we
315  * wouldn't be able to implement exendable ouput.) Note that this function is
316  * not used when the whole input is only 1 chunk long; that's a different
317  * codepath.
318  *
319  * Why not just have the caller split the input on the first update(), instead
320  * of implementing this special rule? Because we don't want to limit SIMD or
321  * multi-threading parallelism for that update().
322  */
323 static size_t blake3_compress_subtree_wide(const blake3_impl_ops_t *ops,
324     const uint8_t *input, size_t input_len, const uint32_t key[8],
325     uint64_t chunk_counter, uint8_t flags, uint8_t *out)
326 {
327 	/*
328 	 * Note that the single chunk case does *not* bump the SIMD degree up
329 	 * to 2 when it is 1. If this implementation adds multi-threading in
330 	 * the future, this gives us the option of multi-threading even the
331 	 * 2-chunk case, which can help performance on smaller platforms.
332 	 */
333 	if (input_len <= (size_t)(ops->degree * BLAKE3_CHUNK_LEN)) {
334 		return (compress_chunks_parallel(ops, input, input_len, key,
335 		    chunk_counter, flags, out));
336 	}
337 
338 
339 	/*
340 	 * With more than simd_degree chunks, we need to recurse. Start by
341 	 * dividing the input into left and right subtrees. (Note that this is
342 	 * only optimal as long as the SIMD degree is a power of 2. If we ever
343 	 * get a SIMD degree of 3 or something, we'll need a more complicated
344 	 * strategy.)
345 	 */
346 	size_t left_input_len = left_len(input_len);
347 	size_t right_input_len = input_len - left_input_len;
348 	const uint8_t *right_input = &input[left_input_len];
349 	uint64_t right_chunk_counter = chunk_counter +
350 	    (uint64_t)(left_input_len / BLAKE3_CHUNK_LEN);
351 
352 	/*
353 	 * Make space for the child outputs. Here we use MAX_SIMD_DEGREE_OR_2
354 	 * to account for the special case of returning 2 outputs when the
355 	 * SIMD degree is 1.
356 	 */
357 	uint8_t cv_array[2 * MAX_SIMD_DEGREE_OR_2 * BLAKE3_OUT_LEN];
358 	size_t degree = ops->degree;
359 	if (left_input_len > BLAKE3_CHUNK_LEN && degree == 1) {
360 
361 		/*
362 		 * The special case: We always use a degree of at least two,
363 		 * to make sure there are two outputs. Except, as noted above,
364 		 * at the chunk level, where we allow degree=1. (Note that the
365 		 * 1-chunk-input case is a different codepath.)
366 		 */
367 		degree = 2;
368 	}
369 	uint8_t *right_cvs = &cv_array[degree * BLAKE3_OUT_LEN];
370 
371 	/*
372 	 * Recurse! If this implementation adds multi-threading support in the
373 	 * future, this is where it will go.
374 	 */
375 	size_t left_n = blake3_compress_subtree_wide(ops, input, left_input_len,
376 	    key, chunk_counter, flags, cv_array);
377 	size_t right_n = blake3_compress_subtree_wide(ops, right_input,
378 	    right_input_len, key, right_chunk_counter, flags, right_cvs);
379 
380 	/*
381 	 * The special case again. If simd_degree=1, then we'll have left_n=1
382 	 * and right_n=1. Rather than compressing them into a single output,
383 	 * return them directly, to make sure we always have at least two
384 	 * outputs.
385 	 */
386 	if (left_n == 1) {
387 		memcpy(out, cv_array, 2 * BLAKE3_OUT_LEN);
388 		return (2);
389 	}
390 
391 	/* Otherwise, do one layer of parent node compression. */
392 	size_t num_chaining_values = left_n + right_n;
393 	return compress_parents_parallel(ops, cv_array,
394 	    num_chaining_values, key, flags, out);
395 }
396 
397 /*
398  * Hash a subtree with compress_subtree_wide(), and then condense the resulting
399  * list of chaining values down to a single parent node. Don't compress that
400  * last parent node, however. Instead, return its message bytes (the
401  * concatenated chaining values of its children). This is necessary when the
402  * first call to update() supplies a complete subtree, because the topmost
403  * parent node of that subtree could end up being the root. It's also necessary
404  * for extended output in the general case.
405  *
406  * As with compress_subtree_wide(), this function is not used on inputs of 1
407  * chunk or less. That's a different codepath.
408  */
409 static void compress_subtree_to_parent_node(const blake3_impl_ops_t *ops,
410     const uint8_t *input, size_t input_len, const uint32_t key[8],
411     uint64_t chunk_counter, uint8_t flags, uint8_t out[2 * BLAKE3_OUT_LEN])
412 {
413 	uint8_t cv_array[MAX_SIMD_DEGREE_OR_2 * BLAKE3_OUT_LEN];
414 	size_t num_cvs = blake3_compress_subtree_wide(ops, input, input_len,
415 	    key, chunk_counter, flags, cv_array);
416 
417 	/*
418 	 * If MAX_SIMD_DEGREE is greater than 2 and there's enough input,
419 	 * compress_subtree_wide() returns more than 2 chaining values. Condense
420 	 * them into 2 by forming parent nodes repeatedly.
421 	 */
422 	uint8_t out_array[MAX_SIMD_DEGREE_OR_2 * BLAKE3_OUT_LEN / 2];
423 	while (num_cvs > 2) {
424 		num_cvs = compress_parents_parallel(ops, cv_array, num_cvs, key,
425 		    flags, out_array);
426 		memcpy(cv_array, out_array, num_cvs * BLAKE3_OUT_LEN);
427 	}
428 	memcpy(out, cv_array, 2 * BLAKE3_OUT_LEN);
429 }
430 
431 static void hasher_init_base(BLAKE3_CTX *ctx, const uint32_t key[8],
432     uint8_t flags)
433 {
434 	memcpy(ctx->key, key, BLAKE3_KEY_LEN);
435 	chunk_state_init(&ctx->chunk, key, flags);
436 	ctx->cv_stack_len = 0;
437 	ctx->ops = blake3_impl_get_ops();
438 }
439 
440 /*
441  * As described in hasher_push_cv() below, we do "lazy merging", delaying
442  * merges until right before the next CV is about to be added. This is
443  * different from the reference implementation. Another difference is that we
444  * aren't always merging 1 chunk at a time. Instead, each CV might represent
445  * any power-of-two number of chunks, as long as the smaller-above-larger
446  * stack order is maintained. Instead of the "count the trailing 0-bits"
447  * algorithm described in the spec, we use a "count the total number of
448  * 1-bits" variant that doesn't require us to retain the subtree size of the
449  * CV on top of the stack. The principle is the same: each CV that should
450  * remain in the stack is represented by a 1-bit in the total number of chunks
451  * (or bytes) so far.
452  */
453 static void hasher_merge_cv_stack(BLAKE3_CTX *ctx, uint64_t total_len)
454 {
455 	size_t post_merge_stack_len = (size_t)popcnt(total_len);
456 	while (ctx->cv_stack_len > post_merge_stack_len) {
457 		uint8_t *parent_node =
458 		    &ctx->cv_stack[(ctx->cv_stack_len - 2) * BLAKE3_OUT_LEN];
459 		output_t output =
460 		    parent_output(parent_node, ctx->key, ctx->chunk.flags);
461 		output_chaining_value(ctx->ops, &output, parent_node);
462 		ctx->cv_stack_len -= 1;
463 	}
464 }
465 
466 /*
467  * In reference_impl.rs, we merge the new CV with existing CVs from the stack
468  * before pushing it. We can do that because we know more input is coming, so
469  * we know none of the merges are root.
470  *
471  * This setting is different. We want to feed as much input as possible to
472  * compress_subtree_wide(), without setting aside anything for the chunk_state.
473  * If the user gives us 64 KiB, we want to parallelize over all 64 KiB at once
474  * as a single subtree, if at all possible.
475  *
476  * This leads to two problems:
477  * 1) This 64 KiB input might be the only call that ever gets made to update.
478  *    In this case, the root node of the 64 KiB subtree would be the root node
479  *    of the whole tree, and it would need to be ROOT finalized. We can't
480  *    compress it until we know.
481  * 2) This 64 KiB input might complete a larger tree, whose root node is
482  *    similarly going to be the the root of the whole tree. For example, maybe
483  *    we have 196 KiB (that is, 128 + 64) hashed so far. We can't compress the
484  *    node at the root of the 256 KiB subtree until we know how to finalize it.
485  *
486  * The second problem is solved with "lazy merging". That is, when we're about
487  * to add a CV to the stack, we don't merge it with anything first, as the
488  * reference impl does. Instead we do merges using the *previous* CV that was
489  * added, which is sitting on top of the stack, and we put the new CV
490  * (unmerged) on top of the stack afterwards. This guarantees that we never
491  * merge the root node until finalize().
492  *
493  * Solving the first problem requires an additional tool,
494  * compress_subtree_to_parent_node(). That function always returns the top
495  * *two* chaining values of the subtree it's compressing. We then do lazy
496  * merging with each of them separately, so that the second CV will always
497  * remain unmerged. (That also helps us support extendable output when we're
498  * hashing an input all-at-once.)
499  */
500 static void hasher_push_cv(BLAKE3_CTX *ctx, uint8_t new_cv[BLAKE3_OUT_LEN],
501     uint64_t chunk_counter)
502 {
503 	hasher_merge_cv_stack(ctx, chunk_counter);
504 	memcpy(&ctx->cv_stack[ctx->cv_stack_len * BLAKE3_OUT_LEN], new_cv,
505 	    BLAKE3_OUT_LEN);
506 	ctx->cv_stack_len += 1;
507 }
508 
509 void
510 Blake3_Init(BLAKE3_CTX *ctx)
511 {
512 	hasher_init_base(ctx, BLAKE3_IV, 0);
513 }
514 
515 void
516 Blake3_InitKeyed(BLAKE3_CTX *ctx, const uint8_t key[BLAKE3_KEY_LEN])
517 {
518 	uint32_t key_words[8];
519 	load_key_words(key, key_words);
520 	hasher_init_base(ctx, key_words, KEYED_HASH);
521 }
522 
523 static void
524 Blake3_Update2(BLAKE3_CTX *ctx, const void *input, size_t input_len)
525 {
526 	/*
527 	 * Explicitly checking for zero avoids causing UB by passing a null
528 	 * pointer to memcpy. This comes up in practice with things like:
529 	 *   std::vector<uint8_t> v;
530 	 *   blake3_hasher_update(&hasher, v.data(), v.size());
531 	 */
532 	if (input_len == 0) {
533 		return;
534 	}
535 
536 	const uint8_t *input_bytes = (const uint8_t *)input;
537 
538 	/*
539 	 * If we have some partial chunk bytes in the internal chunk_state, we
540 	 * need to finish that chunk first.
541 	 */
542 	if (chunk_state_len(&ctx->chunk) > 0) {
543 		size_t take = BLAKE3_CHUNK_LEN - chunk_state_len(&ctx->chunk);
544 		if (take > input_len) {
545 			take = input_len;
546 		}
547 		chunk_state_update(ctx->ops, &ctx->chunk, input_bytes, take);
548 		input_bytes += take;
549 		input_len -= take;
550 		/*
551 		 * If we've filled the current chunk and there's more coming,
552 		 * finalize this chunk and proceed. In this case we know it's
553 		 * not the root.
554 		 */
555 		if (input_len > 0) {
556 			output_t output = chunk_state_output(&ctx->chunk);
557 			uint8_t chunk_cv[32];
558 			output_chaining_value(ctx->ops, &output, chunk_cv);
559 			hasher_push_cv(ctx, chunk_cv, ctx->chunk.chunk_counter);
560 			chunk_state_reset(&ctx->chunk, ctx->key,
561 			    ctx->chunk.chunk_counter + 1);
562 		} else {
563 			return;
564 		}
565 	}
566 
567 	/*
568 	 * Now the chunk_state is clear, and we have more input. If there's
569 	 * more than a single chunk (so, definitely not the root chunk), hash
570 	 * the largest whole subtree we can, with the full benefits of SIMD
571 	 * (and maybe in the future, multi-threading) parallelism. Two
572 	 * restrictions:
573 	 * - The subtree has to be a power-of-2 number of chunks. Only
574 	 *   subtrees along the right edge can be incomplete, and we don't know
575 	 *   where the right edge is going to be until we get to finalize().
576 	 * - The subtree must evenly divide the total number of chunks up
577 	 *   until this point (if total is not 0). If the current incomplete
578 	 *   subtree is only waiting for 1 more chunk, we can't hash a subtree
579 	 *   of 4 chunks. We have to complete the current subtree first.
580 	 * Because we might need to break up the input to form powers of 2, or
581 	 * to evenly divide what we already have, this part runs in a loop.
582 	 */
583 	while (input_len > BLAKE3_CHUNK_LEN) {
584 		size_t subtree_len = round_down_to_power_of_2(input_len);
585 		uint64_t count_so_far =
586 		    ctx->chunk.chunk_counter * BLAKE3_CHUNK_LEN;
587 		/*
588 		 * Shrink the subtree_len until it evenly divides the count so
589 		 * far. We know that subtree_len itself is a power of 2, so we
590 		 * can use a bitmasking trick instead of an actual remainder
591 		 * operation. (Note that if the caller consistently passes
592 		 * power-of-2 inputs of the same size, as is hopefully
593 		 * typical, this loop condition will always fail, and
594 		 * subtree_len will always be the full length of the input.)
595 		 *
596 		 * An aside: We don't have to shrink subtree_len quite this
597 		 * much. For example, if count_so_far is 1, we could pass 2
598 		 * chunks to compress_subtree_to_parent_node. Since we'll get
599 		 * 2 CVs back, we'll still get the right answer in the end,
600 		 * and we might get to use 2-way SIMD parallelism. The problem
601 		 * with this optimization, is that it gets us stuck always
602 		 * hashing 2 chunks. The total number of chunks will remain
603 		 * odd, and we'll never graduate to higher degrees of
604 		 * parallelism. See
605 		 * https://github.com/BLAKE3-team/BLAKE3/issues/69.
606 		 */
607 		while ((((uint64_t)(subtree_len - 1)) & count_so_far) != 0) {
608 			subtree_len /= 2;
609 		}
610 		/*
611 		 * The shrunken subtree_len might now be 1 chunk long. If so,
612 		 * hash that one chunk by itself. Otherwise, compress the
613 		 * subtree into a pair of CVs.
614 		 */
615 		uint64_t subtree_chunks = subtree_len / BLAKE3_CHUNK_LEN;
616 		if (subtree_len <= BLAKE3_CHUNK_LEN) {
617 			blake3_chunk_state_t chunk_state;
618 			chunk_state_init(&chunk_state, ctx->key,
619 			    ctx->chunk.flags);
620 			chunk_state.chunk_counter = ctx->chunk.chunk_counter;
621 			chunk_state_update(ctx->ops, &chunk_state, input_bytes,
622 			    subtree_len);
623 			output_t output = chunk_state_output(&chunk_state);
624 			uint8_t cv[BLAKE3_OUT_LEN];
625 			output_chaining_value(ctx->ops, &output, cv);
626 			hasher_push_cv(ctx, cv, chunk_state.chunk_counter);
627 		} else {
628 			/*
629 			 * This is the high-performance happy path, though
630 			 * getting here depends on the caller giving us a long
631 			 * enough input.
632 			 */
633 			uint8_t cv_pair[2 * BLAKE3_OUT_LEN];
634 			compress_subtree_to_parent_node(ctx->ops, input_bytes,
635 			    subtree_len, ctx->key, ctx-> chunk.chunk_counter,
636 			    ctx->chunk.flags, cv_pair);
637 			hasher_push_cv(ctx, cv_pair, ctx->chunk.chunk_counter);
638 			hasher_push_cv(ctx, &cv_pair[BLAKE3_OUT_LEN],
639 			    ctx->chunk.chunk_counter + (subtree_chunks / 2));
640 		}
641 		ctx->chunk.chunk_counter += subtree_chunks;
642 		input_bytes += subtree_len;
643 		input_len -= subtree_len;
644 	}
645 
646 	/*
647 	 * If there's any remaining input less than a full chunk, add it to
648 	 * the chunk state. In that case, also do a final merge loop to make
649 	 * sure the subtree stack doesn't contain any unmerged pairs. The
650 	 * remaining input means we know these merges are non-root. This merge
651 	 * loop isn't strictly necessary here, because hasher_push_chunk_cv
652 	 * already does its own merge loop, but it simplifies
653 	 * blake3_hasher_finalize below.
654 	 */
655 	if (input_len > 0) {
656 		chunk_state_update(ctx->ops, &ctx->chunk, input_bytes,
657 		    input_len);
658 		hasher_merge_cv_stack(ctx, ctx->chunk.chunk_counter);
659 	}
660 }
661 
662 void
663 Blake3_Update(BLAKE3_CTX *ctx, const void *input, size_t todo)
664 {
665 	size_t done = 0;
666 	const uint8_t *data = input;
667 	const size_t block_max = 1024 * 64;
668 
669 	/* max feed buffer to leave the stack size small */
670 	while (todo != 0) {
671 		size_t block = (todo >= block_max) ? block_max : todo;
672 		Blake3_Update2(ctx, data + done, block);
673 		done += block;
674 		todo -= block;
675 	}
676 }
677 
678 void
679 Blake3_Final(const BLAKE3_CTX *ctx, uint8_t *out)
680 {
681 	Blake3_FinalSeek(ctx, 0, out, BLAKE3_OUT_LEN);
682 }
683 
684 void
685 Blake3_FinalSeek(const BLAKE3_CTX *ctx, uint64_t seek, uint8_t *out,
686     size_t out_len)
687 {
688 	/*
689 	 * Explicitly checking for zero avoids causing UB by passing a null
690 	 * pointer to memcpy. This comes up in practice with things like:
691 	 *   std::vector<uint8_t> v;
692 	 *   blake3_hasher_finalize(&hasher, v.data(), v.size());
693 	 */
694 	if (out_len == 0) {
695 		return;
696 	}
697 	/* If the subtree stack is empty, then the current chunk is the root. */
698 	if (ctx->cv_stack_len == 0) {
699 		output_t output = chunk_state_output(&ctx->chunk);
700 		output_root_bytes(ctx->ops, &output, seek, out, out_len);
701 		return;
702 	}
703 	/*
704 	 * If there are any bytes in the chunk state, finalize that chunk and
705 	 * do a roll-up merge between that chunk hash and every subtree in the
706 	 * stack. In this case, the extra merge loop at the end of
707 	 * blake3_hasher_update guarantees that none of the subtrees in the
708 	 * stack need to be merged with each other first. Otherwise, if there
709 	 * are no bytes in the chunk state, then the top of the stack is a
710 	 * chunk hash, and we start the merge from that.
711 	 */
712 	output_t output;
713 	size_t cvs_remaining;
714 	if (chunk_state_len(&ctx->chunk) > 0) {
715 		cvs_remaining = ctx->cv_stack_len;
716 		output = chunk_state_output(&ctx->chunk);
717 	} else {
718 		/* There are always at least 2 CVs in the stack in this case. */
719 		cvs_remaining = ctx->cv_stack_len - 2;
720 		output = parent_output(&ctx->cv_stack[cvs_remaining * 32],
721 		    ctx->key, ctx->chunk.flags);
722 	}
723 	while (cvs_remaining > 0) {
724 		cvs_remaining -= 1;
725 		uint8_t parent_block[BLAKE3_BLOCK_LEN];
726 		memcpy(parent_block, &ctx->cv_stack[cvs_remaining * 32], 32);
727 		output_chaining_value(ctx->ops, &output, &parent_block[32]);
728 		output = parent_output(parent_block, ctx->key,
729 		    ctx->chunk.flags);
730 	}
731 	output_root_bytes(ctx->ops, &output, seek, out, out_len);
732 }
733