Halide学习笔记----Halide tutorial源码阅读8

Halide入门教程08

// Halide tutorial lesson 8: Scheduling multi-stage pipelines
// Halide入门教程第八课，多阶段流水线调度

// On linux, you can compile and run it like so:
// 在linux上按如下方法编译执行，在仅仅测试的情况下，建议将如下两行写入shell脚本文件，避免每次
// 输入一大段指令
// g++ lesson_08*.cpp -g -std=c++11 -I ../include -L ../bin -lHalide -lpthread -ldl -o lesson_08
// LD_LIBRARY_PATH=../bin ./lesson_08

#include "Halide.h"
#include <stdio.h>

using namespace Halide;

int main(int argc, char **argv) {
// First we'll declare some Vars to use below.
Var x("x"), y("y");

// Let's examine various scheduling options for a simple two stage
// pipeline. We'll start with the default schedule:
{
Func producer("producer_default"), consumer("consumer_default");

// The first stage will be some simple pointwise math similar
// to our familiar gradient function. The value at position x,
// y is the sin of product of x and y.
producer(x, y) = sin(x * y);

// Now we'll add a second stage which averages together multiple
// points in the first stage.
consumer(x, y) = (producer(x, y) +
producer(x, y+1) +
producer(x+1, y) +
producer(x+1, y+1))/4;

// We'll turn on tracing for both functions.
consumer.trace_stores();
producer.trace_stores();

// And evaluate it over a 4x4 box.
printf("\nEvaluating producer-consumer pipeline with default schedule\n");
consumer.realize(4, 4);

// There were no messages about computing values of the
// producer. This is because the default schedule fully
// inlines 'producer' into 'consumer'. It is as if we had
// written the following code instead:
// Halide默认的调度策略是inline策略，所谓的inline策略就是生产者-消费者模型中的消费者需要用到
// 生产者的哪一部分数据，就即时调用生产者计算消费者需要的那一部分数据。
// 即如下面的等价c代码所描述的那样
// for y
//      for x
//          produce what comsumer need
//          consumer compute

// consumer(x, y) = (sin(x * y) +
//                   sin(x * (y + 1)) +
//                   sin((x + 1) * y) +
//                   sin((x + 1) * (y + 1))/4);

// All calls to 'producer' have been replaced with the body of
// 'producer', with the arguments substituted in for the
// variables.

// The equivalent C code is:
float result[
190ef
4][4];
for (int y = 0; y < 4; y++) {
for (int x = 0; x < 4; x++) {
result[y][x] = (sin(x*y) +
sin(x*(y+1)) +
sin((x+1)*y) +
sin((x+1)*(y+1)))/4;
}
}
printf("\n");

// If we look at the loop nest, the producer doesn't appear
// at all. It has been inlined into the consumer.
// 仔细查看上述循环，producer根本没有出现，而是内联到comsumer内部了
printf("Pseudo-code for the schedule:\n");
consumer.print_loop_nest();
printf("\n");
}

// Next we'll examine the next simplest option - computing all
// values required in the producer before computing any of the
// consumer. We call this schedule "root".
// compute_root()调度
// 即在consumer开始消费producer数据之前，计算完所有的producer数据
{
// Start with the same function definitions:
// 本篇幅中所有的算法描述一致，重点的关注的是多阶段pipeline之间的调度关系
Func producer("producer_root"), consumer("consumer_root");
producer(x, y) = sin(x * y);
consumer(x, y) = (producer(x, y) +
producer(x, y+1) +
producer(x+1, y) +
producer(x+1, y+1))/4;

// Tell Halide to evaluate all of producer before any of consumer.
// 告诉Halide在consumer之前计算完所有producer
producer.compute_root();

// Turn on tracing.
consumer.trace_stores();
producer.trace_stores();

// Compile and run.
printf("\nEvaluating producer.compute_root()\n");
consumer.realize(4, 4);

// Reading the output we can see that:
// A) There were stores to producer.
// B) They all happened before any stores to consumer.
// 仔细观察输出跟踪结果可以看出：producer的store全部发生在consumer之前

// See figures/lesson_08_compute_root.gif for a visualization.
// The producer is on the left and the consumer is on the
// right. Stores are marked in orange and loads are marked in
// blue.

// Equivalent C:

float result[4][4];

// Allocate some temporary storage for the producer.
float producer_storage[5][5];

// Compute the producer.
for (int y = 0; y < 5; y++) {
for (int x = 0; x < 5; x++) {
producer_storage[y][x] = sin(x * y);
}
}

// Compute the consumer. Skip the prints this time.
for (int y = 0; y < 4; y++) {
for (int x = 0; x < 4; x++) {
result[y][x] = (producer_storage[y][x] +
producer_storage[y+1][x] +
producer_storage[y][x+1] +
producer_storage[y+1][x+1])/4;
}
}

// Note that consumer was evaluated over a 4x4 box, so Halide
// automatically inferred that producer was needed over a 5x5
// box. This is the same 'bounds inference' logic we saw in
// the previous lesson, where it was used to detect and avoid
// out-of-bounds reads from an input image.

// If we print the loop nest, we'll see something very
// similar to the C above.
printf("Pseudo-code for the schedule:\n");
consumer.print_loop_nest();
printf("\n");
}

// Let's compare the two approaches above from a performance
// perspective.
// 对比分析上述两种方法的性能

// Full inlining (the default schedule):
// - Temporary memory allocated: 0
// - Loads: 0
// - Stores: 16
// - Calls to sin: 64
// 默认的内联调度策略
// 临时内存分配：0
// 数据读取：0
// 数据存储：16
// 调用sin函数：64

// producer.compute_root():
// - Temporary memory allocated: 25 floats
// - Loads: 64
// - Stores: 41
// - Calls to sin: 25
// producer.compute_root()策略
// 临时内存分配：25个float
// 数据读取：64
// 数据存储：41
// 调用sin函数：25

// There's a trade-off here. Full inlining used minimal temporary
// memory and memory bandwidth, but did a whole bunch of redundant
// expensive math (calling sin). It evaluated most points in
// 'producer' four times. The second schedule,
// producer.compute_root(), did the mimimum number of calls to
// sin, but used more temporary memory and more memory bandwidth.
// 这里有一种折中处理。全内联方式用了很小的临时内存和内存带宽，但是在调用sin数学函数上有很大的冗余
// 在每个producer点处做了四次sin计算。第二者compute_root调度策略则是做了最少的sin函数运算，但是
// 用了跟多的临时存储和存储器带宽

// In any given situation the correct choice can be difficult to
// make. If you're memory-bandwidth limited, or don't have much
// memory (e.g. because you're running on an old cell-phone), then
// it can make sense to do redundant math. On the other hand, sin
// is expensive, so if you're compute-limited then fewer calls to
// sin will make your program faster. Adding vectorization or
// multi-core parallelism tilts the scales in favor of doing
// redundant work, because firing up multiple cpu cores increases
// the amount of math you can do per second, but doesn't increase
// your system memory bandwidth or capacity.
// 在任何给定的情形下，做出一个各方面都完美的调度是很难的。如果你收到了内存带宽的限制， 没有太多的
// 内存资源，那么你可以选择做更多的数学计算，用冗余的计算来换取空间上的不足。另一方面，sin函数的
// 计算很消耗计算资源，如果你的计算资源是不足，那么少做运算而用存储资源来换来计算速度。加入向量化和多核
// 的并行的计算资源倾向与做冗余的工作，因为一旦开启了多核的cpu，可以在相同时间内做更多的计算工作，而不
// 增加系统的内存带宽和容量。这里主要考虑所在平台的计算资源和存储资源之间的折中。

// We can make choices in between full inlining and
// compute_root. Next we'll alternate between computing the
// producer and consumer on a per-scanline basis:
// 这里考虑在每一行之间切换计算producer和consumer
{
// Start with the same function definitions:
Func producer("producer_y"), consumer("consumer_y");
producer(x, y) = sin(x * y);
consumer(x, y) = (producer(x, y) +
producer(x, y+1) +
producer(x+1, y) +
producer(x+1, y+1))/4;

// Tell Halide to evaluate producer as needed per y coordinate
// of the consumer:
// 告诉Halide在consumer每个行循环（y循环）计算producer
// 这里的a.compute_at(b, c)理解为在b的c循环里插入a的计算
producer.compute_at(consumer, y);

// This places the code that computes the producer just
// *inside* the consumer's for loop over y, as in the
// equivalent C below.
// 这里的compute_at将producer的计算放在consumer里y的for循环内部

// Turn on tracing.
producer.trace_stores();
consumer.trace_stores();

// Compile and run.
printf("\nEvaluating producer.compute_at(consumer, y)\n");
consumer.realize(4, 4);

// See figures/lesson_08_compute_y.gif for a visualization.

// Reading the log or looking at the figure you should see
// that producer and consumer alternate on a per-scanline
// basis. Let's look at the equivalent C:

float result[4][4];

// There's an outer loop over scanlines of consumer:
for (int y = 0; y < 4; y++) {

// Allocate space and compute enough of the producer to
// satisfy this single scanline of the consumer. This
// means a 5x2 box of the producer.
// 分配存储空间，在consumer的y循环内部计算producer
float producer_storage[2][5];
for (int py = y; py < y + 2; py++) {
for (int px = 0; px < 5; px++) {
producer_storage[py-y][px] = sin(px * py);
}
}

// Compute a scanline of the consumer.
// 使用（消费）producer对应的数据
for (int x = 0; x < 4; x++) {
result[y][x] = (producer_storage[0][x] +
producer_storage[1][x] +
producer_storage[0][x+1] +
producer_storage[1][x+1])/4;
}
}

// Again, if we print the loop nest, we'll see something very
// similar to the C above.
printf("Pseudo-code for the schedule:\n");
consumer.print_loop_nest();
printf("\n");

// The performance characteristics of this strategy are in
// between inlining and compute root. We still allocate some
// temporary memory, but less that compute_root, and with
// better locality (we load from it soon after writing to it,
// so for larger images, values should still be in cache). We
// still do some redundant work, but less than full inlining:
// 这种策略的性能位于inline和compute root之间。我们仍然需要分配临时内存，但是少于compute_root
// 同时有更好的局部性（数据生产出来立即消费掉，在缓存中就消费掉，避免在大图像时，数据太多写回
// 到内存中，需要使用的时候又需要从内存中调入到cpu内）。我们仍然需要做一部分冗余的计算，但是
// 比完全内联少。

// producer.compute_at(consumer, y):
// - Temporary memory allocated: 10 floats
// - Loads: 64
// - Stores: 56
// - Calls to sin: 40
// producer.compute_at(consumer, y)计算策略统计
// 临时内存分配：10个float
// 数据加载：64
// 数据存储：56
// 调用sin函数：40
}

// We could also say producer.compute_at(consumer, x), but this
// would be very similar to full inlining (the default
// schedule). Instead let's distinguish between the loop level at
// which we allocate storage for producer, and the loop level at
// which we actually compute it. This unlocks a few optimizations.
// producer.compute_at(consumer, x)和完全内联类似。
// 相反探索在哪一层循环为producer分配存储空间以及在哪一层循环开始计算producer可以为我们提供优化的
// 可能性。
{
Func producer("producer_root_y"), consumer("consumer_root_y");
producer(x, y) = sin(x * y);
consumer(x, y) = (producer(x, y) +
producer(x, y+1) +
producer(x+1, y) +
producer(x+1, y+1))/4;

// Tell Halide to make a buffer to store all of producer at
// the outermost level:
// 告诉Halide在最外层循环开辟buffer存储所有的producer数据
producer.store_root();
// ... but compute it as needed per y coordinate of the
// consumer.
// 在每一行y循环计算producer
producer.compute_at(consumer, y);

producer.trace_stores();
consumer.trace_stores();

printf("\nEvaluating producer.store_root().compute_at(consumer, y)\n");
consumer.realize(4, 4);

// See figures/lesson_08_store_root_compute_y.gif for a
// visualization.

// Reading the log or looking at the figure you should see
// that producer and consumer again alternate on a
// per-scanline basis. It computes a 5x2 box of the producer
// to satisfy the first scanline of the consumer, but after
// that it only computes a 5x1 box of the output for each new
// scanline of the consumer!
// Halide交替计算producer和consumer，在最开始的时候，计算5x2的producer，
// 随后计算5x1的consumer，下一次producer到consumer计算只需要5x1，可以利用上一次保存下来的数据

// Halide has detected that for all scanlines except for the
// first, it can reuse the values already sitting in the
// buffer we've allocated for producer. Let's look at the
// equivalent C:
// 除了第一行，随后的每一行Halide都会检测，并且重复利用之前已经计算并且保存在缓存中的数据。

float result[4][4];

// producer.store_root() implies that storage goes here:
float producer_storage[5][5];

// There's an outer loop over scanlines of consumer:
for (int y = 0; y < 4; y++) {

// Compute enough of the producer to satisfy this scanline
// of the consumer.
for (int py = y; py < y + 2; py++) {

// Skip over rows of producer that we've already
// computed in a previous iteration.
if (y > 0 && py == y) continue;

for (int px = 0; px < 5; px++) {
producer_storage[py][px] = sin(px * py);
}
}

// Compute a scanline of the consumer.
for (int x = 0; x < 4; x++) {
result[y][x] = (producer_storage[y][x] +
producer_storage[y+1][x] +
producer_storage[y][x+1] +
producer_storage[y+1][x+1])/4;
}
}

printf("Pseudo-code for the schedule:\n");
consumer.print_loop_nest();
printf("\n");

// The performance characteristics of this strategy are pretty
// good! The numbers are similar compute_root, except locality
// is better. We're doing the minimum number of sin calls,
// and we load values soon after they are stored, so we're
// probably making good use of the cache:
// 这种策略的计算性能很好，调用sin函数的次数和compute_root很compute_root一样，但是数据的局部性
// 更好。数据存在缓存上就立即拿出来使用了，充分利用了缓存。

// producer.store_root().compute_at(consumer, y):
// - Temporary memory allocated: 10 floats
// - Loads: 64
// - Stores: 39
// - Calls to sin: 25

// Note that my claimed amount of memory allocated doesn't
// match the reference C code. Halide is performing one more
// optimization under the hood. It folds the storage for the
// producer down into a circular buffer of two
// scanlines. Equivalent C would actually look like this:
// 上面列出的存储空间和上面c语言描述不一致。Halide实际上有更优化的方法。它只开辟出了两行来存储中间
// 变量，充分利用这两行存储空间，不断摒弃前一行不用的数据，接着存入新的数据

{
// Actually store 2 scanlines instead of 5
float producer_storage[2][5];
for (int y = 0; y < 4; y++) {
for (int py = y; py < y + 2; py++) {
if (y > 0 && py == y) continue;
for (int px = 0; px < 5; px++) {
// Stores to producer_storage have their y coordinate bit-masked.
producer_storage[py & 1][px] = sin(px * py);
}
}

// Compute a scanline of the consumer.
for (int x = 0; x < 4; x++) {
// Loads from producer_storage have their y coordinate bit-masked.
result[y][x] = (producer_storage[y & 1][x] +
producer_storage[(y+1) & 1][x] +
producer_storage[y & 1][x+1] +
producer_storage[(y+1) & 1][x+1])/4;
}
}
}
}

// We can do even better, by leaving the storage outermost, but
// moving the computation into the innermost loop:
// 我们可以通过将存储放在最外层，将计算放在最里层，来达到更好的优化
{
Func producer("producer_root_x"), consumer("consumer_root_x");
producer(x, y) = sin(x * y);
consumer(x, y) = (producer(x, y) +
producer(x, y+1) +
producer(x+1, y) +
producer(x+1, y+1))/4;

// Store outermost, compute innermost.
producer.store_root().compute_at(consumer, x);

producer.trace_stores();
consumer.trace_stores();

printf("\nEvaluating producer.store_root().compute_at(consumer, x)\n");
consumer.realize(4, 4);

// See figures/lesson_08_store_root_compute_x.gif for a
// visualization.

// You should see that producer and consumer now alternate on
// a per-pixel basis. Here's the equivalent C:

float result[4][4];

// producer.store_root() implies that storage goes here, but
// we can fold it down into a circular buffer of two
// scanlines:
float producer_storage[2][5];

// For every pixel of the consumer:
for (int y = 0; y < 4; y++) {
for (int x = 0; x < 4; x++) {

// Compute enough of the producer to satisfy this
// pixel of the consumer, but skip values that we've
// already computed:
if (y == 0 && x == 0)
producer_storage[y & 1][x] = sin(x*y);
if (y == 0)
producer_storage[y & 1][x+1] = sin((x+1)*y);
if (x == 0)
producer_storage[(y+1) & 1][x] = sin(x*(y+1));
producer_storage[(y+1) & 1][x+1] = sin((x+1)*(y+1));

result[y][x] = (producer_storage[y & 1][x] +
producer_storage[(y+1) & 1][x] +
producer_storage[y & 1][x+1] +
producer_storage[(y+1) & 1][x+1])/4;
}
}

printf("Pseudo-code for the schedule:\n");
consumer.print_loop_nest();
printf("\n");

// The performance characteristics of this strategy are the
// best so far. One of the four values of the producer we need
// is probably still sitting in a register, so I won't count
// it as a load:
// producer.store_root().compute_at(consumer, x):
// - Temporary memory allocated: 10 floats
// - Loads: 48
// - Stores: 56
// - Calls to sin: 40
}

// So what's the catch? Why not always do
// producer.store_root().compute_at(consumer, x) for this type of
// code?
// 这种方法在理论行分析是最好的策略，但是我们为什么不经常用这种方法呢？
// 原因在于当今的cpu多为多核结构，考虑到程序的并行化。在前面的两个调度策略中，我们假设当前循环计算
// 数据依赖它周围的数据。

// The answer is parallelism. In both of the previous two
// strategies we've assumed that values computed on previous
// iterations are lying around for us to reuse. This assumes that
// previous values of x or y happened earlier in time and have
// finished. This is not true if you parallelize or vectorize
// either loop. Darn. If you parallelize, Halide won't inject the
// optimizations that skip work already done if there's a parallel
// loop in between the store_at level and the compute_at level,
// and won't fold the storage down into a circular buffer either,
// which makes our store_root pointless.
// 在向量化和并行执行的时候，这样做就不对了。当你并行的时候，如果并行循环位于store_at和compute_at之间
// Halide将不会注入已经做过的优化，并且不会将存储折叠成循环buffer，这使得store_root变得没有意义了。

// We're running out of options. We can make new ones by
// splitting. We can store_at or compute_at at the natural
// variables of the consumer (x and y), or we can split x or y
// into new inner and outer sub-variables and then schedule with
// respect to those. We'll use this to express fusion in tiles:
// 我们可以start_at或者compute_at在消费者的自然变量x/y，或者将x/y差分成外部和内部子循环变量，然后
// 对子循环变量进行调度。这里的tile理解成拆分出来的小数据块
{
Func producer("producer_tile"), consumer("consumer_tile");
producer(x, y) = sin(x * y);
consumer(x, y) = (producer(x, y) +
producer(x, y+1) +
producer(x+1, y) +
producer(x+1, y+1))/4;

// We'll compute 8x8 of the consumer, in 4x4 tiles.
Var x_outer, y_outer, x_inner, y_inner;
consumer.tile(x, y, x_outer, y_outer, x_inner, y_inner, 4, 4);

// Compute the producer per tile of the consumer
// 在每个小的tile级别上进行producer计算
producer.compute_at(consumer, x_outer);

// Notice that I wrote my schedule starting from the end of
// the pipeline (the consumer). This is because the schedule
// for the producer refers to x_outer, which we introduced
// when we tiled the consumer. You can write it in the other
// order, but it tends to be harder to read.
// 通常，我们是从pipeline的尾部开始不断调度，反向推进。其他的方法亦可，不过阅读上带来不便

// Turn on tracing.
producer.trace_stores();
consumer.trace_stores();

printf("\nEvaluating:\n"
"consumer.tile(x, y, x_outer, y_outer, x_inner, y_inner, 4, 4);\n"
"producer.compute_at(consumer, x_outer);\n");
consumer.realize(8, 8);

// See figures/lesson_08_tile.gif for a visualization.

// The producer and consumer now alternate on a per-tile
// basis. Here's the equivalent C:
// 这样调度之后，生产者和消费者在每个tile级别上不断交替进行

float result[8][8];

// For every tile of the consumer:
for (int y_outer = 0; y_outer < 2; y_outer++) {
for (int x_outer = 0; x_outer < 2; x_outer++) {
// Compute the x and y coords of the start of this tile.
int x_base = x_outer*4;
int y_base = y_outer*4;

// Compute enough of producer to satisfy this tile. A
// 4x4 tile of the consumer requires a 5x5 tile of the
// producer.
float producer_storage[5][5];
for (int py = y_base; py < y_base + 5; py++) {
for (int px = x_base; px < x_base + 5; px++) {
producer_storage[py-y_base][px-x_base] = sin(px * py);
}
}

// Compute this tile of the consumer
for (int y_inner = 0; y_inner < 4; y_inner++) {
for (int x_inner = 0; x_inner < 4; x_inner++) {
int x = x_base + x_inner;
int y = y_base + y_inner;
result[y][x] =
(producer_storage[y - y_base][x - x_base] +
producer_storage[y - y_base + 1][x - x_base] +
producer_storage[y - y_base][x - x_base + 1] +
producer_storage[y - y_base + 1][x - x_base + 1])/4;
}
}
}
}

printf("Pseudo-code for the schedule:\n");
consumer.print_loop_nest();
printf("\n");

// Tiling can make sense for problems like this one with
// stencils that reach outwards in x and y. Each tile can be
// computed independently in parallel, and the redundant work
// done by each tile isn't so bad once the tiles get large
// enough.
// 每个tile可以独立并行地进行计算，当tile变得足够大的时候，冗余计算变现的可以忽略了。
}

// Let's try a mixed strategy that combines what we have done with
// splitting, parallelizing, and vectorizing. This is one that
// often works well in practice for large images. If you
// understand this schedule, then you understand 95% of scheduling
// in Halide.
// 综合拆分，并行，向量化，一起来完成调度。这在常用的图像处理算法中经常使用。
{
Func producer("producer_mixed"), consumer("consumer_mixed");
producer(x, y) = sin(x * y);
consumer(x, y) = (producer(x, y) +
producer(x, y+1) +
producer(x+1, y) +
producer(x+1, y+1))/4;

// Split the y coordinate of the consumer into strips of 16 scanlines:
// 拆分consumer的y坐标
Var yo, yi;
consumer.split(y, yo, yi, 16);
// Compute the strips using a thread pool and a task queue.
// 在yo层级上多线程并行
consumer.parallel(yo);
// Vectorize across x by a factor of four.
// 在x方向，采用simd向量化
consumer.vectorize(x, 4);

// Now store the producer per-strip. This will be 17 scanlines
// of the producer (16+1), but hopefully it will fold down
// into a circular buffer of two scanlines:
// 存下producer的每个步长（每个yo更行）迭代。这需要17行，但是采用循环buffer可以减少到两行
producer.store_at(consumer, yo);
// Within each strip, compute the producer per scanline of the
// consumer, skipping work done on previous scanlines.
// 在每次迭代中，consumer会充分利用之前缓存下的数据，避免不必要的计算
producer.compute_at(consumer, yi);
// Also vectorize the producer (because sin is vectorizable on x86 using SSE).
// 充分利用cpu的向量化指令
producer.vectorize(x, 4);

// Let's leave tracing off this time, because we're going to
// evaluate over a larger image.
// consumer.trace_stores();
// producer.trace_stores();

Buffer<float> halide_result = consumer.realize(160, 160);

// See figures/lesson_08_mixed.mp4 for a visualization.

// Here's the equivalent (serial) C:

float c_result[160][160];

// For every strip of 16 scanlines (this loop is parallel in
// the Halide version)
for (int yo = 0; yo < 160/16 + 1; yo++) {

// 16 doesn't divide 160, so push the last slice upwards
// to fit within [0, 159] (see lesson 05).
int y_base = yo * 16;
if (y_base > 160-16) y_base = 160-16;

// Allocate a two-scanline circular buffer for the producer
float producer_storage[2][161];

// For every scanline in the strip of 16:
for (int yi = 0; yi < 16; yi++) {
int y = y_base + yi;

for (int py = y; py < y+2; py++) {
// Skip scanlines already computed *within this task*
if (yi > 0 && py == y) continue;

// Compute this scanline of the producer in 4-wide vectors
for (int x_vec = 0; x_vec < 160/4 + 1; x_vec++) {
int x_base = x_vec*4;
// 4 doesn't divide 161, so push the last vector left
// (see lesson 05).
if (x_base > 161 - 4) x_base = 161 - 4;
// If you're on x86, Halide generates SSE code for this part:
int x[] = {x_base, x_base + 1, x_base + 2, x_base + 3};
float vec[4] = {sinf(x[0] * py), sinf(x[1] * py),
sinf(x[2] * py), sinf(x[3] * py)};
producer_storage[py & 1][x[0]] = vec[0];
producer_storage[py & 1][x[1]] = vec[1];
producer_storage[py & 1][x[2]] = vec[2];
producer_storage[py & 1][x[3]] = vec[3];
}
}

// Now compute consumer for this scanline:
for (int x_vec = 0; x_vec < 160/4; x_vec++) {
int x_base = x_vec * 4;
// Again, Halide's equivalent here uses SSE.
int x[] = {x_base, x_base + 1, x_base + 2, x_base + 3};
float vec[] = {
(producer_storage[y & 1][x[0]] +
producer_storage[(y+1) & 1][x[0]] +
producer_storage[y & 1][x[0]+1] +
producer_storage[(y+1) & 1][x[0]+1])/4,
(producer_storage[y & 1][x[1]] +
producer_storage[(y+1) & 1][x[1]] +
producer_storage[y & 1][x[1]+1] +
producer_storage[(y+1) & 1][x[1]+1])/4,
(producer_storage[y & 1][x[2]] +
producer_storage[(y+1) & 1][x[2]] +
producer_storage[y & 1][x[2]+1] +
producer_storage[(y+1) & 1][x[2]+1])/4,
(producer_storage[y & 1][x[3]] +
producer_storage[(y+1) & 1][x[3]] +
producer_storage[y & 1][x[3]+1] +
producer_storage[(y+1) & 1][x[3]+1])/4};

c_result[y][x[0]] = vec[0];
c_result[y][x[1]] = vec[1];
c_result[y][x[2]] = vec[2];
c_result[y][x[3]] = vec[3];
}

}
}
printf("Pseudo-code for the schedule:\n");
consumer.print_loop_nest();
printf("\n");

// Look on my code, ye mighty, and despair!

// Let's check the C result against the Halide result. Doing
// this I found several bugs in my C implementation, which
// should tell you something.
for (int y = 0; y < 160; y++) {
for (int x = 0; x < 160; x++) {
float error = halide_result(x, y) - c_result[y][x];
// It's floating-point math, so we'll allow some slop:
if (error < -0.001f || error > 0.001f) {
printf("halide_result(%d, %d) = %f instead of %f\n",
x, y, halide_result(x, y), c_result[y][x]);
return -1;
}
}
}

}

// This stuff is hard. We ended up in a three-way trade-off
// between memory bandwidth, redundant work, and
// parallelism. Halide can't make the correct choice for you
// automatically (sorry). Instead it tries to make it easier for
// you to explore various options, without messing up your
// program. In fact, Halide promises that scheduling calls like
// compute_root won't change the meaning of your algorithm -- you
// should get the same bits back no matter how you schedule
// things.
// 这些工作是非常困难的。Halide在内存带宽，冗余工作量，和并行之间做一个这种。Halide本身并不会自动
// 给出最有的策略。它会给你在做不同的尝试带来变量，不同的调度并不会将代码弄得一团糟。
// 实际上的各种调度策略的调整并不会影响代码的正确性。

// So be empirical! Experiment with various schedules and keep a
// log of performance. Form hypotheses and then try to prove
// yourself wrong. Don't assume that you just need to vectorize
// your code by a factor of four and run it on eight cores and
// you'll get 32x faster. This almost never works. Modern systems
// are complex enough that you can't predict performance reliably
// without running your code.
// 可行的方法是不断做实验，并且记录下性能日志。不要想着简单的向量化并行化就能达到性能上的飞跃。

// We suggest you start by scheduling all of your non-trivial
// stages compute_root, and then work from the end of the pipeline
// upwards, inlining, parallelizing, and vectorizing each stage in
// turn until you reach the top.

// Halide is not just about vectorizing and parallelizing your
// code. That's not enough to get you very far. Halide is about
// giving you tools that help you quickly explore different
// trade-offs between locality, redundant work, and parallelism,
// without messing up the actual result you're trying to compute.

printf("Success!\n");
return 0;
}

在终端中编译，并执行代码

$ g++ lesson_08*.cpp -g -std=c++11 -I ../include -L ../bin -lHalide -lpthread -ldl -o lesson_08
$ ./lesson_08

本课要点提炼：

默认的调度策略是内联模式，即消费者需要什么数据，生产者立即计算对应的数据，这样会有大量的计算冗余

producer.compute_root()，生产者在消费者之间将所有数据计算完毕，然后消费者才开始计算，这样做计算冗余性小，需要更多的存储空间



producer.compute_at(consumer, y), 在y循环层面缓存producer数据，是计算冗余和存储空间二者的一个折中



producer.store_root().compute_at(consumer, y),缓存所有数据，计算冗余性降低，同时生产之后立即被消费者消费，数据的locality好



与4向对应的producer.sore_root().compute_at(consumer, x)在x层级进行计算和消费并不是很好的选择，尤其是针对多核的计算平台，因此并不经常采用



consumer.tile(x, y, x_outer, y_outer, x_inner, y_inner, 4, 4);

producer.compute_at(consumer, x_outer);



将图像拆分成更小的数据块进行调度，x_outer, y_outer可以合并（fuse），然后在合并的层级上进行并行调度。

7.综合调度策略

consumer.split(y, yo, yi, 16);

consumer.parallel(yo);

consumer.vectorize(x, 4);

producer.store_at(consumer, yo);

producer.compute_at(consumer, yi);

producer.vectorize(x, 4);

消费者y循环拆分并行调度，x循环线量化，生产者在yo层级上缓存重复利用数据，在yi层级上生产数据，消费者x方向

上向量化执行。