Halide
tutorial/lesson_09_update_definitions.cpp
// Halide tutorial lesson 9: Multi-pass Funcs, update definitions, and reductions
// On linux, you can compile and run it like so:
// g++ lesson_09*.cpp -g -std=c++11 -I ../include -I ../tools -L ../bin -lHalide `libpng-config --cflags --ldflags` -ljpeg -lpthread -ldl -fopenmp -o lesson_09
// LD_LIBRARY_PATH=../bin ./lesson_09
// On os x (will only work if you actually have g++, not Apple's pretend g++ which is actually clang):
// g++ lesson_09*.cpp -g -std=c++11 -I ../include -I ../tools -L ../bin -lHalide `libpng-config --cflags --ldflags` -ljpeg -fopenmp -o lesson_09
// DYLD_LIBRARY_PATH=../bin ./lesson_09
// If you have the entire Halide source tree, you can also build it by
// running:
// make tutorial_lesson_09_update_definitions
// in a shell with the current directory at the top of the halide
// source tree.
#include "Halide.h"
#include <stdio.h>
// We're going to be using x86 SSE intrinsics later on in this lesson.
#ifdef __SSE2__
#include <emmintrin.h>
#endif
// We'll also need a clock to do performance testing at the end.
#include "clock.h"
using namespace Halide;
// Support code for loading pngs.
#include "halide_image_io.h"
using namespace Halide::Tools;
int main(int argc, char **argv) {
// Declare some Vars to use below.
Var x("x"), y("y");
// Load a grayscale image to use as an input.
Buffer<uint8_t> input = load_image("images/gray.png");
// You can define a Func in multiple passes. Let's see a toy
// example first.
{
// The first definition must be one like we have seen already
// - a mapping from Vars to an Expr:
Func f;
f(x, y) = x + y;
// We call this first definition the "pure" definition.
// But the later definitions can include computed expressions on
// both sides. The simplest example is modifying a single point:
f(3, 7) = 42;
// We call these extra definitions "update" definitions, or
// "reduction" definitions. A reduction definition is an
// update definition that recursively refers back to the
// function's current value at the same site:
f(x, y) = f(x, y) + 17;
// If we confine our update to a single row, we can
// recursively refer to values in the same column:
f(x, 3) = f(x, 0) * f(x, 10);
// Similarly, if we confine our update to a single column, we
// can recursively refer to other values in the same row.
f(0, y) = f(0, y) / f(3, y);
// The general rule is: Each Var used in an update definition
// must appear unadorned in the same position as in the pure
// definition in all references to the function on the left-
// and right-hand sides. So the following definitions are
// legal updates:
f(x, 17) = x + 8;
f(0, y) = y * 8;
f(x, x + 1) = x + 8;
f(y/2, y) = f(0, y) * 17;
// But these ones would cause an error:
// f(x, 0) = f(x + 1, 0);
// First argument to f on the right-hand-side must be 'x', not 'x + 1'.
// f(y, y + 1) = y + 8;
// Second argument to f on the left-hand-side must be 'y', not 'y + 1'.
// f(y, x) = y - x;
// Arguments to f on the left-hand-side are in the wrong places.
// f(3, 4) = x + y;
// Free variables appear on the right-hand-side but not the left-hand-side.
// We'll realize this one just to make sure it compiles. The
// second-to-last definition forces us to realize over a
// domain that is taller than it is wide.
f.realize(100, 101);
// For each realization of f, each step runs in its entirety
// before the next one begins. Let's trace the loads and
// stores for a simpler example:
Func g("g");
g(x, y) = x + y; // Pure definition
g(2, 1) = 42; // First update definition
g(x, 0) = g(x, 1); // Second update definition
g.realize(4, 4);
// See figures/lesson_09_update.gif for a visualization.
// Reading the log, we see that each pass is applied in
// turn. The equivalent C is:
int result[4][4];
// Pure definition
for (int y = 0; y < 4; y++) {
for (int x = 0; x < 4; x++) {
result[y][x] = x + y;
}
}
// First update definition
result[1][2] = 42;
// Second update definition
for (int x = 0; x < 4; x++) {
result[0][x] = result[1][x];
}
}
// Putting update passes inside loops.
{
// Starting with this pure definition:
Func f;
f(x, y) = (x + y)/100.0f;
// Say we want an update that squares the first fifty rows. We
// could do this by adding 50 update definitions:
// f(x, 0) = f(x, 0) * f(x, 0);
// f(x, 1) = f(x, 1) * f(x, 1);
// f(x, 2) = f(x, 2) * f(x, 2);
// ...
// f(x, 49) = f(x, 49) * f(x, 49);
// Or equivalently using a compile-time loop in our C++:
// for (int i = 0; i < 50; i++) {
// f(x, i) = f(x, i) * f(x, i);
// }
// But it's more manageable and more flexible to put the loop
// in the generated code. We do this by defining a "reduction
// domain" and using it inside an update definition:
RDom r(0, 50);
f(x, r) = f(x, r) * f(x, r);
Buffer<float> halide_result = f.realize(100, 100);
// See figures/lesson_09_update_rdom.mp4 for a visualization.
// The equivalent C is:
float c_result[100][100];
for (int y = 0; y < 100; y++) {
for (int x = 0; x < 100; x++) {
c_result[y][x] = (x + y)/100.0f;
}
}
for (int x = 0; x < 100; x++) {
for (int r = 0; r < 50; r++) {
// The loop over the reduction domain occurs inside of
// the loop over any pure variables used in the update
// step:
c_result[r][x] = c_result[r][x] * c_result[r][x];
}
}
// Check the results match:
for (int y = 0; y < 100; y++) {
for (int x = 0; x < 100; x++) {
if (fabs(halide_result(x, y) - c_result[y][x]) > 0.01f) {
printf("halide_result(%d, %d) = %f instead of %f\n",
x, y, halide_result(x, y), c_result[y][x]);
return -1;
}
}
}
}
// Now we'll examine a real-world use for an update definition:
// computing a histogram.
{
// Some operations on images can't be cleanly expressed as a pure
// function from the output coordinates to the value stored
// there. The classic example is computing a histogram. The
// natural way to do it is to iterate over the input image,
// updating histogram buckets. Here's how you do that in Halide:
Func histogram("histogram");
// Histogram buckets start as zero.
histogram(x) = 0;
// Define a multi-dimensional reduction domain over the input image:
RDom r(0, input.width(), 0, input.height());
// For every point in the reduction domain, increment the
// histogram bucket corresponding to the intensity of the
// input image at that point.
histogram(input(r.x, r.y)) += 1;
Buffer<int> halide_result = histogram.realize(256);
// The equivalent C is:
int c_result[256];
for (int x = 0; x < 256; x++) {
c_result[x] = 0;
}
for (int r_y = 0; r_y < input.height(); r_y++) {
for (int r_x = 0; r_x < input.width(); r_x++) {
c_result[input(r_x, r_y)] += 1;
}
}
// Check the answers agree:
for (int x = 0; x < 256; x++) {
if (c_result[x] != halide_result(x)) {
printf("halide_result(%d) = %d instead of %d\n",
x, halide_result(x), c_result[x]);
return -1;
}
}
}
// Scheduling update steps
{
// The pure variables in an update step and can be
// parallelized, vectorized, split, etc as usual.
// Vectorizing, splitting, or parallelize the variables that
// are part of the reduction domain is trickier. We'll cover
// that in a later lesson.
// Consider the definition:
Func f;
f(x, y) = x*y;
// Set row zero to each row 8
f(x, 0) = f(x, 8);
// Set column 1 equal to column 8 plus 2
f(0, y) = f(8, y) + 2;
// The pure variables in each stage can be scheduled
// independently. To control the pure definition, we schedule
// as we have done in the past. The following code vectorizes
// and parallelizes the pure definition only.
f.vectorize(x, 4).parallel(y);
// We use Func::update(int) to get a handle to an update step
// for the purposes of scheduling. The following line
// vectorizes the first update step across x. We can't do
// anything with y for this update step, because it doesn't
// use y.
f.update(0).vectorize(x, 4);
// Now we parallelize the second update step in chunks of size
// 4.
Var yo, yi;
f.update(1).split(y, yo, yi, 4).parallel(yo);
Buffer<int> halide_result = f.realize(16, 16);
// See figures/lesson_09_update_schedule.mp4 for a visualization.
// Here's the equivalent (serial) C:
int c_result[16][16];
// Pure step. Vectorized in x and parallelized in y.
for (int y = 0; y < 16; y++) { // Should be a parallel for loop
for (int x_vec = 0; x_vec < 4; x_vec++) {
int x[] = {x_vec*4, x_vec*4+1, x_vec*4+2, x_vec*4+3};
c_result[y][x[0]] = x[0] * y;
c_result[y][x[1]] = x[1] * y;
c_result[y][x[2]] = x[2] * y;
c_result[y][x[3]] = x[3] * y;
}
}
// First update. Vectorized in x.
for (int x_vec = 0; x_vec < 4; x_vec++) {
int x[] = {x_vec*4, x_vec*4+1, x_vec*4+2, x_vec*4+3};
c_result[0][x[0]] = c_result[8][x[0]];
c_result[0][x[1]] = c_result[8][x[1]];
c_result[0][x[2]] = c_result[8][x[2]];
c_result[0][x[3]] = c_result[8][x[3]];
}
// Second update. Parallelized in chunks of size 4 in y.
for (int yo = 0; yo < 4; yo++) { // Should be a parallel for loop
for (int yi = 0; yi < 4; yi++) {
int y = yo*4 + yi;
c_result[y][0] = c_result[y][8] + 2;
}
}
// Check the C and Halide results match:
for (int y = 0; y < 16; y++) {
for (int x = 0; x < 16; x++) {
if (halide_result(x, y) != c_result[y][x]) {
printf("halide_result(%d, %d) = %d instead of %d\n",
x, y, halide_result(x, y), c_result[y][x]);
return -1;
}
}
}
}
// That covers how to schedule the variables within a Func that
// uses update steps, but what about producer-consumer
// relationships that involve compute_at and store_at? Let's
// examine a reduction as a producer, in a producer-consumer pair.
{
// Because an update does multiple passes over a stored array,
// it's not meaningful to inline them. So the default schedule
// for them does the closest thing possible. It computes them
// in the innermost loop of their consumer. Consider this
// trivial example:
Func producer, consumer;
producer(x) = x*2;
producer(x) += 10;
consumer(x) = 2 * producer(x);
Buffer<int> halide_result = consumer.realize(10);
// See figures/lesson_09_inline_reduction.gif for a visualization.
// The equivalent C is:
int c_result[10];
for (int x = 0; x < 10; x++) {
int producer_storage[1];
// Pure step for producer
producer_storage[0] = x * 2;
// Update step for producer
producer_storage[0] = producer_storage[0] + 10;
// Pure step for consumer
c_result[x] = 2 * producer_storage[0];
}
// Check the results match
for (int x = 0; x < 10; x++) {
if (halide_result(x) != c_result[x]) {
printf("halide_result(%d) = %d instead of %d\n",
x, halide_result(x), c_result[x]);
return -1;
}
}
// For all other compute_at/store_at options, the reduction
// gets placed where you would expect, somewhere in the loop
// nest of the consumer.
}
// Now let's consider a reduction as a consumer in a
// producer-consumer pair. This is a little more involved.
{
{
// Case 1: The consumer references the producer in the pure step only.
Func producer, consumer;
// The producer is pure.
producer(x) = x*17;
consumer(x) = 2 * producer(x);
consumer(x) += 50;
// The valid schedules for the producer in this case are
// the default schedule - inlined, and also:
//
// 1) producer.compute_at(x), which places the computation of
// the producer inside the loop over x in the pure step of the
// consumer.
//
// 2) producer.compute_root(), which computes all of the
// producer ahead of time.
//
// 3) producer.store_root().compute_at(x), which allocates
// space for the consumer outside the loop over x, but fills
// it in as needed inside the loop.
//
// Let's use option 1.
producer.compute_at(consumer, x);
Buffer<int> halide_result = consumer.realize(10);
// See figures/lesson_09_compute_at_pure.gif for a visualization.
// The equivalent C is:
int c_result[10];
// Pure step for the consumer
for (int x = 0; x < 10; x++) {
// Pure step for producer
int producer_storage[1];
producer_storage[0] = x * 17;
c_result[x] = 2 * producer_storage[0];
}
// Update step for the consumer
for (int x = 0; x < 10; x++) {
c_result[x] += 50;
}
// All of the pure step is evaluated before any of the
// update step, so there are two separate loops over x.
// Check the results match
for (int x = 0; x < 10; x++) {
if (halide_result(x) != c_result[x]) {
printf("halide_result(%d) = %d instead of %d\n",
x, halide_result(x), c_result[x]);
return -1;
}
}
}
{
// Case 2: The consumer references the producer in the update step only
Func producer, consumer;
producer(x) = x * 17;
consumer(x) = 100 - x * 10;
consumer(x) += producer(x);
// Again we compute the producer per x coordinate of the
// consumer. This places producer code inside the update
// step of the consumer, because that's the only step that
// uses the producer.
producer.compute_at(consumer, x);
// Note however, that we didn't say:
//
// producer.compute_at(consumer.update(0), x).
//
// Scheduling is done with respect to Vars of a Func, and
// the Vars of a Func are shared across the pure and
// update steps.
Buffer<int> halide_result = consumer.realize(10);
// See figures/lesson_09_compute_at_update.gif for a visualization.
// The equivalent C is:
int c_result[10];
// Pure step for the consumer
for (int x = 0; x < 10; x++) {
c_result[x] = 100 - x * 10;
}
// Update step for the consumer
for (int x = 0; x < 10; x++) {
// Pure step for producer
int producer_storage[1];
producer_storage[0] = x * 17;
c_result[x] += producer_storage[0];
}
// Check the results match
for (int x = 0; x < 10; x++) {
if (halide_result(x) != c_result[x]) {
printf("halide_result(%d) = %d instead of %d\n",
x, halide_result(x), c_result[x]);
return -1;
}
}
}
{
// Case 3: The consumer references the producer in
// multiple steps that share common variables
Func producer, consumer;
producer(x) = x * 17;
consumer(x) = 170 - producer(x);
consumer(x) += producer(x)/2;
// Again we compute the producer per x coordinate of the
// consumer. This places producer code inside both the
// pure and the update step of the consumer. So there end
// up being two separate realizations of the producer, and
// redundant work occurs.
producer.compute_at(consumer, x);
Buffer<int> halide_result = consumer.realize(10);
// See figures/lesson_09_compute_at_pure_and_update.gif for a visualization.
// The equivalent C is:
int c_result[10];
// Pure step for the consumer
for (int x = 0; x < 10; x++) {
// Pure step for producer
int producer_storage[1];
producer_storage[0] = x * 17;
c_result[x] = 170 - producer_storage[0];
}
// Update step for the consumer
for (int x = 0; x < 10; x++) {
// Another copy of the pure step for producer
int producer_storage[1];
producer_storage[0] = x * 17;
c_result[x] += producer_storage[0]/2;
}
// Check the results match
for (int x = 0; x < 10; x++) {
if (halide_result(x) != c_result[x]) {
printf("halide_result(%d) = %d instead of %d\n",
x, halide_result(x), c_result[x]);
return -1;
}
}
}
{
// Case 4: The consumer references the producer in
// multiple steps that do not share common variables
Func producer, consumer;
producer(x, y) = (x * y) / 10 + 8;
consumer(x, y) = x + y;
consumer(x, 0) = producer(x, x);
consumer(0, y) = producer(y, 9-y);
// In this case neither producer.compute_at(consumer, x)
// nor producer.compute_at(consumer, y) will work, because
// either one fails to cover one of the uses of the
// producer. So we'd have to inline producer, or use
// producer.compute_root().
// Let's say we really really want producer to be
// compute_at the inner loops of both consumer update
// steps. Halide doesn't allow multiple different
// schedules for a single Func, but we can work around it
// by making two wrappers around producer, and scheduling
// those instead:
// Attempt 2:
Func producer_1, producer_2, consumer_2;
producer_1(x, y) = producer(x, y);
producer_2(x, y) = producer(x, y);
consumer_2(x, y) = x + y;
consumer_2(x, 0) += producer_1(x, x);
consumer_2(0, y) += producer_2(y, 9-y);
// The wrapper functions give us two separate handles on
// the producer, so we can schedule them differently.
producer_1.compute_at(consumer_2, x);
producer_2.compute_at(consumer_2, y);
Buffer<int> halide_result = consumer_2.realize(10, 10);
// See figures/lesson_09_compute_at_multiple_updates.mp4 for a visualization.
// The equivalent C is:
int c_result[10][10];
// Pure step for the consumer
for (int y = 0; y < 10; y++) {
for (int x = 0; x < 10; x++) {
c_result[y][x] = x + y;
}
}
// First update step for consumer
for (int x = 0; x < 10; x++) {
int producer_1_storage[1];
producer_1_storage[0] = (x * x) / 10 + 8;
c_result[0][x] += producer_1_storage[0];
}
// Second update step for consumer
for (int y = 0; y < 10; y++) {
int producer_2_storage[1];
producer_2_storage[0] = (y * (9-y)) / 10 + 8;
c_result[y][0] += producer_2_storage[0];
}
// Check the results match
for (int y = 0; y < 10; y++) {
for (int x = 0; x < 10; x++) {
if (halide_result(x, y) != c_result[y][x]) {
printf("halide_result(%d, %d) = %d instead of %d\n",
x, y, halide_result(x, y), c_result[y][x]);
return -1;
}
}
}
}
{
// Case 5: Scheduling a producer under a reduction domain
// variable of the consumer.
// We are not just restricted to scheduling producers at
// the loops over the pure variables of the consumer. If a
// producer is only used within a loop over a reduction
// domain (RDom) variable, we can also schedule the
// producer there.
Func producer, consumer;
RDom r(0, 5);
producer(x) = x % 8;
consumer(x) = x + 10;
consumer(x) += r + producer(x + r);
producer.compute_at(consumer, r);
Buffer<int> halide_result = consumer.realize(10);
// See figures/lesson_09_compute_at_rvar.gif for a visualization.
// The equivalent C is:
int c_result[10];
// Pure step for the consumer.
for (int x = 0; x < 10; x++) {
c_result[x] = x + 10;
}
// Update step for the consumer.
for (int x = 0; x < 10; x++) {
// The loop over the reduction domain is always the inner loop.
for (int r = 0; r < 5; r++) {
// We've schedule the storage and computation of
// the producer here. We just need a single value.
int producer_storage[1];
// Pure step of the producer.
producer_storage[0] = (x + r) % 8;
// Now use it in the update step of the consumer.
c_result[x] += r + producer_storage[0];
}
}
// Check the results match
for (int x = 0; x < 10; x++) {
if (halide_result(x) != c_result[x]) {
printf("halide_result(%d) = %d instead of %d\n",
x, halide_result(x), c_result[x]);
return -1;
}
}
}
}
// A real-world example of a reduction inside a producer-consumer chain.
{
// The default schedule for a reduction is a good one for
// convolution-like operations. For example, the following
// computes a 5x5 box-blur of our grayscale test image with a
// clamp-to-edge boundary condition:
// First add the boundary condition.
// Define a 5x5 box that starts at (-2, -2)
RDom r(-2, 5, -2, 5);
// Compute the 5x5 sum around each pixel.
Func local_sum;
local_sum(x, y) = 0; // Compute the sum as a 32-bit integer
local_sum(x, y) += clamped(x + r.x, y + r.y);
// Divide the sum by 25 to make it an average
Func blurry;
blurry(x, y) = cast<uint8_t>(local_sum(x, y) / 25);
Buffer<uint8_t> halide_result = blurry.realize(input.width(), input.height());
// The default schedule will inline 'clamped' into the update
// step of 'local_sum', because clamped only has a pure
// definition, and so its default schedule is fully-inlined.
// We will then compute local_sum per x coordinate of blurry,
// because the default schedule for reductions is
// compute-innermost. Here's the equivalent C:
Buffer<uint8_t> c_result(input.width(), input.height());
for (int y = 0; y < input.height(); y++) {
for (int x = 0; x < input.width(); x++) {
int local_sum[1];
// Pure step of local_sum
local_sum[0] = 0;
// Update step of local_sum
for (int r_y = -2; r_y <= 2; r_y++) {
for (int r_x = -2; r_x <= 2; r_x++) {
// The clamping has been inlined into the update step.
int clamped_x = std::min(std::max(x + r_x, 0), input.width()-1);
int clamped_y = std::min(std::max(y + r_y, 0), input.height()-1);
local_sum[0] += input(clamped_x, clamped_y);
}
}
// Pure step of blurry
c_result(x, y) = (uint8_t)(local_sum[0] / 25);
}
}
// Check the results match
for (int y = 0; y < input.height(); y++) {
for (int x = 0; x < input.width(); x++) {
if (halide_result(x, y) != c_result(x, y)) {
printf("halide_result(%d, %d) = %d instead of %d\n",
x, y, halide_result(x, y), c_result(x, y));
return -1;
}
}
}
}
// Reduction helpers.
{
// There are several reduction helper functions provided in
// Halide.h, which compute small reductions and schedule them
// innermost into their consumer. The most useful one is
// "sum".
Func f1;
RDom r(0, 100);
f1(x) = sum(r + x) * 7;
// Sum creates a small anonymous Func to do the reduction. It's equivalent to:
Func f2;
Func anon;
anon(x) = 0;
anon(x) += r + x;
f2(x) = anon(x) * 7;
// So even though f1 references a reduction domain, it is a
// pure function. The reduction domain has been swallowed to
// define the inner anonymous reduction.
Buffer<int> halide_result_1 = f1.realize(10);
Buffer<int> halide_result_2 = f2.realize(10);
// The equivalent C is:
int c_result[10];
for (int x = 0; x < 10; x++) {
int anon[1];
anon[0] = 0;
for (int r = 0; r < 100; r++) {
anon[0] += r + x;
}
c_result[x] = anon[0] * 7;
}
// Check they all match.
for (int x = 0; x < 10; x++) {
if (halide_result_1(x) != c_result[x]) {
printf("halide_result_1(%d) = %d instead of %d\n",
x, halide_result_1(x), c_result[x]);
return -1;
}
if (halide_result_2(x) != c_result[x]) {
printf("halide_result_2(%d) = %d instead of %d\n",
x, halide_result_2(x), c_result[x]);
return -1;
}
}
}
// A complex example that uses reduction helpers.
{
// Other reduction helpers include "product", "minimum",
// "maximum", "argmin", and "argmax". Using argmin and argmax
// requires understanding tuples, which come in a later
// lesson. Let's use minimum and maximum to compute the local
// spread of our grayscale image.
// First, add a boundary condition to the input.
Func clamped;
Expr x_clamped = clamp(x, 0, input.width()-1);
Expr y_clamped = clamp(y, 0, input.height()-1);
clamped(x, y) = input(x_clamped, y_clamped);
RDom box(-2, 5, -2, 5);
// Compute the local maximum minus the local minimum:
Func spread;
spread(x, y) = (maximum(clamped(x + box.x, y + box.y)) -
minimum(clamped(x + box.x, y + box.y)));
// Compute the result in strips of 32 scanlines
Var yo, yi;
spread.split(y, yo, yi, 32).parallel(yo);
// Vectorize across x within the strips. This implicitly
// vectorizes stuff that is computed within the loop over x in
// spread, which includes our minimum and maximum helpers, so
// they get vectorized too.
spread.vectorize(x, 16);
// We'll apply the boundary condition by padding each scanline
// as we need it in a circular buffer (see lesson 08).
clamped.store_at(spread, yo).compute_at(spread, yi);
Buffer<uint8_t> halide_result = spread.realize(input.width(), input.height());
// The C equivalent is almost too horrible to contemplate (and
// took me a long time to debug). This time I want to time
// both the Halide version and the C version, so I'll use sse
// intrinsics for the vectorization, and openmp to do the
// parallel for loop (you'll need to compile with -fopenmp or
// similar to get correct timing).
#ifdef __SSE2__
// Don't include the time required to allocate the output buffer.
Buffer<uint8_t> c_result(input.width(), input.height());
#ifdef _OPENMP
double t1 = current_time();
#endif
// Run this one hundred times so we can average the timing results.
for (int iters = 0; iters < 100; iters++) {
#pragma omp parallel for
for (int yo = 0; yo < (input.height() + 31)/32; yo++) {
int y_base = std::min(yo * 32, input.height() - 32);
// Compute clamped in a circular buffer of size 8
// (smallest power of two greater than 5). Each thread
// needs its own allocation, so it must occur here.
int clamped_width = input.width() + 4;
uint8_t *clamped_storage = (uint8_t *)malloc(clamped_width * 8);
for (int yi = 0; yi < 32; yi++) {
int y = y_base + yi;
uint8_t *output_row = &c_result(0, y);
// Compute clamped for this scanline, skipping rows
// already computed within this slice.
int min_y_clamped = (yi == 0) ? (y - 2) : (y + 2);
int max_y_clamped = (y + 2);
for (int cy = min_y_clamped; cy <= max_y_clamped; cy++) {
// Figure out which row of the circular buffer
// we're filling in using bitmasking:
uint8_t *clamped_row =
clamped_storage + (cy & 7) * clamped_width;
// Figure out which row of the input we're reading
// from by clamping the y coordinate:
int clamped_y = std::min(std::max(cy, 0), input.height()-1);
uint8_t *input_row = &input(0, clamped_y);
// Fill it in with the padding.
for (int x = -2; x < input.width() + 2; x++) {
int clamped_x = std::min(std::max(x, 0), input.width()-1);
*clamped_row++ = input_row[clamped_x];
}
}
// Now iterate over vectors of x for the pure step of the output.
for (int x_vec = 0; x_vec < (input.width() + 15)/16; x_vec++) {
int x_base = std::min(x_vec * 16, input.width() - 16);
// Allocate storage for the minimum and maximum
// helpers. One vector is enough.
__m128i minimum_storage, maximum_storage;
// The pure step for the maximum is a vector of zeros
maximum_storage = _mm_setzero_si128();
// The update step for maximum
for (int max_y = y - 2; max_y <= y + 2; max_y++) {
uint8_t *clamped_row =
clamped_storage + (max_y & 7) * clamped_width;
for (int max_x = x_base - 2; max_x <= x_base + 2; max_x++) {
__m128i v = _mm_loadu_si128(
(__m128i const *)(clamped_row + max_x + 2));
maximum_storage = _mm_max_epu8(maximum_storage, v);
}
}
// The pure step for the minimum is a vector of
// ones. Create it by comparing something to
// itself.
minimum_storage = _mm_cmpeq_epi32(_mm_setzero_si128(),
_mm_setzero_si128());
// The update step for minimum.
for (int min_y = y - 2; min_y <= y + 2; min_y++) {
uint8_t *clamped_row =
clamped_storage + (min_y & 7) * clamped_width;
for (int min_x = x_base - 2; min_x <= x_base + 2; min_x++) {
__m128i v = _mm_loadu_si128(
(__m128i const *)(clamped_row + min_x + 2));
minimum_storage = _mm_min_epu8(minimum_storage, v);
}
}
// Now compute the spread.
__m128i spread = _mm_sub_epi8(maximum_storage, minimum_storage);
// Store it.
_mm_storeu_si128((__m128i *)(output_row + x_base), spread);
}
}
free(clamped_storage);
}
}
// Skip the timing comparison if we don't have openmp
// enabled. Otherwise it's unfair to C.
#ifdef _OPENMP
double t2 = current_time();
// Now run the Halide version again without the
// jit-compilation overhead. Also run it one hundred times.
for (int iters = 0; iters < 100; iters++) {
spread.realize(halide_result);
}
double t3 = current_time();
// Report the timings. On my machine they both take about 3ms
// for the 4-megapixel input (fast!), which makes sense,
// because they're using the same vectorization and
// parallelization strategy. However I find the Halide easier
// to read, write, debug, modify, and port.
printf("Halide spread took %f ms. C equivalent took %f ms\n",
(t3 - t2)/100, (t2 - t1)/100);
#endif // _OPENMP
// Check the results match:
for (int y = 0; y < input.height(); y++) {
for (int x = 0; x < input.width(); x++) {
if (halide_result(x, y) != c_result(x, y)) {
printf("halide_result(%d, %d) = %d instead of %d\n",
x, y, halide_result(x, y), c_result(x, y));
return -1;
}
}
}
#endif // __SSE2__
}
printf("Success!\n");
return 0;
}