Halide示例学习五
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- 函数重定义操作
- RDom下标使用
- 使用RDom计算直方图
- 调度函数更新
- 多阶段重定义操作
- 使用归化函数进行模糊操作
- 内置的归化函数sum
- 使用归约函数的一个复杂例子
函数重定义操作
#include "Halide.h"
#include
using namespace Halide;
int main(int argc, char **argv) {
Var x("x"), y("y");
Buffer input = load_image("../../images/gray.png");
{
// 这里有着var到expr的转换，是一种原始定义
Func f;
f(x, y) = x + y;
// 这里是单点定义
f(3, 7) = 42;
// 递归定义：表示在同一位置递归调用本身值f(x,y) = g(f(x,y), var)
f(x, y) = f(x, y) + 17;
// 特指某一行进行更新
f(x, 3) = f(x, 0) * f(x, 10);
// 特指某一列进行更新
f(0, y) = f(0, y) / f(3, y);
// 更新定义的规则是：左右两边的var参数必须是(x,其他)，或者是(其他,y)，这跟之前的f定义是一致的。
// 以下部分是合法的
f(x, 17) = x + 8; // y >= 18
f(0, y) = y * 8; // x >= 4
f(x, x + 1) = x + 8; // y >= x + 1
f(y / 2, y) = f(0, y) * 17; // x >= y / 2
// 下面的部分是非法的
// 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.
// 这里需要根据前面的重定义函数来评估x,y的尺寸关系
f.realize({100, 101});
// 每一次的重定义，都会产生多次更新动作，那么在realize的时候将会体现每次的重定义
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.trace_loads();//跟踪读数据
g.trace_stores();//跟踪写数据
g.realize({4, 4});
// 下述是等效C代码
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];
}
}
return 0;
}
Store g.0(0, 0) = 0
Store g.0(1, 0) = 1
Store g.0(2, 0) = 2
Store g.0(3, 0) = 3
Store g.0(0, 1) = 1
Store g.0(1, 1) = 2
Store g.0(2, 1) = 3
Store g.0(3, 1) = 4
Store g.0(0, 2) = 2
Store g.0(1, 2) = 3
Store g.0(2, 2) = 4
Store g.0(3, 2) = 5
Store g.0(0, 3) = 3
Store g.0(1, 3) = 4
Store g.0(2, 3) = 5
Store g.0(3, 3) = 6
Store g.0(2, 1) = 42
Load g.0(0, 1) = 1
Store g.0(0, 0) = 1
Load g.0(1, 1) = 2
Store g.0(1, 0) = 2
Load g.0(2, 1) = 42
Store g.0(2, 0) = 42
Load g.0(3, 1) = 4
Store g.0(3, 0) = 4
RDom下标使用
#include "Halide.h"
#include
using namespace Halide;
int main(int argc, char **argv) {
Var x("x"), y("y");
Buffer input = load_image("../../images/gray.png");
// 把更新放置到内部循环中
{
// 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);
// }
// 使用RDom变量，我们可以把限定范围的更新组织成循环
RDom r(0, 50);
f(x, r) = f(x, r) * f(x, r);
Buffer 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++) {
// RDom这个限定轴，被内嵌为内部循环
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;
}
}
}
}
return 0;
}
使用RDom计算直方图
#include "Halide.h"
#include
using namespace Halide;
int main(int argc, char **argv) {
Var x("x"), y("y");
Buffer input = load_image("../../images/gray.png");
// 现在使用重定义来实现直方图
{
Func histogram("histogram");
// 直方图置零
histogram(x) = 0;
// 定义一个多维的归化维度
RDom r(0, input.width(), 0, input.height());
// 统计input的直方图(0,255)灰度值之间的个数
histogram(input(r.x, r.y)) += 1;
// 只保留[0,255]之间的值
Buffer halide_result = histogram.realize({256});
// 等效C代码如下所示
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;
}
}
}
return 0;
}
调度函数更新
#include "Halide.h"
#include
using namespace Halide;
int main(int argc, char **argv) {
Var x("x"), y("y");
Buffer input = load_image("../../images/gray.png");
// 调度函数更新顺序
{
// 对于重定义的函数，我们该如何实现向量化、并行化等操作呢
// 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 zero equal to column 8 plus 2
f(0, y) = f(8, y) + 2;
// 原始的 f(x,y) = x * y向量化后再并行化
f.vectorize(x, 4).parallel(y);
// 使用Func::update(int upid)来指定第upid次的函数更新的后续优化手段，这里upid=0只能对x那个轴进行向量化
f.update(0).vectorize(x, 4);
// upid = 1,表示在第二update公式上进行操作
Var yo, yi;
f.update(1).split(y, yo, yi, 4).parallel(yo);//分割后向量化，再接并行
Buffer 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;
}
}
}
}
return 0;
}
多阶段重定义操作
#include "Halide.h"
#include
using namespace Halide;
int main(int argc, char **argv) {
Var x("x"), y("y");
Buffer input = load_image("../../images/gray.png");
// 当涉及到多个计算图的时候，重定义操作又是如何进行的呢
{
// 对于默认的多计算内敛模式而言，重定义似乎没有意义
Func producer, consumer;
producer(x) = x * 2;//纯定义
producer(x) += 10;//更新
consumer(x) = 2 * producer(x);//内敛
Buffer 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;
}
}
}
// 对comsumer这个后计算图进行重定义操作
// 情况1: consumer只引用producer的纯定义
{
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 halide_result = consumer.realize({10});
// See figures/lesson_09_compute_at_pure.gif for a visualization.
// The equivalent C is:
int c_result[10];
// compute_at使得prudcer先算，又因为内建的原因，consumer的内建操作使得完成consumer的纯定义的计算
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];
}
// 然后重定义计算
for (int x = 0; x < 10; x++) {
c_result[x] += 50;
}
// 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;
}
}
}
{
// 情况 2: consumer只在重定义的时候引用producer
Func producer, consumer;
producer(x) = x * 17;
consumer(x) = 100 - x * 10;
consumer(x) += producer(x);
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 halide_result = consumer.realize({10});
// See figures/lesson_09_compute_at_update.gif for a visualization.
// The equivalent C is:
int c_result[10];
// 纯定义
for (int x = 0; x < 10; x++) {
c_result[x] = 100 - x * 10;
}
// 重定义部分
for (int x = 0; x < 10; x++) {
// producer的纯定义
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;
}
}
}
{
// 情况 3： 在纯定义和重定义的时候都使用了producer的情况
Func producer, consumer;
producer(x) = x * 17;
consumer(x) = 170 - producer(x);
consumer(x) += producer(x) / 2;
producer.compute_at(consumer, x);
Buffer halide_result = consumer.realize({10});
// See figures/lesson_09_compute_at_pure_and_update.gif for a visualization.
// 不管如何，中间计算过程都没有专门为了保存producer而设置的中间Buffer
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;
}
}
}
{
// 情况 4：在多次重定义的过程中才使用producer
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 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: 当使用了RDom限定下标的时候，使用compute_at
// 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 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;
}
}
}
return 0;
}
使用归化函数进行模糊操作
#include "Halide.h"
#include
using namespace Halide;
int main(int argc, char **argv) {
Var x("x"), y("y");
Buffer input = load_image("../../images/gray.png");
// 下面是一个模糊的归化函数
{
// 下述是一个边界限定的函数
Func clamped = BoundaryConditions::repeat_edge(input);
// 定义一个从该点开始的相对位置为-2的点
RDom r(-2, 5, -2, 5);
// 计算每个像素值的附近的和
Func local_sum;
local_sum(x, y) = 0;// 纯定义
local_sum(x, y) += clamped(x + r.x, y + r.y); // 归约求和
// 求平均值，然后int32->uint8
Func blurry;
blurry(x, y) = cast(local_sum(x, y) / 25);
Buffer halide_result = blurry.realize({input.width(), input.height()});
// 默认的调度就是把clamped进行了内联
// 等效C代码如下
Buffer 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;
}
}
}
}
return 0;
}
内置的归化函数sum
#include "Halide.h"
#include
using namespace Halide;
int main(int argc, char **argv) {
Var x("x"), y("y");
Buffer input = load_image("../../images/gray.png");
// 归约函数sum
{
// Halide.h中包含了sum归约函数
Func f1;
RDom r(0, 100);
f1(x) = sum(r + x) * 7;
// 上述操作等下等效
Func f2;
Func anon;
anon(x) = 0;
anon(x) += r + x;//固定x,那么对于每一个r[0,100],我们必须计算anon[x] = [x + r for r in [0,100]]
f2(x) = anon(x) * 7;
// 所以尽管f1使用了sum函数，但是这个依然是个纯定义函数
Buffer halide_result_1 = f1.realize({10});
Buffer 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;
}
}
}
return 0;
}
使用归约函数的一个复杂例子
#include "Halide.h"
#include
using namespace Halide;
int main(int argc, char **argv) {
Var x("x"), y("y");
Buffer input = load_image("../../images/gray.png");
// 一个使用归约函数的复杂例子
{
// 其他的归约函数如下所示：
// 累乘"product", 最小值"minimum",
// 最大值"maximum", 反解最小"argmin", 反解最大"argmax".
// 首先进行边界限定
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);
// 然后进行如下归约
Func spread;
spread(x, y) = (maximum(clamped(x + box.x, y + box.y)) -
minimum(clamped(x + box.x, y + box.y)));
// 循环拆分、并行化
Var yo, yi;
spread.split(y, yo, yi, 32).parallel(yo);
// 使用向量化，maximum和minimum归约函数也会受到向量化影响
spread.vectorize(x, 16);
// 为了放置边界被多次计算，这里使用了store_at和compute_at
clamped.store_at(spread, yo).compute_at(spread, yi);
Buffer halide_result = spread.realize({input.width(), input.height()});
// 这里使用优化的C代码和Halide代码进行比较，来看两者的速度差别
#define __SSE2__
#define _OPENMP
#ifdef __SSE2__
// Don't include the time required to allocate the output buffer.
Buffer 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);//中间计算结果，store_at
for (int yi = 0; yi < 32; yi++) {//循环拆分的内层循环
int y = y_base + yi;
uint8_t *output_row = &c_result(0, y);
// compute_at的效用，使得在向量化前计算了clamped
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++) { //compute_at,用于进行计算操作
// 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];
}
}
// 现在进行最终的计算
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();
// maximum操作
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);
}
}
// 定义一个0值的minimum
minimum_storage = _mm_cmpeq_epi32(_mm_setzero_si128(),
_mm_setzero_si128());
// minmum操作
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);
}
}
// sse减法
__m128i spread = _mm_sub_epi8(maximum_storage, minimum_storage);
// sse存储
_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();
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__
}
return 0;
}
Halide spread took 0.668330 ms. C equivalent took 68.161500 ms
