为了实现一些图像分析算法而不必过多担心数据类型（即没有太多重复代码），我正在为 Java 中的原始数组设置访问者模式。
在下面的示例中，我定义了两种类型的访问者
- 原始类型，其中
visit方法的签名是visit(int, int double) - 泛型类型，其中
visit方法的签名是visit(int, int Double).
Appart 由此，两个访问者执行完全相同的操作。我的想法是尝试衡量装箱/拆箱的成本。
所以这是完整的程序
public class VisitorsBenchmark {
public interface Array2DGenericVisitor<TYPE, RET> {
void begin(int width, int height);
RET end();
void visit(int x, int y, TYPE value);
}
public interface Array2DPrimitiveVisitor<RET> {
void begin(final int width, final int height);
RET end();
void visit(final int x, final int y, final double value);
}
public static <RET>
RET
accept(final int width,
final int height,
final double[] data,
final Array2DGenericVisitor<Double, RET> visitor) {
final int size = width * height;
visitor.begin(width, height);
for (int i = 0, x = 0, y = 0; i < size; i++) {
visitor.visit(x, y, data[i]);
x++;
if (x == width) {
x = 0;
y++;
if (y == height) {
y = 0;
}
}
}
return visitor.end();
}
public static <RET> RET accept(final int width,
final int height,
final double[] data,
final Array2DPrimitiveVisitor<RET> visitor) {
final int size = width * height;
visitor.begin(width, height);
for (int i = 0, x = 0, y = 0; i < size; i++) {
visitor.visit(x, y, data[i]);
x++;
if (x == width) {
x = 0;
y++;
if (y == height) {
y = 0;
}
}
}
return visitor.end();
}
private static final Array2DGenericVisitor<Double, double[]> generic;
private static final Array2DPrimitiveVisitor<double[]> primitive;
static {
generic = new Array2DGenericVisitor<Double, double[]>() {
private double[] sum;
@Override
public void begin(final int width, final int height) {
final int length = (int) Math.ceil(Math.hypot(WIDTH, HEIGHT));
sum = new double[length];
}
@Override
public void visit(final int x, final int y, final Double value) {
final int r = (int) Math.round(Math.sqrt(x * x + y * y));
sum[r] += value;
}
@Override
public double[] end() {
return sum;
}
};
primitive = new Array2DPrimitiveVisitor<double[]>() {
private double[] sum;
@Override
public void begin(final int width, final int height) {
final int length = (int) Math.ceil(Math.hypot(WIDTH, HEIGHT));
sum = new double[length];
}
@Override
public void visit(final int x, final int y, final double value) {
final int r = (int) Math.round(Math.sqrt(x * x + y * y));
sum[r] += value;
}
@Override
public double[] end() {
return sum;
}
};
}
private static final int WIDTH = 300;
private static final int HEIGHT = 300;
private static final int NUM_ITERATIONS_PREHEATING = 10000;
private static final int NUM_ITERATIONS_BENCHMARKING = 10000;
public static void main(String[] args) {
final double[] data = new double[WIDTH * HEIGHT];
for (int i = 0; i < data.length; i++) {
data[i] = Math.random();
}
/*
* Pre-heating.
*/
for (int i = 0; i < NUM_ITERATIONS_PREHEATING; i++) {
accept(WIDTH, HEIGHT, data, generic);
}
for (int i = 0; i < NUM_ITERATIONS_PREHEATING; i++) {
accept(WIDTH, HEIGHT, data, primitive);
}
/*
* Benchmarking proper.
*/
double[] sumPrimitive = null;
double[] sumGeneric = null;
double aux = System.nanoTime();
for (int i = 0; i < NUM_ITERATIONS_BENCHMARKING; i++) {
sumGeneric = accept(WIDTH, HEIGHT, data, generic);
}
final double timeGeneric = System.nanoTime() - aux;
aux = System.nanoTime();
for (int i = 0; i < NUM_ITERATIONS_BENCHMARKING; i++) {
sumPrimitive = accept(WIDTH, HEIGHT, data, primitive);
}
final double timePrimitive = System.nanoTime() - aux;
System.out.println("prim = " + timePrimitive);
System.out.println("generic = " + timeGeneric);
System.out.println("generic / primitive = "
+ (timeGeneric / timePrimitive));
}
}
我知道 JIT 非常聪明，所以当两位访问者的表现同样出色时，我并没有太惊讶。更令人惊讶的是，通用访问者的执行速度似乎比原始访问者稍快，这是出乎意料的。我知道基准测试有时会很困难，所以我一定做错了什么。你能发现错误吗？
非常感谢你的帮助！！！塞巴斯蒂安
[编辑]我更新了代码以说明预热阶段（为了让 JIT 编译器完成它的工作）。这不会改变始终低于 1 (0.95 - 0.98) 的结果。