就对象大小而言，Dalvik是否比HotSpot更需要内存？
2022-09-01 17:48:58
我一直在想一个对象在Android上占用了多少内存。有许多与HotSpot JVM相关的资源（像这样）告诉一个空对象需要8个字节，一个空数组需要12个字节，并且所有对象都与8字节边界对齐。因此，没有额外字段的对象应该占用8个字节，最小的对象至少有一个额外的字段 - 16个字节，一个空数组 - 16个字节，对吧？
在这个问题上，我没有找到关于Dalvik的具体信息，并决定通过测试来弄清楚。运行测试的结果令人惊讶。
关于计算方法的几句话。Android的Object.hashCode（）实现只是返回指向投射到int.的对象的指针（看起来很明显和通用，但事实证明，[另一个惊喜]，它不在HotSpot JVM上 - 例如，使用HotSpot运行MemTest并查看）。因此，我使用Dalvik上hashCode（）的简单性来计算Android上的对象大小，方法是连续分配两个测试类的实例，并且分配的空间量应该等于它们的hashCode（）值的差异（假设Dalvik在完全随机的地址上分配这些实例没有多大意义）。只是为了确保我在每个测试类的行中始终分配4个对象，这总是提供相同的hashCode（）差异。因此，我相信该方法的正确性几乎没有疑问。
以下是测试的源代码：
public class MemTest {
public static void run() {
Object o1 = new Object();
Object o2 = new Object();
Object o3 = new Object();
Object o4 = new Object();
EmptyObject eo1 = new EmptyObject();
EmptyObject eo2 = new EmptyObject();
EmptyObject eo3 = new EmptyObject();
EmptyObject eo4 = new EmptyObject();
ObjectWithBoolean ob1 = new ObjectWithBoolean();
ObjectWithBoolean ob2 = new ObjectWithBoolean();
ObjectWithBoolean ob3 = new ObjectWithBoolean();
ObjectWithBoolean ob4 = new ObjectWithBoolean();
ObjectWithBooleanAndInt obi1 = new ObjectWithBooleanAndInt();
ObjectWithBooleanAndInt obi2 = new ObjectWithBooleanAndInt();
ObjectWithBooleanAndInt obi3 = new ObjectWithBooleanAndInt();
ObjectWithBooleanAndInt obi4 = new ObjectWithBooleanAndInt();
ObjectWithLong ol1 = new ObjectWithLong();
ObjectWithLong ol2 = new ObjectWithLong();
ObjectWithLong ol3 = new ObjectWithLong();
ObjectWithLong ol4 = new ObjectWithLong();
ObjectWith4Ints o4i1 = new ObjectWith4Ints();
ObjectWith4Ints o4i2 = new ObjectWith4Ints();
ObjectWith4Ints o4i3 = new ObjectWith4Ints();
ObjectWith4Ints o4i4 = new ObjectWith4Ints();
ObjectWith4IntsAndByte o4ib1 = new ObjectWith4IntsAndByte();
ObjectWith4IntsAndByte o4ib2 = new ObjectWith4IntsAndByte();
ObjectWith4IntsAndByte o4ib3 = new ObjectWith4IntsAndByte();
ObjectWith4IntsAndByte o4ib4 = new ObjectWith4IntsAndByte();
ObjectWith5Ints o5i1 = new ObjectWith5Ints();
ObjectWith5Ints o5i2 = new ObjectWith5Ints();
ObjectWith5Ints o5i3 = new ObjectWith5Ints();
ObjectWith5Ints o5i4 = new ObjectWith5Ints();
ObjectWithArrayRef oar1 = new ObjectWithArrayRef();
ObjectWithArrayRef oar2 = new ObjectWithArrayRef();
ObjectWithArrayRef oar3 = new ObjectWithArrayRef();
ObjectWithArrayRef oar4 = new ObjectWithArrayRef();
byte[] a0b1 = new byte[0];
byte[] a0b2 = new byte[0];
byte[] a0b3 = new byte[0];
byte[] a0b4 = new byte[0];
byte[] a1b1 = new byte[1];
byte[] a1b2 = new byte[1];
byte[] a1b3 = new byte[1];
byte[] a1b4 = new byte[1];
byte[] a5b1 = new byte[5];
byte[] a5b2 = new byte[5];
byte[] a5b3 = new byte[5];
byte[] a5b4 = new byte[5];
byte[] a9b1 = new byte[9];
byte[] a9b2 = new byte[9];
byte[] a9b3 = new byte[9];
byte[] a9b4 = new byte[9];
byte[] a12b1 = new byte[12];
byte[] a12b2 = new byte[12];
byte[] a12b3 = new byte[12];
byte[] a12b4 = new byte[12];
byte[] a13b1 = new byte[13];
byte[] a13b2 = new byte[13];
byte[] a13b3 = new byte[13];
byte[] a13b4 = new byte[13];
print("java.lang.Object", o1, o2, o3, o4);
print("Empty object", eo1, eo2, eo3, eo4);
print("Object with boolean", ob1, ob2, ob3, ob4);
print("Object with boolean and int", obi1, obi2, obi3, obi4);
print("Object with long", ol1, ol2, ol3, ol4);
print("Object with 4 ints", o4i1, o4i2, o4i3, o4i4);
print("Object with 4 ints and byte", o4ib1, o4ib2, o4ib3, o4ib4);
print("Object with 5 ints", o5i1, o5i2, o5i3, o5i4);
print("Object with array ref", new Object[]{oar1, oar2, oar3, oar4});
print("new byte[0]", a0b1, a0b2, a0b3, a0b4);
print("new byte[1]", a1b1, a1b2, a1b3, a1b4);
print("new byte[5]", a5b1, a5b2, a5b3, a5b4);
print("new byte[9]", a9b1, a9b2, a9b3, a9b4);
print("new byte[12]", a12b1, a12b2, a12b3, a12b4);
print("new byte[13]", a13b1, a13b2, a13b3, a13b4);
}
static void print(String title, Object... objects) {
StringBuilder buf = new StringBuilder(title).append(":");
int prevHash = objects[0].hashCode();
int prevDiff = -1;
for (int i = 1; i < objects.length; i++) {
int hash = objects[i].hashCode();
int diff = Math.abs(hash - prevHash);
if (prevDiff == -1 || prevDiff != diff) {
buf.append(' ').append(diff);
}
prevDiff = diff;
prevHash = hash;
}
System.out.println(buf.toString());
}
/******** Test classes ******/
public static class EmptyObject {
}
public static class ObjectWith4Ints {
int i1;
int i2;
int i3;
int i4;
}
public static class ObjectWith4IntsAndByte {
int i1;
int i2;
int i3;
int i4;
byte b;
}
public static class ObjectWith5Ints {
int i1;
int i2;
int i3;
int i4;
int i5;
}
public static class ObjectWithArrayRef {
byte[] b;
}
public static class ObjectWithBoolean {
boolean b;
}
public static class ObjectWithBooleanAndInt {
boolean b;
int i;
}
public static class ObjectWithLong {
long l;
}
}
以下是结果：
java.lang.Object: 16
Empty object: 16
Object with boolean: 16
Object with boolean and int: 24
Object with long: 24
Object with 4 ints: 32
Object with 4 ints and byte: 32
Object with 5 ints: 32
Object with array ref: 16
new byte[0]: 24
new byte[1]: 24
new byte[5]: 32
new byte[9]: 32
new byte[12]: 32
new byte[13]: 40
总结结果：
8字节边界对齐与HotSpot相同，这是唯一相同的东西。
普通对象至少为 16 个字节（热点上为 8 个字节）
显然，一个空对象本身占用了12个字节（而在HotSpot上为8个字节），并且有4个额外字节的空间，直到对象大小从16个字节“跳转到”下一个24字节的边界。
空数组的最小值为 24 个字节（而 HotSpot 上为 12 个字节）
类似地，数组本身占用 20 个字节（HotSpot 上为 12 个字节），并且有空间容纳 4 个额外的数组数据字节，直到对象大小从 24 个字节“跳转到”下一个边界 32 字节。
补充：（回应Louis的建议）另一个压力测试表明，即使分配一百万个对象实例，任何两个实例之间的距离也永远不会小于16字节。这证明了对象之间潜在的8字节孔肯定是进一步分配的死空间，否则当大约一半的内存被分配给对象时，dalvik肯定也应该将它们中的一些放入“孔”中，并且压力测试将返回8，而不是16。
public static void run2() {
int count = 1024 * 1024;
Object[] arr = new Object[count];
for (int i = 0; i < count; i++) {
arr[i] = new Object();
}
int[] hashes = new int[count];
for (int i = 0; i < count; i++) {
hashes[i] = arr[i].hashCode();
}
Arrays.sort(hashes);
int minDist = Integer.MAX_VALUE;
for (int i = 1; i < count; i++) {
int dist = Math.abs(hashes[i] - hashes[i - 1]);
if (dist < minDist) {
minDist = dist;
}
}
System.out.println("Allocated "+ count + " Objects, minimum distance is "+ minDist);
}
与 HotSpot 相比，Dalvik 的对象最多需要 8 个字节，数组需要 8-12 个字节，这是否正确？
