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SparseArray
cheyiliu edited this page Jan 22, 2015
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- 版本5.0 http://androidxref.com/5.0.0_r2/xref/frameworks/base/core/java/android/util/SparseArray.java
- 代码走起
/*
* Copyright (C) 2006 The Android Open Source Project
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
package android.util;
import com.android.internal.util.ArrayUtils;
import com.android.internal.util.GrowingArrayUtils;
import libcore.util.EmptyArray;
/**
* SparseArrays map integers to Objects. Unlike a normal array of Objects,
* there can be gaps in the indices. It is intended to be more memory efficient
* than using a HashMap to map Integers to Objects, both because it avoids
* auto-boxing keys and its data structure doesn't rely on an extra entry object
* for each mapping.
*
* SparseArray用于映射integers到object。但不像普通数组那样，sparseArray的元素间没有无用元素。
* 在映射integers到object的过程中，SparseArray由于采用避免自动装箱的keys和它的数据结构不依赖额外
* 的对象来存储映射关系的实现，因此它比hashMap的内存使用更高效一些。
*
*
* <p>Note that this container keeps its mappings in an array data structure,
* using a binary search to find keys. The implementation is not intended to be appropriate for
* data structures
* that may contain large numbers of items. It is generally slower than a traditional
* HashMap, since lookups require a binary search and adds and removes require inserting
* and deleting entries in the array. For containers holding up to hundreds of items,
* the performance difference is not significant, less than 50%.</p>
*
* 注意：SparseArray在查找keys的过程中采用了二分查找， 这种实现不适合数据量大的情况。由于查找时要用到二分查找，
* 添加删除时涉及到数组其他元素的挪动，因此通常SparseArray会比hashMap慢。当处理上百的数据量，这种性能差异不是特别
* 明显，性能差异不超过50%。
*
*
* <p>To help with performance, the container includes an optimization when removing
* keys: instead of compacting its array immediately, it leaves the removed entry marked
* as deleted. The entry can then be re-used for the same key, or compacted later in
* a single garbage collection step of all removed entries. This garbage collection will
* need to be performed at any time the array needs to be grown or the the map size or
* entry values are retrieved.</p>
*
* 为了优化性能，SparseArray针对remove case作了优化，remove时它不是立即挤压数组空间，而是标记为delete。
* 这个被标记的元素要么被重复利用，要么在多次remove之后通过一次gc操作中被挤压出去。
* gc需要在下列情况之前被执行：数组要扩容；get map size；get values；
*
*
* <p>It is possible to iterate over the items in this container using
* {@link #keyAt(int)} and {@link #valueAt(int)}. Iterating over the keys using
* <code>keyAt(int)</code> with ascending values of the index will return the
* keys in ascending order, or the values corresponding to the keys in ascending
* order in the case of <code>valueAt(int)</code>.</p>
*
* 可以用keyAt valueAt实现遍历
*/
public class SparseArray<E> implements Cloneable {
//巧妙的flag
//当有元素被remove delete时，先暂时设置该对象为DELETED，并置mGarbage=true
//用以提升性能，体现在哪里？问题1
private static final Object DELETED = new Object();
private boolean mGarbage = false;
//存储的数据结构，两个数组加一个size
//特殊在哪里？或者怎么用的呢？问题2
private int[] mKeys;
private Object[] mValues;
private int mSize;
/**
* Creates a new SparseArray containing no mappings.
*/
public SparseArray() {
this(10);//默认容量是10个元素
}
/**
* Creates a new SparseArray containing no mappings that will not
* require any additional memory allocation to store the specified
* number of mappings. If you supply an initial capacity of 0, the
* sparse array will be initialized with a light-weight representation
* not requiring any additional array allocations.
*/
public SparseArray(int initialCapacity) {
if (initialCapacity == 0) {
//看EmptyArray的实现便知，mKeys的初值等于new int[0], 其他同理
mKeys = EmptyArray.INT;
mValues = EmptyArray.OBJECT;
} else {
//newUnpaddedObjectArray有啥特性？问题3
mValues = ArrayUtils.newUnpaddedObjectArray(initialCapacity);
mKeys = new int[mValues.length];
}
mSize = 0;
}
@Override
@SuppressWarnings("unchecked")
public SparseArray<E> clone() {
SparseArray<E> clone = null;
try {
//java深拷贝
clone = (SparseArray<E>) super.clone();
clone.mKeys = mKeys.clone();
clone.mValues = mValues.clone();
} catch (CloneNotSupportedException cnse) {
/* ignore */
}
return clone;
}
/**
* Gets the Object mapped from the specified key, or <code>null</code>
* if no such mapping has been made.
*/
public E get(int key) {
return get(key, null);
}
/**
* Gets the Object mapped from the specified key, or the specified Object
* if no such mapping has been made.
*/
@SuppressWarnings("unchecked")
public E get(int key, E valueIfKeyNotFound) {
//二分查找, 返回值含义是什么？问题4
int i = ContainerHelpers.binarySearch(mKeys, mSize, key);
if (i < 0 || mValues[i] == DELETED) {
// 没找到对应的key，或者找到了Key，但该元素被标记为delete
// 则返回默认值
return valueIfKeyNotFound;
} else {
// i>0 且该位置的元素未被标记为待删除，返回该值
return (E) mValues[i];
}
}
/**
* Removes the mapping from the specified key, if there was any.
*/
public void delete(int key) {
//二分查找
int i = ContainerHelpers.binarySearch(mKeys, mSize, key);
//若找到了
if (i >= 0) {
//若之前未被标记delete
if (mValues[i] != DELETED) {
//标记为delete，garbage=true
mValues[i] = DELETED;
mGarbage = true;
}
}
}
/**
* Alias for {@link #delete(int)}.
*/
public void remove(int key) {
delete(key);
}
/**
* Removes the mapping at the specified index.
*/
public void removeAt(int index) {
//移除特定位置的元素，注意传入的是mValues的index不是Key
if (mValues[index] != DELETED) {
mValues[index] = DELETED;
mGarbage = true;
}
}
/**
* Remove a range of mappings as a batch.
*
* @param index Index to begin at
* @param size Number of mappings to remove
*/
public void removeAtRange(int index, int size) {
//确定结束位置
final int end = Math.min(mSize, index + size);
//从起点开始循环 remove
for (int i = index; i < end; i++) {
removeAt(i);
}
}
private void gc() {
// Log.e("SparseArray", "gc start with " + mSize);
//'挤'空间了， 提供空间利用率
int n = mSize;
int o = 0;
int[] keys = mKeys;
Object[] values = mValues;
//循环整个元素区间
/*
* 模拟一遍就明白了
*i = 0 1 2 3 4 5 6
*key = 1 2 6 8 9 10 11
*value= x y z D u D w
*value D代表delete的对象
*
*i=0时，val==x, val != DELETED，i==o, o++, o == 1
*i=1时，val==y, val != DELETED，i==o, o++, o == 2
*i=2时，val==z, val != DELETED，i==o, o++, o == 3
*i=3时，val==D, val == DELETED，o == 3
*i=4时，val==u, val != DELETED，i!=o, keys[3] = keys[4], values[3] = values[4], values[4] = null, o++, o == 4
* (i=4之后，key= 1 2 6 9 9 10 11, value= x y z u null D w)
*i=5时，val==D, val == DELETED，o == 4
* (i=5之后，key= 1 2 6 9 9 10 11, value= x y z u null D w)较上一步没变化
*i=6时，val==w, val != DELETED，i!=o, keys[4] = keys[6], values[4] = values[6], values[6] = null, o++, o == 5
* (i=6之后，key= 1 2 6 9 11 10 11, value= x y z u w D null)
*循环结束
*mSize = 5
*/
for (int i = 0; i < n; i++) {
Object val = values[i];
if (val != DELETED) {
if (i != o) {
keys[o] = keys[i];
values[o] = val;
values[i] = null;
}
o++;
}
}
mGarbage = false;
mSize = o;
// Log.e("SparseArray", "gc end with " + mSize);
}
/**
* Adds a mapping from the specified key to the specified value,
* replacing the previous mapping from the specified key if there
* was one.
*/
public void put(int key, E value) {
//先二分查找，确定插入位置，保证了key数组的有序性
int i = ContainerHelpers.binarySearch(mKeys, mSize, key);
if (i >= 0) {
//找到了，直接替换
mValues[i] = value;
} else {
//一点小技巧，跟二分查找的返回值有关
//没找到的情况下i的意义是 i = -insertPoint -1，比如i=-2，则insertPoint=1
//而-2在内存中存的是补码，对他取反刚好得insertPoint，跟上面计算结果一样，但位操作更高效
i = ~i;
//若i在size范围内，且刚好对应位置标记为delete了，直接放入
if (i < mSize && mValues[i] == DELETED) {
mKeys[i] = key;
mValues[i] = value;
return;
}
//若前面if不成立，即i超出了size范围，或者对应的位置的元素是有效的
if (mGarbage && mSize >= mKeys.length) {
//压缩空间
gc();
//重新查找插入点
// Search again because indices may have changed.
i = ~ContainerHelpers.binarySearch(mKeys, mSize, key);
}
//插入，若空间不够则会重新分配数组
mKeys = GrowingArrayUtils.insert(mKeys, mSize, i, key);
mValues = GrowingArrayUtils.insert(mValues, mSize, i, value);
mSize++;
}
}
/**
* Returns the number of key-value mappings that this SparseArray
* currently stores.
*/
public int size() {
if (mGarbage) {
gc();
}
return mSize;
}
/**
* Given an index in the range <code>0...size()-1</code>, returns
* the key from the <code>index</code>th key-value mapping that this
* SparseArray stores.
*
* <p>The keys corresponding to indices in ascending order are guaranteed to
* be in ascending order, e.g., <code>keyAt(0)</code> will return the
* smallest key and <code>keyAt(size()-1)</code> will return the largest
* key.</p>
*/
public int keyAt(int index) {
if (mGarbage) {
gc();
}
return mKeys[index];
}
/**
* Given an index in the range <code>0...size()-1</code>, returns
* the value from the <code>index</code>th key-value mapping that this
* SparseArray stores.
*
* <p>The values corresponding to indices in ascending order are guaranteed
* to be associated with keys in ascending order, e.g.,
* <code>valueAt(0)</code> will return the value associated with the
* smallest key and <code>valueAt(size()-1)</code> will return the value
* associated with the largest key.</p>
*/
@SuppressWarnings("unchecked")
public E valueAt(int index) {
if (mGarbage) {
gc();
}
return (E) mValues[index];
}
/**
* Given an index in the range <code>0...size()-1</code>, sets a new
* value for the <code>index</code>th key-value mapping that this
* SparseArray stores.
*/
public void setValueAt(int index, E value) {
if (mGarbage) {
gc();
}
mValues[index] = value;
}
/**
* Returns the index for which {@link #keyAt} would return the
* specified key, or a negative number if the specified
* key is not mapped.
*/
public int indexOfKey(int key) {
if (mGarbage) {
gc();
}
return ContainerHelpers.binarySearch(mKeys, mSize, key);
}
/**
* Returns an index for which {@link #valueAt} would return the
* specified key, or a negative number if no keys map to the
* specified value.
* <p>Beware that this is a linear search, unlike lookups by key,
* and that multiple keys can map to the same value and this will
* find only one of them.
* <p>Note also that unlike most collections' {@code indexOf} methods,
* this method compares values using {@code ==} rather than {@code equals}.
*/
public int indexOfValue(E value) {
if (mGarbage) {
gc();
}
for (int i = 0; i < mSize; i++)
if (mValues[i] == value)
return i;
return -1;
}
/**
* Removes all key-value mappings from this SparseArray.
*/
public void clear() {
int n = mSize;
Object[] values = mValues;
for (int i = 0; i < n; i++) {
//值空，利于jvm gc
values[i] = null;
}
mSize = 0;
mGarbage = false;
}
/**
* Puts a key/value pair into the array, optimizing for the case where
* the key is greater than all existing keys in the array.
*/
public void append(int key, E value) {
//若key小于等于已有的最大key，直接Put
if (mSize != 0 && key <= mKeys[mSize - 1]) {
put(key, value);
return;
}
if (mGarbage && mSize >= mKeys.length) {
gc();
}
//若key大于了现有的所有key，就不用走put的二分查找过程了，直接append
//即comments说的优化
mKeys = GrowingArrayUtils.append(mKeys, mSize, key);
mValues = GrowingArrayUtils.append(mValues, mSize, value);
mSize++;
}
/**
* {@inheritDoc}
*
* <p>This implementation composes a string by iterating over its mappings. If
* this map contains itself as a value, the string "(this Map)"
* will appear in its place.
*/
@Override
public String toString() {
if (size() <= 0) {
return "{}";
}
StringBuilder buffer = new StringBuilder(mSize * 28);
buffer.append('{');
for (int i=0; i<mSize; i++) {
if (i > 0) {
buffer.append(", ");
}
int key = keyAt(i);
buffer.append(key);
buffer.append('=');
Object value = valueAt(i);
if (value != this) {
buffer.append(value);
} else {
buffer.append("(this Map)");
}
}
buffer.append('}');
return buffer.toString();
}
}
问题1，mGarbage标记结合DELETED，用以提升性能，体现在哪里？
将可能的多次gc操作变为一次完成。比如用户调用了多次removeAt操作，若不用这种策略，则会每次remove都对应一次gc即用一次循环来'挤压'空间。
采用这种flag优化后，多次remove仅多次设置了标志，在gc触发时，仅需要一次循环就可以将空间'挤压'好。
问题2，sparseArray的存储结构，两个数组加一个size有的特性？
特性一，key数组是有序的，key在数组的index和vale在数组的index一样
特性二，sparseArray每次gc过程，保证了他的有效值(size)区间内没有无效值，或者说'无缝隙'，有效值是连续的
特性三，如文档所说，当map Integers to Objects时，相对hashMap，sparseArray被设计成更加的内存高效，
sparseArray避免了自动装箱机制和用额外的对象来存储每个映射。
问题3，ArrayUtils.newUnpaddedObjectArray(initialCapacity);有什么特性？
最终调用的是VMRuntime.getRuntime().newUnpaddedArray(Object.class, minLen);
参考 http://androidxref.com/5.0.0_r2/xref/libcore/libart/src/main/java/dalvik/system/VMRuntime.java#260
问题4, 二分查找的返回值含义，若值存在则返回该值的Index，否则返回和该值的插入点相关的一个值(-(insertion point) - 1)
(the index of the search key, if it is contained in the list; otherwise, (-(insertion point) - 1).)
- SparseArray的特性和一些内部实现见上面代码中的comments
- SparseArray get put都用了二分查找，时间复杂的是O(log n), vs HashMap的get put的时间负责度是O(1)
- SparseArray remove delete put时都有空间挪动的开销
- SparseArray相对于hashMap优化在哪里？避免了AutoBox和不用额外的空间来存映射关系，故它的唯一有点就是针对小数据量时的空间效率更高，尤其适合手机这种设备
- 在开发android时，若写这句话
HashMap<Integer, String> hashMap = new HashMap<Integer, String>();时，IDE会警告Use new SparseArray<String>(...) instead for better performance你得视你的数据量情况而定了
Just build something.