/* * 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 * [email protected] #keyAt(int)} and [email protected] #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 [email protected] #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 [email protected] #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 [email protected] #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' [email protected] indexOf} methods, * this method compares values using [email protected] ==} rather than [email protected] 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++; } /** * [email protected]} * * <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).)
原文地址：https://github.com/cheyiliu/All-in-One/wiki/SparseArray