Question

Android是否维护应用程序可绘制资源的内存缓存并重用它们，或者预先加载可动态分配给不同窗口小部件的所有可绘制内容是一种好习惯？

例如：

public static final int[] SETS = {
        R.drawable.set0, R.drawable.set1, R.drawable.set2,
        R.drawable.set3, R.drawable.set4, R.drawable.set5, R.drawable.set6,
        R.drawable.set7, R.drawable.set8, R.drawable.set9, R.drawable.set10};
public Drawable[] sets;

void init() {
    load(sets, SETS);
}

public void load(Drawable[] d, int[] ids) {
    for (int i = 0; i < ids.length; i++) {
        if (ids[i] == 0)
            d[i] = null;
        else
            d[i] = context.getResources().getDrawable(ids[i]);
    }
}

Answer 1

这闻起来像是不必要的预优化。但是，android确实缓存了drawables，因此您不必预先加载它们。来自ApplicationContext

的相关代码

  /*package*/ Drawable loadDrawable(TypedValue value, int id)
            throws NotFoundException {
        .
        .
        .

        final long key = (((long) value.assetCookie) << 32) | value.data;
        Drawable dr = getCachedDrawable(key);

        if (dr != null) {
            return dr;
        }

        .
        .
        .

        if (dr != null) {
            dr.setChangingConfigurations(value.changingConfigurations);
            cs = dr.getConstantState();
            if (cs != null) {
                if (mPreloading) {
                    sPreloadedDrawables.put(key, cs);
                } else {
                    synchronized (mTmpValue) {
                        //Log.i(TAG, "Saving cached drawable @ #" +
                        //        Integer.toHexString(key.intValue())
                        //        + " in " + this + ": " + cs);
                        mDrawableCache.put(key, new WeakReference<Drawable.ConstantState>(cs));
                    }
                }
            }
        }

        return dr;
    }

    private Drawable getCachedDrawable(long key) {
        synchronized (mTmpValue) {
            WeakReference<Drawable.ConstantState> wr = mDrawableCache.get(key);
            if (wr != null) {   // we have the key
                Drawable.ConstantState entry = wr.get();
                if (entry != null) {
                    //Log.i(TAG, "Returning cached drawable @ #" +
                    //        Integer.toHexString(((Integer)key).intValue())
                    //        + " in " + this + ": " + entry);
                    return entry.newDrawable(this);
                }
                else {  // our entry has been purged
                    mDrawableCache.delete(key);
                }
            }
        }
        return null;
    }

有没有理由从资源中预加载drawable？

1 个答案: