android 中毛玻璃效果的实现
最近在做一款叫叽叽的App(男银懂的),其中有一个功能需要对图片处理实现毛玻璃的特效
进过一番预研,找到了3中实现方案,其中各有优缺点:
1、如果系统的api在16以上,可以使用系统提供的方法直接处理图片
if (VERSION.SDK_INT > 16) {
Bitmap bitmap = sentBitmap.copy(sentBitmap.getConfig(), true);
final RenderScript rs = RenderScript.create(context);
final Allocation input = Allocation.createFromBitmap(rs, sentBitmap, Allocation.MipmapControl.MIPMAP_NONE,
Allocation.USAGE_SCRIPT);
final Allocation output = Allocation.createTyped(rs, input.getType());
final ScriptIntrinsicBlur script = ScriptIntrinsicBlur.create(rs, Element.U8_4(rs));
script.setRadius(radius /* e.g. 3.f */);
script.setInput(input);
script.forEach(output);
output.copyTo(bitmap);
return bitmap;
}
2、 如果Api条件不满足,可以使用如下方法
@SuppressLint("NewApi")
public static Bitmap fastblur(Context context, Bitmap sentBitmap, int radius) {
Bitmap bitmap = sentBitmap.copy(sentBitmap.getConfig(), true);
if (radius < 1) {
return (null);
}
int w = bitmap.getWidth();
int h = bitmap.getHeight();
int[] pix = new int[w * h];
// Log.e("pix", w + " " + h + " " + pix.length);
bitmap.getPixels(pix, 0, w, 0, 0, w, h);
int wm = w - 1;
int hm = h - 1;
int wh = w * h;
int div = radius + radius + 1;
int r[] = new int[wh];
int g[] = new int[wh];
int b[] = new int[wh];
int rsum, gsum, bsum, x, y, i, p, yp, yi, yw;
int vmin[] = new int[Math.max(w, h)];
int divsum = (div + 1) >> 1;
divsum *= divsum;
int temp = 256 * divsum;
int dv[] = new int[temp];
for (i = 0; i < temp; i++) {
dv[i] = (i / divsum);
}
yw = yi = 0;
int[][] stack = new int[div][3];
int stackpointer;
int stackstart;
int[] sir;
int rbs;
int r1 = radius + 1;
int routsum, goutsum, boutsum;
int rinsum, ginsum, binsum;
for (y = 0; y < h; y++) {
rinsum = ginsum = binsum = routsum = goutsum = boutsum = rsum = gsum = bsum = 0;
for (i = -radius; i <= radius; i++) {
p = pix[yi + Math.min(wm, Math.max(i, 0))];
sir = stack[i + radius];
sir[0] = (p & 0xff0000) >> 16;
sir[1] = (p & 0x00ff00) >> 8;
sir[2] = (p & 0x0000ff);
rbs = r1 - Math.abs(i);
rsum += sir[0] * rbs;
gsum += sir[1] * rbs;
bsum += sir[2] * rbs;
if (i > 0) {
rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];
} else {
routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];
}
}
stackpointer = radius;
for (x = 0; x < w; x++) {
r[yi] = dv[rsum];
g[yi] = dv[gsum];
b[yi] = dv[bsum];
rsum -= routsum;
gsum -= goutsum;
bsum -= boutsum;
stackstart = stackpointer - radius + div;
sir = stack[stackstart % div];
routsum -= sir[0];
goutsum -= sir[1];
boutsum -= sir[2];
if (y == 0) {
vmin[x] = Math.min(x + radius + 1, wm);
}
p = pix[yw + vmin[x]];
sir[0] = (p & 0xff0000) >> 16;
sir[1] = (p & 0x00ff00) >> 8;
sir[2] = (p & 0x0000ff);
rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];
rsum += rinsum;
gsum += ginsum;
bsum += binsum;
stackpointer = (stackpointer + 1) % div;
sir = stack[(stackpointer) % div];
routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];
rinsum -= sir[0];
ginsum -= sir[1];
binsum -= sir[2];
yi++;
}
yw += w;
}
for (x = 0; x < w; x++) {
rinsum = ginsum = binsum = routsum = goutsum = boutsum = rsum = gsum = bsum = 0;
yp = -radius * w;
for (i = -radius; i <= radius; i++) {
yi = Math.max(0, yp) + x;
sir = stack[i + radius];
sir[0] = r[yi];
sir[1] = g[yi];
sir[2] = b[yi];
rbs = r1 - Math.abs(i);
rsum += r[yi] * rbs;
gsum += g[yi] * rbs;
bsum += b[yi] * rbs;
if (i > 0) {
rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];
} else {
routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];
}
if (i < hm) {
yp += w;
}
}
yi = x;
stackpointer = radius;
for (y = 0; y < h; y++) {
// Preserve alpha channel: ( 0xff000000 & pix[yi] )
pix[yi] = (0xff000000 & pix[yi]) | (dv[rsum] << 16) | (dv[gsum] << 8) | dv[bsum];
rsum -= routsum;
gsum -= goutsum;
bsum -= boutsum;
stackstart = stackpointer - radius + div;
sir = stack[stackstart % div];
routsum -= sir[0];
goutsum -= sir[1];
boutsum -= sir[2];
if (x == 0) {
vmin[y] = Math.min(y + r1, hm) * w;
}
p = x + vmin[y];
sir[0] = r[p];
sir[1] = g[p];
sir[2] = b[p];
rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];
rsum += rinsum;
gsum += ginsum;
bsum += binsum;
stackpointer = (stackpointer + 1) % div;
sir = stack[stackpointer];
routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];
rinsum -= sir[0];
ginsum -= sir[1];
binsum -= sir[2];
yi += w;
}
}
// Log.e("pix", w + " " + h + " " + pix.length);
bitmap.setPixels(pix, 0, w, 0, 0, w, h);
return (bitmap);
}
3、以上方法都存在一个问题,性能较低,下面提供一个C实现
static int* StackBlur(int* pix, int w, int h, int radius) {
int wm = w - 1;
int hm = h - 1;
int wh = w * h;
int div = radius + radius + 1;
int *r = (int *)malloc(wh * sizeof(int));
int *g = (int *)malloc(wh * sizeof(int));
int *b = (int *)malloc(wh * sizeof(int));
int rsum, gsum, bsum, x, y, i, p, yp, yi, yw;
int *vmin = (int *)malloc(MAX(w,h) * sizeof(int));
int divsum = (div + 1) >> 1;
divsum *= divsum;
int *dv = (int *)malloc(256 * divsum * sizeof(int));
for (i = 0; i < 256 * divsum; i++) {
dv[i] = (i / divsum);
}
yw = yi = 0;
int(*stack)[3] = (int(*)[3])malloc(div * 3 * sizeof(int));
int stackpointer;
int stackstart;
int *sir;
int rbs;
int r1 = radius + 1;
int routsum, goutsum, boutsum;
int rinsum, ginsum, binsum;
for (y = 0; y < h; y++) {
rinsum = ginsum = binsum = routsum = goutsum = boutsum = rsum = gsum = bsum = 0;
for (i = -radius; i <= radius; i++) {
p = pix[yi + (MIN(wm, MAX(i, 0)))];
sir = stack[i + radius];
sir[0] = (p & 0xff0000) >> 16;
sir[1] = (p & 0x00ff00) >> 8;
sir[2] = (p & 0x0000ff);
rbs = r1 - ABS(i);
rsum += sir[0] * rbs;
gsum += sir[1] * rbs;
bsum += sir[2] * rbs;
if (i > 0) {
rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];
}
else {
routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];
}
}
stackpointer = radius;
for (x = 0; x < w; x++) {
r[yi] = dv[rsum];
g[yi] = dv[gsum];
b[yi] = dv[bsum];
rsum -= routsum;
gsum -= goutsum;
bsum -= boutsum;
stackstart = stackpointer - radius + div;
sir = stack[stackstart % div];
routsum -= sir[0];
goutsum -= sir[1];
boutsum -= sir[2];
if (y == 0) {
vmin[x] = MIN(x + radius + 1, wm);
}
p = pix[yw + vmin[x]];
sir[0] = (p & 0xff0000) >> 16;
sir[1] = (p & 0x00ff00) >> 8;
sir[2] = (p & 0x0000ff);
rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];
rsum += rinsum;
gsum += ginsum;
bsum += binsum;
stackpointer = (stackpointer + 1) % div;
sir = stack[(stackpointer) % div];
routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];
rinsum -= sir[0];
ginsum -= sir[1];
binsum -= sir[2];
yi++;
}
yw += w;
}
for (x = 0; x < w; x++) {
rinsum = ginsum = binsum = routsum = goutsum = boutsum = rsum = gsum = bsum = 0;
yp = -radius * w;
for (i = -radius; i <= radius; i++) {
yi = MAX(0, yp) + x;
sir = stack[i + radius];
sir[0] = r[yi];
sir[1] = g[yi];
sir[2] = b[yi];
rbs = r1 - ABS(i);
rsum += r[yi] * rbs;
gsum += g[yi] * rbs;
bsum += b[yi] * rbs;
if (i > 0) {
rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];
}
else {
routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];
}
if (i < hm) {
yp += w;
}
}
yi = x;
stackpointer = radius;
for (y = 0; y < h; y++) {
// Preserve alpha channel: ( 0xff000000 & pix[yi] )
pix[yi] = (0xff000000 & pix[yi]) | (dv[rsum] << 16) | (dv[gsum] << 8) | dv[bsum];
rsum -= routsum;
gsum -= goutsum;
bsum -= boutsum;
stackstart = stackpointer - radius + div;
sir = stack[stackstart % div];
routsum -= sir[0];
goutsum -= sir[1];
boutsum -= sir[2];
if (x == 0) {
vmin[y] = MIN(y + r1, hm) * w;
}
p = x + vmin[y];
sir[0] = r[p];
sir[1] = g[p];
sir[2] = b[p];
rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];
rsum += rinsum;
gsum += ginsum;
bsum += binsum;
stackpointer = (stackpointer + 1) % div;
sir = stack[stackpointer];
routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];
rinsum -= sir[0];
ginsum -= sir[1];
binsum -= sir[2];
yi += w;
}
}
free(r);
free(g);
free(b);
free(vmin);
free(dv);
free(stack);
return(pix);
}
android 中毛玻璃效果的实现的更多相关文章
- CSS3中毛玻璃效果的使用方法
今天在使用icloud的时候看到苹果icloud官网的毛玻璃效果非常赞,仔细研究了一下它的实现方式,是使用js配合background-image: -webkit-canvas的形式绘制出的毛玻璃背 ...
- android 开发 - 对图片进行虚化(毛玻璃效果,模糊)
概述 IPAD,IPHONE上首页背景的模糊效果是不是很好看,那么在 Android中如何实现呢.我通过一种方式实现了这样的效果. 开源库名称:anroid-image-blur 一个android ...
- Android如何实现毛玻璃效果之Android高级模糊技术
自从iOS系统引入了Blur效果,也就是所谓的毛玻璃.模糊化效果,磨砂效果,各大系统就开始竞相模仿,这是怎样的一个效果呢,我们先来看一下,如下面的图片: 效果我们知道了,如何在Android中实现呢, ...
- Android怎样实现毛玻璃效果之Android高级模糊技术
自从iOS系统引入了Blur效果,也就是所谓的毛玻璃.模糊化效果.磨砂效果.各大系统就開始竞相模仿,这是如何的一个效果呢,我们先来看一下,如以下的图片: 效果我们知道了,怎样在Android中实现呢. ...
- Android中使用ViewPager实现屏幕页面切换和页面切换效果
之前关于如何实现屏幕页面切换,写过一篇博文<Android中使用ViewFlipper实现屏幕切换>,相比ViewFlipper,ViewPager更适用复杂的视图切换,而且Viewpag ...
- Android中使用ImageViewSwitcher实现图片切换轮播导航效果
前面写过了使用ViewFlipper和ViewPager实现屏幕中视图切换的效果(ViewPager未实现轮播)附链接: Android中使用ViewFlipper实现屏幕切换 Android中使用V ...
- Android应用开发中半透明效果实现方案
下面是自定义Activity半透明的效果例子:res/values/styles.xml<resources> <stylename="Transparent " ...
- Android中xml设置Animation动画效果详解
在 Android 中, Animation 动画效果的实现可以通过两种方式进行实现,一种是 tweened animation 渐变动画,另一种是 frame by frame animation ...
- 【Android 界面效果19】Android中shape的使用
Android中常常使用shape来定义控件的一些显示属性,今天看了一些shape的使用,对shape有了大体的了解,稍作总结: 先看下面的代码: <shape> ...
随机推荐
- oracle 如何查看oracle数据库版本
select * from v$version 写在最后 哪位大佬如若发现文章存在纰漏之处或需要补充更多内容,欢迎留言!!! 相关推荐: 个人主页 oracle专题
- java第七节 IO
/* *第七讲 IO/输入与输出 * * File类 * RandomAccessFile类 * 各种节点流类 * 字符编码 * 各种过滤流与包装类 * *File类 * File类是IO包中唯一代表 ...
- Light OJ 1406 Assassin`s Creed 状态压缩DP+强连通缩点+最小路径覆盖
题目来源:Light OJ 1406 Assassin`s Creed 题意:有向图 派出最少的人经过全部的城市 而且每一个人不能走别人走过的地方 思路:最少的的人能够走全然图 明显是最小路径覆盖问题 ...
- ContentResolver.query()—>buildQueryString()
Cursor cursor = context.getContentResolver().query(Sms.CONTENT_URI, new String[]{"threa ...
- ios中布局(推荐一)
- (void)viewDidLoad { [super viewDidLoad]; NSArray *data=@[@"标题一",@"标题二",@" ...
- [转载]Ubuntu安装配置 git 服务器和客户端
原文地址:Ubuntu安装配置 git 服务器和客户端作者:ding404 1.配置前准备 服务器:安装ssh server另外还装了gitosis做git的权限管理 sudo apt-get ins ...
- ROS学习(十二)—— 编写简单的消息发布器和订阅器(C++)
一.创建发布器节点 1 节点功能: 不断的在ROS网络中广播消息 2 创建节点 (1)打开工作空间目录 cd ~/catkin_ws/src/beginner_tutorials 创建一个发布器节点( ...
- React(0.13) 定义一个checked组件
<!DOCTYPE html> <html> <head> <title>React JS</title> <script src=& ...
- Mac XMPP Openfire 服务器配置
前言 Openfire 是免费的.开源的.基于可拓展通讯和表示协议(XMPP).采用 Java 编程语言开发的实时协作服务器.Openfire 安装和使用都非常简单,并利用 Web 进行管理.单台服务 ...
- golang 学习 ---- channel
把一个loop放在一个goroutine里跑,我们可以使用关键字go来定义并启动一个goroutine: package main import "fmt" func loop() ...