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Serializing an Image

LukeW — Thu, 11 Sep 2008 09:09:00 GMT

Serializing an Image

To create a byte-array of data from an image, we can use the getRGB(..) method in the image class in MIDP 2.0. From the getRGB method we get an int-array of data containing all the ARGB values of each pixel in the image. Integers in java are four bytes, so we need to split each int value into four byte values.

To mask out each of the bytes in the int we can use the 'AND &' operator and the 'shift-right >>' operator. Here's an example:

int ARGB = 0xFFFFFFFF;  // AARRGGBB 

int a = (ARGB & 0xFF000000); 

int r = (ARGB & 0x00FF0000); 

int g = (ARGB & 0x0000FF00); 

int b = (ARGB & 0x000000FF);

// Here we move each bit to the right.

a = (a >> 24); // a = 0x000000FF 

r = (r >> 16); // r = 0x000000FF 

g = (g >> 8); // g = 0x000000FF

b = b;        // b = 0x000000FF

When we convert the integer to a byte, there are some problems since there are only signed bytes in Java. A byte may contain values between –128 and 127 and from our integer we'll have a value between 0 and 255.

Here are some conversion examples:

Integer val 127 = byte val 127. 

Integer val 128 = byte val –128. 

Integer val 255 = byte val –1.

byte ba = (byte)(a);  // if a=0x000000FF (255), then ba = -1

byte br = (byte)(r); 

byte bg = (byte)(g); 

byte bb = (byte)(b);

So, when we convert the byte back to an integer we have to check if the byte value is less than zero.

Int a, r, g, b;

If(ba<0)

     a = 256 – a;

Now our new integer value will be between 0 and 255 and we just have to use the 'shift-left <<' operator and add the values together to get our new int-array.

a = (a << 24);

r = (r << 16);

g = (g << 8);

b = (b);

  0xFF000000 (a)

+ 0x00FF0000 (r)

+ 0x0000FF00 (g)

+ 0x000000FF (b)

= 0xFFFFFFFF (argb)

int ARGB = a+r+g+b;

After the integer-array is recreated we can use the createRGBImage(..) method in the Image class to create our image.

Be aware that the byte-array is quite large as it contains all of the ARGB values for each pixel. If an image is 100*60 px, where each pixel is four bytes, the byte array will be 24kb.

ImageBytearryConvert.rar

LukeW 2008-09-11 17:09 发表评论

J2me的List�ȝ��

LukeW — Fri, 15 Aug 2008 07:20:00 GMT

List控�g是��用频率非帔R��的显�C�控件之一了。但是最�q�发现它的一点不��?br /> 那就是getSelectedIndex()函数�Q�一般情况下它都能正常工作。唯有在List处于复选模式（MULTIPLE�Q�时�Q��用该函数无法获得当前高亮条选中索引�Q�而��L��q�回-1�?br />
参考List控�g的源代码�Q�可以发玎ͼ�List�c�d��C��Choice接口�Q��ƈ且包含一个ChoiceGroup成员。它才是实现List大部分功能的大功臣�?br /> List�cȝ��getSelectedIndex()�Ҏ��Q�实际上��是ChoiceGroup的getSelectedIndex()�Ҏ��。大家可以参考Doc中关于ChoiceGroup的getSelectedIndex()�Ҏ��的如下部分�?br /> “For ChoiceGroup objects of type MULTIPLE, this always returns -1 because no single value can in general represent the state of such a ChoiceGroup.”

��是��_��List在多选模式（MULTIPLE�Q�下�Q�我们是无法获取当前高亮条所在项的烦引值的�?br /> 当然�Q�你可以自己数数�?br />
也许你觉得这个问题不��严重，也许吧。但我觉得对于程序员来说�Q�最严重的问题就是理解发生偏差�?br /> 比如我，在发现这个问题之前。凭借多�q�的�~�程�l�验�Q�我很确定很��定的认为，�q�个getSelectedIndex()永远能够获得当前光标所指项的烦引倹{�?br /> ……当错误来临时�Q�我百思不得其解。最后，�׃��好些旉��去调试才发现�Q�那个值�ȝ��?1。操�Q�花了太多时间去猜测原本正确的代码�?br /> 我只惛_��诉大�Ӟ��真正耗费旉��最多，让�h最恶心的错误，往往��是�q�样的问题。再操！

那么有什么简单的解决办法么？
我可以很负责�ȝ��告诉你，自己写一个多选的List控�g吧，��C��不要让getSelectedIndex()总返�?1。即使是多选，有时候也是需要这个值的�?br />
也许你想�l�承List,然后重蝲getSelectedIndex()�Ҏ��。但你无法重载List的keyPressed�Ҏ��Q�其实List�{�Screen控�g都用Canvas写的�Q��?br /> ……或许�q�能惛_��别的什么好办法。不�q�我的选择是自己写一个List控�g代替它（如果对它感兴��，请回复，改天我在弄出来吧�Q��?br />
至于List的其它功能，��没什么重复的必要的，看看文档吧�?

LukeW 2008-08-15 15:20 发表评论

Vector �ȝ��

LukeW — Fri, 15 Aug 2008 07:19:00 GMT

Vector是在java�~�程中比较常用的动态数�l�。一直以为它是个数组的链表，当内存不够用了，��新甌��一个capacityIncrement大小的数�l�，�q�到原来的链表上�?br /> 在仔�l�阅��L��代码后发玎ͼ�Vector�q�没有�Q何链表的性质。它是一个纯�_�的数组。当内存不够用时�Q�就重新初始化一个容量较大新数组�Q�然后��用System.arraycopy()函数��原有的数组copy到新的数�l�当中�?br />
System.arraycopy()是一个由�pȝ��q�_��来实现的函数�Q�这��L��pȝ��调用性能是比较高的�?br /> 即��如此�Q�我们在写程序时�Q�注意initialCapacity�Q�初始容量）和capacityIncrement�Q�增量）的设�|�，��会有效的减��重新定义数�l��ƈ且拷贝数�l�的�ơ数�?br />
例如�Q?br />

Vector v = new Vector(10, 10);//初始定w��?0�Q�增�?0
Integer[] ints = new Integer11;
for (int i = 0; i < ints.length; i++) {
ints = new Integer(i);
v.addElement(ints);
}

在这里，当v��d��W?1个Integer的时候，v��׃��自动创徏一个长度�ؓ20的数�l�（当前定w��+增量�Q�，以后每次装满数组�Q�都会重新按增量�q�加数组长度�?br /> 所以，讄��一个合适的初始定w��和增量，��会提高Vector的效率。千万别像我一��P��把它当做链表。因为链表在增长�I�间时是不会影响��C��前��用的�I�间和数据的�?br />
我想指出一点，因�ؓVector中存储的�Q�实际上都是Object变量�Q�大家可以把它理解�ؓ指针。重新初始化Vector内置的数�l��ƈ且拷贝，相当于对一个指针数�l�进行操作，�q�不�{�同于对输入�c�d��的重新分配内存�?br /> ��是说Vector的重新分配，不论�q�行多少�ơ都不涉及到Integer对象的创建，拯��{�工作。它只是重新创徏�q�拷�?#8220;指针”数组�Q�也��是Object数组�?br /> 此外�Q�java中有一个容易被忽视的基本概��c��当某个对象再没有指向该对象的引用时�Q�垃圑֛�收器才会自动��其自动释放。不��心使用�Q�很�Ҏ��l�程序造成内存问题�?br /> 例如�Q?br />

Vector v = new Vector(10, 10);//初始定w��?0�Q�增�?0
Integer[] ints = new Integer11;
for (int i = 0; i < ints.length; i++) {
ints = new Integer(i);
v.addElement(ints);
}
ints = null;//对它的释攑ְ�造成数组元素的释放。但是由于v中还保存有元素的引用�Q�所以这些Integer对象�q�不会被回收�?/span>

我强调这个问题，是因为曾�l�写�q�一�D�代码：
Image img = Image.createImage(”/xxx.png”);
Image img2 = img;
�q�张囄��非常大，在��用完img之后�Q�我释放掉它�Q�然后重新申请另一个图片。于是，内存爆了。因为img2仍然持有xxx.png的引用，所以无法释放�?br /> Vectro中存储的也是引用�Q�所以在使用时应该更加注意编�E�规范，以免发生�c�M��的问题。其实应该尽量避免��用多个引用�?br />
上面说了�q�些题外话，正是惌��告各位程序员�Q�Vector是一个可变的Object数组�Q�一个引用数�l�。所以请大家使用时要��心�Q�别惛_��然的以�ؓ它是一个容器，它里面存储的可不是对象，而是引用�?br /> 而且�Q�在释放曄��加入到vector的对象时�Q�对象本�w��ƈ不会被真正释放，得到回收。只是原有的引用无法再��用�Ş了�?br />
最后还有一�Ҏ��巧�?br /> 应该��量使用索引获取对象�Q�避免��用IndexOf()�Ҏ��。把它当做堆栈来使用时更应该注意�Q�要避免使用insertElementAt()�Ҏ��?br /> 此外�Q�j2me中的Stack�c�L��Z��Vector实现的，使用时也要留心�?/em>

LukeW 2008-08-15 15:19 发表评论

LukeW — Fri, 15 Aug 2008 07:15:00 GMT

摘要: 在网上参考了一些汉字�{换到拼音的资料。思�\应该只有以下两种�?1�Q�查表法。这样做需要一个庞大的映射表，在j2me环境下不大合适。不�q�效果好�Q�有些还支持多音字�?2�Q��用GB字库的映��关�p�R��因为GB2312及其扩展GBK的汉字编码都�Ҏ��Z��于拼韛_��在映��关�p�R�?实际上网�l�上的大部分文章都是�Ҏ��W�二�U�方法来实现的�? 我也是采用这�U�方法，因�ؓ它基本上可以利用GB2312字库�Q�直接映��成拼音�?.. 阅读全文

LukeW 2008-08-15 15:15 发表评论

GB2312转换Unicode

LukeW — Fri, 15 Aug 2008 06:58:00 GMT

之前的文章介�l�了在j2me环境下GB2312转换为UTF-8的方法�?br /> 后来�l�箋对编码及char�c�d��q�行学习�Q�发��C��些有��的问题�?br /> 首先java环境下的char�c�d��变量�Q�实际上��是以unicode方式存储的�?br /> 所以以下方法有效：

输入unicode�~�码的byte数组�Q�即可两两拼接成一个char�?br /> 而String�c�d��实际上就是在char数组的基��上衍生出来的。大家可以参考cldc的源代码�?br />

public static String read_Uni(byte[] word_unicode) {
        StringBuffer stringbuffer = new StringBuffer("");
        for (int j = 0; j < word_unicode.length;) {
            int l = word_unicode[j++];
            int h = word_unicode[j++];
            char c = (char) ((l & 0xff) | ((h << 8) & 0xff00));
            stringbuffer.append(c);
        }
        return stringbuffer.toString();
    }

j2me环境下也是如此�?br /> 所以在�W�一�ơ给出的转换�c�M��Q�提供的gb2312到utf-8直接转换的快速方法。现在看来是画蛇添��了�?br />
�Ҏ��以上�l�验�Q�更新�{换类如下�Q?br />

public class HGB2312 {

        private byte[] map = new byte[15228];

        public HGB2312() throws Exception {
            InputStream is = getClass().getResourceAsStream("/gb2u.dat");
            is.read(map);
            is.close();
        }

        public String gb2utf8(byte[] gb) {
            StringBuffer sb = new StringBuffer();
            int c, h, l, ind;
            for (int i = 0; i < gb.length;) {
                if (gb[i] >= 0) {
                    sb.append((char) gb[i++]);
                } else {
                    h = 256 + gb[i++];
                    l = 256 + gb[i++];
                    h = h - 0xA0 - 1;
                    l = l - 0xA0 - 1;
                    if (h < 9) {
                        ind = (h * 94 + l) << 1;
                        c = (byte2Int(map[ind]) << 8 | byte2Int(map[ind + 1]));
                        sb.append((char) c);
                    } else if (h >= 9 && h <= 14) {
                        sb.append((char) 0);
                    } else if (h > 14) {
                        h -= 6;
                        ind = (h * 94 + l) << 1;
                        c = (byte2Int(map[ind]) << 8 | byte2Int(map[ind + 1]));
                        sb.append((char) c);

                    } else {
                        sb.append((char) 0);
                    }
                }
            }
            return sb.toString();
        }

        private int byte2Int(byte b) {
            if (b < 0) {
                return 256 + b;
            } else {
                return b;
            }
        }
    }

�q�个�Ҏ��明显要比�W�一�ơ快很多了，直接查表�Q�然后拼接成String�Q�不需要�{换成utf-8�~�码�?br />
数据文�g请在http://download.csdn.net/source/263609获取

��M��Q�java中的char�c�d��实际上是存储了unicode�~�码。目前在nokia 5300上测试通过�?br /> 我觉得其它机器也应该是这栗��如果哪位大侠知道这斚w��的资料，误��教�?

LukeW 2008-08-15 14:58 发表评论

MIDP2.0及MIDP数字�{�֐�

LukeW — Fri, 15 Aug 2008 06:53:00 GMT

本文档是 WoTrust �Ҏ�� Forum Nokia 提供的技术文档《MIDP 2.0: Tutorial On Signed MIDlets》翻译整理的�Q�请同时参考此英文原文文档。请用户在编�?MIDlet 和签�?MIdlet 之前阅读此文档，以便�?MIDP2.0 的安全机制有一个深�ȝ��理解�Q�有助于用户能用�?MIDlet 代码�{�֐�证书�?br />
一、概�q?

MIDP2.0 采用了全新的安全机制�Q�这对于需要调用一个敏感的(重要�?函数�?API �?MIDlet 开发者来讲是必须了解的，如：�|�络�q�接 API 、消�?API 和推 (Push) 函数�{�，�q�有一些可选的 MIDP 包也有许多受限制�?API �?

虽然购买代码�{�֐�证书需要费用，但签�?MIDlet 对开发者来讲是收益非浅的，因�ؓ许多受保护的 API 都是需要签名的�Q�以保护开发者和用户的利益。当�Ӟ��有些应用是不需要签名的�Q�如有些不需要联�|�的仅用��C��些图�?API 的小游戏软�g。但一些重要的应用�Q�如�Q�连接网�l�、发送短消息 ( 短信和彩�?) 或访问移动终�?( ��手机�?PDA �{�，以下��U�Cؓ手机 ) 上的 PIM( 个�h信息��理 ) 数据�{�等都需要签名�?

数字�{�֐� MIDlet 的好处包括：

(1) ��Z�� MIDlet 的安全策略，某些功能是必��ȝ��名才能��用的�Q�而有些功能虽然不�{�֐�也可以��用，但必��要求用户在使用时确认和修改其安全策略，如：写用��h��据缺省是不允许没有签名的 MIDlet 操作的；

(2) ��Z��手机的系�l�安全和�U�d��|�络的安全考虑�Q�某些手机制造商、移动运营商�{�可能拒�l�没有签名的 MIDlet 在手��Z��安装和运行；

(3) 大大改善用户体验�Q�让用户使用方便�Q��得用户不会遭遇调用受保护 API 时的安全警告的烦��|��

(4) ��Z��安全考虑�Q�安装没有签名的 MIDlet 是会有安全警告的�Q�而相反，安装已经�{�֐��?MIDlet 则不会出现烦人的警告�Q�手��Z��自动验证�{�֐�而顺利地安装成功�Q?

(5) 已经�{�֐��?MIDlet ��得用戯��改善其低安全�{�略讄��Q�提高手机的安全性；

(6) ��保已经�{�֐��?MIDlet 不会被非法篡改和非法盗用�?

二�?MIDP 2.0 安全机制

MIDP 是一个开攄��q�_��Q��得�Q何�h都可以�ؓ支持 MIDP 的设备开发各�U�应用��Y�Ӟ��一般都是移动终端设备�?MIDlet 套�g可以以匿名方式通过�|�络下蝲�Q�非常方便，但这也会带来许多安全问题和隐�U�信息保护问题，用户会问�Q?MIDlet 能把用户的个��Z��息发�l�不知道的服务器吗？会自动��生没有授权的呼叫或短消息而给用户带来费用吗？恶意软�g会破坏手机？�{�等�?

除了 Java 语言的安全特性外�Q?MIDP �q�增加了许多安全考虑�?MIDP 2.0 �?MIDP 1.0 增强了安全策略，�?API 分�ؓ普�?API 和敏�?API �Q�如�Q�通过 HTTP 协议讉K��U�d��|�络�Q�由于会�l�用户��生费用，所以被列�ؓ 敏感 API �?MIDlet 2.0 推出了可信�Q MIDlet(trusted) 和不可信�?MIDlet(untrusted) 的概念，一个不可信�?MIDlet 只能讉K��有限�?API �Q�同时还需要用��h��动确认�ƈ修改其安全策略；而可信�Q MIDlet 则自动��承系�l�中的安全策略而获得访问许可�?

许可 (Permissions) 用于需要��n份认证的敏感 API �?MIDP 2.0 要求调用敏感 API 之前必须获得必要的许可，�q�些许可包的命名�?J2SE 许可�Q�如�Q?HTTP �q�接许可同样�U�Cؓ�Q? javax.microedition.io.Connector.http �?有关许可的文档同意归�c�d��受保�?API 中�?

2.1 Protection Domains( 保护�?)

保护域是 MIDP 2.0 中一个非帔R��要的安全概念�Q�一个保护域��是一个许可集和一�U�交互模式，�q�些许可既可以是自己�l�承的，也可能是用户讄��的，前者称为允�? (allowed) �Q�而后者称为用户允�?(user permission) 。当一�?MIDlet 被安装后�Q�它被分配到一个指定的保护域而获得它的许可和交互模式�?

而用户允许则需要用戯��己决定是否同意，用户既拒�l�一个许可，也可以同意。用户允许有 3 �U�交互模式： blanket( 普遍适用 ) �? session( 短期适用 ) �?oneshot( 本次适用 ) �Q?普遍适用模式��是 MIDlet 安装时获得的许可一直有效，除非用户取消�q�些许可�Q��?短期适用模式则是指第一�ơ调�?API 旉��要用户允许，有效期到�?MIDlet 套�g�q�行�l�束�Q��?本次适用模式则在每次调用 API 旉��要求用户允许。保护域为用戯��可定义了�~�省的交互模式�?

一�?MIDlet 套�g使用 MIDlet-Permissions �?MIDlet-Permissions-Opt 属性来明确地定义其许可�Q�可以是�?JAD 文�g中定义，也可以在 manifest 文�g中定义。其中： MIDlet-Permissions 定义�?MIDlet 套�g中必��d��有的许可�Q��?MIDlet-Permissions-Opt 则定义希望具有的许可。如�Q�一个应用��Y件的基本要求是要�?http �q�接才能正常工作�Q�同�Ӟ��也可以��?https �q�接 ( 服务器部�|�了 SSL 证书 ) 来增强安全性，但不是必��ȝ��Q�这��P��q�个应用软�g的应用描�q�可以是�q�样�Q?

MIDlet-Permissions: javax.microedition.io.Connector.http

MIDlet-Permissions-Opt: javax.microedition.io.Connector.https

��h��意：一�?MIDlet 所要求的许可必��L��安装时分配的保护域所��h��的许可的子集。如�Q?Nokia S60 MIDP Emulator Prototype 2.0 (SDK) 有一个叫�?#8220; minimum ”的域�Q�此域没有�Q何许可。所以，如果一个含有许多许可的已经�{�֐��? MIDlet 如果被安装到此域�Q�则会安装失败，因�ؓ此域不支持这些许可。同��P��如果一个许可的名称有拼写错误，则一样会��D��安装��p�|�Q�因为域中没有此拼写错误的许可�?

MIDP 2.0 �?GSM/UTMS 讑֤�定义�?4 �U�保护域�Q?manufacturer( 讑֤�刉��商 ) , operator( �U�d��q�营�?) , trusted third party( 可信�ȝ��W�三�?) , and untrusted( 不受信�Q�?) �Q?除了 untrusted 域外�Q�每个保护域都对应一�l�根证书�Q�用于签�?MIDlet 的签名证书的根证书必��d��含在�q�些根证书中�Q��用不同的�{�֐�证书�{�֐��?MIDlet ��被自动归类予根证书所属的保护域，根证书与保护域的关系是：一个保护域可以有许多个根证书，而一个根证书只能对应于一个保护域�?

具体来讲�Q?manufacturer 域属于设备制造商�Q�其根证书是讑֤�刉��商自己的根证书�Q��?operator 域运营商�Q�一般��用其 SIM 卡中的根证书�Q��?trusted third party 域则预置了全球知名的数字证书颁发机构 (CA) 的根证书�Q�用于验证由 CA 颁发�? MIDlet �{�֐�证书�Q��?untrusted 域没有根证书�Q�将用于没有�{�֐��?MIDlet �?MIDP 1.0 �?

Thawte �?VeriSign 的根证书已经预置�?trusted third party �?中，�?Java 代码�{�֐�证书可以用于�{�֐� MIDlet 。当�Ӟ��用户也可以选择使用讑֤�刉��商和移动运营商颁发的证书，只要其根证书已经包含在手机的 4 个保护域中。据 WoTrust 了解�Q�大多数摩托�|�拉 (Motorola) 手机只支持设备制造商域，所以，只能�?Motorola 甌��{�֐�服务了�?

��h��意：�׃�� MIDP 2.0 也在不断��C��改和增补�Q�所以，可能不用的移动网�l�运营商有不同的保护域和许可�Q�用户可能需要向�U�d��q�营商了解详�l�信息。而最��单的�Ҏ��是检查目标用��h��使用的手机的根证书是否有计划购买�?MIDlet �{�֐�证书的根证书�?

2.2 Untrusted MIDlet ( 不受信�Q�?MIDlet)

MIDP 2.0 定义了那�?API �?untrusted 的，�q�些 Jar 文�g的来源和完整性是不能被手机验证的。但�q��ƈ不意味着�q�些 MIDlet 不能被安装和�q�行�Q�而是�q�行�q�些 MIDlet 需要用户�h工确认允许。而所�?MIDP 1.0 �?MIDlets 都被定义�? untrusted �?

untrusted �?MIDlets 只能调用一个不需要许可保护的 API �Q�如�Q?
java.util
java.lang
java.io
javax.microedition.rms
javax.microedition.midlet
javax.microedition.lcdui
javax.microedition.lcdui.game
javax.microedition.media
javax.microedition.media.control

如果 untrusted MIDlet 套�g试图调用一个被保护�?API 而且没有被�h工允许，则会产生一�?SecurityException 而被 MIDlet 按照安全�{�略处理。请注意�Q?Nokia �?UI API 是不被保护的�Q�包括类�Q?com.nokia.mid.sound �?com.nokia.mid.ui �?

2.3 Trusted MIDlets ( 可信�ȝ�� MIDlets)

如果手机能验�?MIDlet 的��n份和完整�?( 也就是已�l�数字签�?) �Q�则会自动分配一个合适的保护域这�U?MIDlet 套�g��q��为可信�Q�?MIDlet 。一个可信�Q�?MIDlet 套�g所要求的许可将被准许，只要所属的保护域拥有这�U�许可，假如许可�Q? javax.microedition.io.Connector.http 已经在所属保护域中是允许的，�?MIDlet 在打开一�?http �q�接时是不需要用��L��认的�?

请不要�؜淆了可信�ȝ�� MIDlet 套�g和可信�Q的保护域的不同，每个可信�ȝ�� MIDlet 套�g依据安全�{�略被分配到一个特定的保护域�?

您需要��用一个手��Z��已经预置的根证书的证书颁发机构颁发的代码�{�֐�证书来签�?MIDlet �Q�否则将不能通过�w�䆾验证。成功签名后�?JAD 文�g中一定会包含有整个签名证书的证书链，属性名�U�Cؓ�Q?MIDlet-Certificate-1-1 ��是您的�{�֐�证书�Q��? MIDlet-Certificate-1-2 ��是 CA 的中�U�根证书�Q��?MIDlet-Certificate-1-3 ��是 CA 的顶�U�根证书。同时还会有一�?MIDlet-Jar-RSA-SHA1 属性就�?JAR 文�g的摘要�?

当一�?MIDlet 被下载或被安装时�Q?MIDlet 应用��理器首先会��?JAD 文�g中是否包含了 MIDlet-Jar-RSA-SHA1 �?性，如果有，则启动如下验证过�E�：首先会读�?MIDlet-Certificate-1-1 �?MIDlet-Certificate-1-2 �?MIDlet-Certificate-1-3 属性中的证书，�q�与已经预置的根证书相比较，如果证书链能被根证书验证�Q�则表明开发者��n份已�l�被验证。接着��׃��使用用户证书来解�? MIDlet-Jar-RSA-SHA1 �?性的摘要�Q�再计算出已�l�下载的 Jar 文�g的摘要，比较两个摘要是否相等�Q�如果相�{�，则表�? MIDlet 代码自签名后没有被修攏V��这��P��既验证了�w�䆾又检查了完整性的 MIDlet 会被分配到所属根证书所对应的保护域中。但是，如果 MIDlet 中的许可属�?( MIDlet-Permissions ) 中有一个或多个不属于所属的保护域，则仍然不允许安装。而如�? MIDlet 中的可选许可属�?( MIDlet-Permissions-Opt ) 中有一个或多个不属于所属的保护域，会允许安装。可见，正确讄��许可属性和可选许可属性非帔R��要�?

2.4 Function Groups ( 功能分组 )

��Z��化用��L��理操作， MIDlet 把一些类似功能分�l�，�q�样�Q�用户只需对功能组讄��许可卛_��。如�Q�许�?“Net Access”( �|�络讉K�� ) �l�来代替许可 javax.microedition.io.Connector.http �Q�这对于��化手机的交互操作非常有用�?

MIDP 2.0 �?JTWI 定义了如�?7 个功能组�Q?

(1) Net Access: 包括所有网�l�连接许可；

(2) Messaging: 包括所有与发送和接收短消�?( 短信和彩�?�{?) 相关的许可；

(3) Auto Invocation : 包括与自动启�?MIDlet 相关的许可，如： Push Registration

(4) Local Connectivity : 包括与本地连接相关的许可�Q�如�Q?IrDA �?蓝牙�Q?

(5) Multimedia Recording : 包括与允许录韟뀁照相、摄像等相关的许可；

(6) Read User Data : 包括��d��用户数据相关的许可，如：通讯录、日�E�表�{�；

(7) Write User Data : 包括写用��h��据相关的许可�?

不同的手机支持不同的功能�l�，如： Multimedia Recording ��׃��会包含在没有摄录装置的手��Z��。当�Ӟ��也有可能��来会增加更多的功能�l��?

功能�l�也同时定义了不同的域的不同交互方式�Q�如�Q�在不信��d��Q?“Net Access” ( �|�络讉K�� ) 被设�|��ؓ session( 短期适用 ) �?denied( 拒绝 ) �Q�而在可信��d��则可以设�|��ؓ oneshot �?blanket �?denied 的�?

三、仿真器和手机的�~�省安全讄��

让我们来看看具体的��?Thawte �?VeriSign 代码�{�֐�证书�{�֐�后的 MIDlet �?trusted third party 域中的所有缺省许可，如下�?1 所�C�，点击 NDS 3.0 �?#8220; Config Emulators ”��可以看��C��真器�?trusted third party 域的�~�省安全讄��?#8220; Ask first time ”�Q�即�W?1 �ơ��用是需要确认：

如下�?2 所�C�，您可以下拉所有功能组的许可设�|�，�?#8220; Network Access ”��有 4 个选项可以修改�Q?Ask first time �?Ask every time �?Always allowed �?Not allowed �Q?

而如下图 3 所�C�，�?#8220; Real Life ”模式�Q�也��是实际手机的运行模式，可以看出�Q�定义的 7 个功能组都是“ Always allowed ” ( ��L��允许 ) �Q�这��显�C�出 MIDlet �{�֐�对于开发商来讲是多么的重要�Q�将大大方便了用��L��使用�Q�再也不需要用��h��作烦人的�p�d��认了�?

LukeW 2008-08-15 14:53 发表评论

树�Ş�l�构

LukeW — Fri, 15 Aug 2008 06:51:00 GMT

树�Ş�l�构�Q�tree�Q�是比较常用的数据结构了�Q�MIDP中没有它的��n影，不然我就不用写这��文章了�?br /> 代码如下:

/**
*
* @author hunhun1981
*/
public class HTree {

private HNode root;

private HNode current;

private int currDepth;

private int maxDepth;

public HTree(Object rootValue) {
  root = new HNode(null, rootValue);
  current = root;
}

public void goRoot() {
  current = root;
  currDepth = 0;
}

public boolean goChild(int index) {
  if (current.childList != null) {
   if (current.childList.size() > 0
     && index < current.childList.size()) {
    current = (HNode) current.childList.elementAt(index);
    currDepth++;
    if (currDepth > maxDepth) {
     maxDepth = currDepth;
    }
    return true;
   }
  }
  return false;
}

public void goBack() {
  if (current.father != null) {
   current = current.father;
   currDepth–;
  }
}

public Object getCurrent() {
  return current.value;
}

public int getCurrentDepth() {
  return currDepth;
}

public int getMaxDepth() {
  return maxDepth;
}

public Object[] getChilds() {
  if (current.childList != null) {
   if (current.childList.size() > 0) {
    Object[] ret = new Object[current.childList.size()];
    for (int i = 0; i < ret.length; i++) {
     ret[i] = ((HNode) current.childList.elementAt(i)).value;
    }
    return ret;
   }
  }
  return null;
}

public Object getChild(int index) {
  if (current.childList != null) {
   if (current.childList.size() > 0
     && index < current.childList.size()) {
    return ((HNode) current.childList.elementAt(index)).value;
   }
  }
  return null;
}

public void addChild(Object obj) {
  if (current.childList == null) {
   current.childList = new Vector();
  }
  current.childList.addElement(new HNode(current, obj));
}

public void addChilds(Object[] objs) {
  if (current.childList == null) {
   current.childList = new Vector();
  }
  for (int i = 0; i < objs.length; i++) {
   current.childList.addElement(new HNode(current, objs[i]));
  }
}

public int hasChild() {
  if (current.childList == null || current.childList.size() <= 0) {
   return 0;
  } else {
   return current.childList.size();
  }
}

private class HNode {

  public Vector childList;

  public HNode father;

  public Object value;

  public HNode(HNode father, Object value) {
   this.value = value;
   this.father = father;
   this.childList = null;
  }
}
}

�q�个�c�d��现简单，没有包含复杂的功能，仅仅用来做树形数据的存储�q�是不错的。比如游戏中用来��理场景�Q�管理资源；应用中用来作分类数据的表现等�{�。完全��以胜仅R�?br /> 使用�Ҏ��如下�Q?br />

HTree tree = new HTree(”root”);//会自动创��Z��个默认的根节�?/span>
tree.addChild(”天才”);//在根节点��d��新的节点
tree.addChild(”白痴”);
tree.goChild(0);//�q�入到当前节点的�W�一个节点（天才�Q��?/span>
tree.addChild(”天才1�?#8221;);//在当前节点（天才�Q�添加新的节�?/span>
tree.addChild(”天才2�?#8221;);
tree.goBack();//�q�回当前节点�Q�天才）的父节点�Q�根�Q?/span>
tree.goChild(1);//�q�入到当前节点的�W�二个节点（白痴�Q��?/span>
tree.addChild(”白痴1�?#8221;);//在当前节点（白痴�Q�添加新的节�?/span>
tree.addChild(”白痴2�?#8221;);
tree.goRoot();//完成创徏后将当前节点讄��为根节点�?/span>
…

上面的代码创��Z��一��完整的树，当然�Q�您可以使用��M��对象代替�q�里存储的String对象�?br /> �q�有一些方法，一看函数名大概都能明白�Q�就不再唠叨了�?br /> 遍历的方法于上面创徏树的�Ҏ��怼��Q��M��Q�要注意当前节点的位�|�，以免下次使用时处在错误的位置�?br /> 有兴��的朋友可以扩展一下遍历方法。不�q�我觉得没必要。因为J2ME环境下更需要的是树形结构，而不是强大的tree对象�?br />
��M��Q�我比较們֐�于简单实玎ͼ�希望它不太让��得简陋就好。从实用出发�Q�它�q�是能够满��大部分受限��^台的需求的�?

LukeW 2008-08-15 14:51 发表评论

URLEncoding

LukeW — Fri, 15 Aug 2008 06:49:00 GMT

URLEncoding是用于解决链接字�W�串中包含中文字�W�的一�U��{换编码。各�U�编�E�环境下几乎带有它的库函数�?br />
不过�Q�J2ME除外�?br />
好在JAVA的源代码中带有这个类�Q�我们把它拷贝到J2ME环境下编译到我们的应用当中就可以了�?br />
该文件位于JDK的目录下src.zip文�g中，名叫URLEncoder.java�?br />
但是�Q�这个文件还需要做很多修改才能使用在J2ME环境中�?br />
先警告大�Ӟ��有几个真机（其中一个就是烦��q��Q�好像是k500c�Q�，不管输入什么样的Encodeing都会出错�Q�甚��x��“UTF-8”。所以我一怒之�? 去除了Encodeing参数。（�q�可是在实际应用中得出的�l�论�Q�不��L��的话可以在大部分情况下正�怋�用，但是�Q�现实��L��有点�~�陷�Q?br />
修改后的代码如下�Q�大家请攑ֿ�使用。如果有兴趣�Q�可以比较两个代码，看看我改动了什么地斏V�?br />

import java.io.ByteArrayOutputStream;
import java.io.DataOutputStream;
public class HURLEncoder {

private static boolean[] dontNeedEncoding;

static {
  dontNeedEncoding = new boolean[256];

  for (int i = 0; i < 256; i++) {
   boolean b = ((i >= ‘0′) && (i <= ‘9′))
     || ((i >= ‘A’) && (i <= ‘Z’)) || ((i >= ‘a’) && (i <= ‘z’));

   dontNeedEncoding[i] = b;
  }

  dontNeedEncoding[’ ‘] = true;
  dontNeedEncoding[’-'] = true;
  dontNeedEncoding[’_'] = true;
  dontNeedEncoding[’.'] = true;
  dontNeedEncoding[’*'] = true;
}

public static String encode(String s) {

  boolean wroteUnencodedChar = false;

  StringBuffer writer = new StringBuffer();

  StringBuffer out = new StringBuffer(s.length());

  for (int i = 0; i < s.length(); i++) {
   char c = s.charAt(i);

   if ((c < 256) && dontNeedEncoding[c]) {
    if (c == ‘ ‘) {
     c = ‘+’;
    }

    out.append((char) c);
    wroteUnencodedChar = true;
   } else {
    try {
     if (wroteUnencodedChar) {
      writer = new StringBuffer();
      wroteUnencodedChar = false;
     }

     writer.append(c);

     if (c >= 0xD800 && c <= 0xDBFF) {
      if ((i + 1) < s.length()) {
       int d = (int) (s.charAt(i + 1));

       if (d >= 0xDC00 && d <= 0xDFFF) {
        writer.append(d);
        i++;
       }
      }
     }

    } catch (Exception e) {
     writer = new StringBuffer();
     continue;
    }

    String str = writer.toString();

    ByteArrayOutputStream baos = new ByteArrayOutputStream();
    DataOutputStream dos = new DataOutputStream(baos);
    try {
     dos.writeUTF(str);
     dos.flush();
    } catch (Exception e) {
     e.printStackTrace();
    }

    byte[] temp = baos.toByteArray();
    byte[] ba = new byte[temp.length - 2];
    for (int ix = 0; ix < ba.length; ix++) {
     ba[ix] = temp[ix + 2];
    }

    for (int j = 0; j < ba.length; j++) {
     out.append(’%');

     char ch = forDigit((ba[j] >> 4) & 0xF, 16);
     out.append(ch);

     ch = forDigit(ba[j] & 0xF, 16);
     out.append(ch);
    }

    writer = new StringBuffer();
    try {
     dos.close();
     baos.close();
    } catch (Exception e) {
     e.printStackTrace();
    }
   }
  }

  return out.toString();
}

private static char forDigit(int digit, int radix) {
  if ((digit >= radix) || (digit < 0)) {
   return ‘0′;
  }
  if (digit < 10) {
   return (char) (’0′ + digit);
  }
  return (char) (’A’ + digit - 10);
}
}

LukeW 2008-08-15 14:49 发表评论

GB2312转换为UTF-8

LukeW — Fri, 15 Aug 2008 06:47:00 GMT

摩托�|�拉的部分手机（a1200,e60�{�）�Q�不支持gb2312�~�码。曾�l�给我造成了不��麻烦。现在，大家可以分��n解决�q�个问题的一些经验�?br />
关于gb2312,unicode,utf-8的一些资料，大家误��行搜索。一下列丑և�个比较好的资源网址�?br /> http://baike.baidu.com/view/25492.htm
http://www.utf.com.cn/article/s45
http://www.utf.com.cn/article/s74
http://www.haiyan.com/steelk/navigator/ref/gb2312/gbindex.htm

要点�Q?br /> 1,gb2312于unicode或者utf-8之间�q�不存在直接的映��关�p�R��所以我们只能通过查表法来�q�行转换�?br /> 2,utf-8是unicode用于�|�络传输的一�U��Ş式，它与unicode之间是可以通过�q�算来进行�{换的�?br /> 3,j2me环境使用的都是utf-8�~�码�Q�但是请注意�Q�j2me中的utf-8�~�码比较�Ҏ��Q�在整个�~�码前面对了两个字节�Q�用于存攑֭��W�串的长度�?br />
�q�程�Q?br /> 1�Q�制作映��表gb2312-unicode�Q�应为汉字的unicode比utf-8要小�Q�这样做出的表也会小一些，而且对于unicode的可扩展性也��Z��些�?br /> 2�Q�先��gb2312�~�码串通过查表�Q��{换�ؓunicode�?br /> 3�Q�然后通过�q�算�Q�将unicode转换为utf-8,以便在j2me环境下��用�?br />
我修改了Herong Yang大侠的一个映��表生成函数�Q�原文请参�?a >http://www.herongyang.com/gb2312/gb2312_unicode.html
它的作用是生成一个二�q�制的gb2312到unicode的查找表�Q�它按照gb2312的分区，分块�Ҏ��，��其对应的unicode按顺序存入指定的位置�?br /> �q�样我们只需要根据gb2312的编码，计算出烦引就可以获取�~�码对应的unicode了�?br /> �׃��是修改的代码�Q�没脸脓出来�Q�大家有需求可以直接参考Herong Yang的文章，然后�Ҏ��自己需求修改�ƈ生成自己的映��表�?br />
�q�里我把自己�q�个转换表文件以及访问代码公开�?br /> http://download.csdn.net/source/263609
转帖��h��明。这是个�ȝ��化的代码�Q�在java中给它gb2312的byte数组�Q�它��q��你构造出字符丌Ӏ?br /> 用在不支持gb2312的手��Z��非常方便。这个�{换表的大��是15228byte�Q�对j2me来说�q�是可以接受的�?br />
如果有朋友需要沟通，可以发邮件到hunhun1981@hotmail.com

import java.io.InputStream;

public class HGB2312 {

    private byte[] map = new byte[15228];

    private byte[] buffer;
    private int index;

    public HGB2312() throws Exception {
        InputStream is = getClass().getResourceAsStream("/gb2u.dat");
        is.read(map);
        is.close();
    }

    public String gb2utf8(byte[] gb) throws Exception {
        buffer = new byte[gb.length + gb.length / 2 + 3];
        index = 0;
        int c, h, l, ind;
        for (int i = 0; i < gb.length;) {
            if (gb[i] >= 0) {
                fillBuffer(gb[i++]);
            } else {
                h = 256 + gb[i++];
                l = 256 + gb[i++];
                h = h - 0xA0 - 1;
                l = l - 0xA0 - 1;
                if (h < 9) {
                    ind = (h * 94 + l) << 1;
                    c = (byte2Int(map[ind]) << 8 | byte2Int(map[ind + 1]));
                    fillBuffer(c);
                } else if (h >= 9 && h <= 14) {
                    fillBuffer(0);
                } else if (h > 14) {
                    h -= 6;
                    ind = (h * 94 + l) << 1;
                    c = (byte2Int(map[ind]) << 8 | byte2Int(map[ind + 1]));
                    fillBuffer(c);
                } else {
                    fillBuffer(0);
                }
            }
        }
        // ind = index - 2;
        // h = (byte) ((ind >> 8) & 0x7F);
        // l = (byte) (ind & 0xFF);
        // buffer[0] = h;
        // buffer[1] = l;

        return new String(buffer, 0, index, "UTF-8");
    }

    private void fillBuffer(int value) {
        if (value <= 0x0000007F) {
            buffer[index++] = (byte) value;
        } else if (value >= 0x00000080 && value <= 0x000007FF) {
            byte b1 = (byte) (0x60 | (value >> 6));
            byte b2 = (byte) (0x80 | (value & 0x3F));
            buffer[index++] = b1;
            buffer[index++] = b2;
        } else if (value >= 0x00000800 && value <= 0x0000FFFF) {
            byte b1 = (byte) (0xE0 | (value >> 12));
            byte b2 = (byte) (0x80 | ((value >> 6) & 0x3F));
            byte b3 = (byte) (0x80 | (value & 0x3F));
            buffer[index++] = b1;
            buffer[index++] = b2;
            buffer[index++] = b3;
        } else if (value >= 0x00010000 && value <= 0x001FFFFF) {
            byte b1 = (byte) (0x1E | (value >> 18));
            byte b2 = (byte) (0x80 | ((value >> 12) & 0x3F));
            byte b3 = (byte) (0x80 | ((value >> 6) & 0x3F));
            byte b4 = (byte) (0x80 | (value & 0x3F));
            buffer[index++] = b1;
            buffer[index++] = b2;
            buffer[index++] = b3;
            buffer[index++] = b4;
        } else if (value >= 0x00200000 && value <= 0x03FFFFFF) {
            byte b1 = (byte) (0x3E | (value >> 24));
            byte b2 = (byte) (0x80 | ((value >> 18) & 0x3F));
            byte b3 = (byte) (0x80 | ((value >> 12) & 0x3F));
            byte b4 = (byte) (0x80 | ((value >> 6) & 0x3F));
            byte b5 = (byte) (0x80 | (value & 0x3F));
            buffer[index++] = b1;
            buffer[index++] = b2;
            buffer[index++] = b3;
            buffer[index++] = b4;
            buffer[index++] = b5;
        } else if (value >= 0x04000000 && value <= 0x7FFFFFFF) {
            byte b1 = (byte) (0x7E | (value >> 30));
            byte b2 = (byte) (0x80 | ((value >> 24) & 0x3F));
            byte b3 = (byte) (0x80 | ((value >> 18) & 0x3F));
            byte b4 = (byte) (0x80 | ((value >> 12) & 0x3F));
            byte b5 = (byte) (0x80 | ((value >> 6) & 0x3F));
            byte b6 = (byte) (0x80 | (value & 0x3F));
            buffer[index++] = b1;
            buffer[index++] = b2;
            buffer[index++] = b3;
            buffer[index++] = b4;
            buffer[index++] = b5;
            buffer[index++] = b6;
        }
    }

    private int byte2Int(byte b) {
        if (b < 0) {
            return 256 + b;
        } else {
            return b;
        }
    }
}

LukeW 2008-08-15 14:47 发表评论

LukeW — Fri, 15 Aug 2008 06:34:00 GMT

今天在硬盘上挖出�q�个存放了几�q�的代码。又回忆�?�q�前的那个j2me手机游戏�E�序�?#8230;…

�q�个��法是参考一位高人的文章�Q�直接读取�ƈ修改png格式囄��的调色板�Q�然后生成新的调色板替代原来的�?br /> �q�样可以实现游戏中常见的变色效果�Q�可以解��x��戏容量有限，不能存放太多�_��囄��的问题�?br />
具体�q�程其实�q�不复杂�Q�大家可以先搜烦资料�Q�先看看png囄��的格式定义。这个算法正是找到调色板区，�Ҏ��原有格式修改之后�Q�生成新的crc校验码，然后替换原来的调色板。这样就可以用一个png囄��Q�创建多个变色副本�?br />

public class PalettedImage {

    public Image getPalettedImage(byte[] data, int[] originalColors,
            int[] palettedColors) {
        byte[] tempData = new byte[data.length];
        System.arraycopy(data, 0, tempData, 0, data.length);
        Image img = null;
        int[] parameter = { 0, 0, 0 };
        analyze(tempData, parameter);
        for (int i = 0; i < originalColors.length; i++) {
            replaceColor(tempData, parameter, originalColors[i],
                    palettedColors[i]);
        }
        fillData(tempData, parameter);
        try {
            img = Image.createImage(tempData, 0, data.length);
        } catch (Exception e) {
            System.out.println("getPalettedImage  &&  " + e.toString());
        }
        return img;
    }

    private void analyze(byte[] data, int[] para) {
        int offset = 8;
        int chunkLen = 0;
        while (data[offset + 4] != 0x50 || data[offset + 5] != 0x4c
                || data[offset + 6] != 0x54 || data[offset + 7] != 0x45) {
            chunkLen = readInt(data, offset);
            offset += (4 + 4 + chunkLen + 4);
        }
        chunkLen = readInt(data, offset);
        para[2] = chunkLen / 3;
        para[0] = offset + 8;
        para[1] = offset + 8 + chunkLen;
    }

    private int readInt(byte[] data, int offset) {
        return ((data[offset] & 0xFF) << 24)
                | ((data[offset + 1] & 0xFF) << 16)
                | ((data[offset + 2] & 0xFF) << 8) | (data[offset + 3] & 0xFF);
    }

    private void replaceColor(byte[] data, int[] para, int oldColor,
            int newColor) {
        byte rr = (byte) ((oldColor >> 16) & 0xff);
        byte gg = (byte) ((oldColor >> 8) & 0xff);
        byte bb = (byte) (oldColor & 0xff);
        for (int i = 0, offset = para[0], temp = 0; i < para[2]; i++, offset += 3) {
            if (rr == data[offset] && gg == data[offset + 1]
                    && bb == data[offset + 2]) {
                data[offset] = (byte) ((newColor >> 16) & 0xff);
                data[offset + 1] = (byte) ((newColor >> 8) & 0xff);
                data[offset + 2] = (byte) (newColor & 0xff);
                break;
            }
        }
    }

    private void fillData(byte[] data, int[] para) {
        int checksum = update_crc(data, para[0] - 4, para[2] * 3 + 4);
        data[para[1]] = (byte) ((checksum >> 24) & 0xff);
        data[para[1] + 1] = (byte) ((checksum >> 16) & 0xff);
        data[para[1] + 2] = (byte) ((checksum >> 8) & 0xff);
        data[para[1] + 3] = (byte) ((checksum) & 0xff);
    }

    private int update_crc(byte[] buf, int off, int len) {
        int c = 0xffffffff;
        int n, k;
        int xx;
        int[] crc_table = new int[256];
        for (n = 0; n < 256; n++) {
            xx = n;
            for (k = 0; k < 8; k++) {
                if ((xx & 1) == 1) {
                    xx = 0xedb88320 ^ (xx >>> 1);
                } else {
                    xx = xx >>> 1;
                }
            }
            crc_table[n] = xx;
        }

        for (n = off; n < len + off; n++) {
            c = crc_table[(c ^ buf[n]) & 0xff] ^ (c >>> 8);
        }
        return (c ^ 0xffffffff);
    }

}

接口��是getPalettedImage()函数�Q�只需要输入原始图片的byte数组�Q�以及需要替换颜色的颜色��D��有目标颜色值就行了。因为可以同时替换多个颜�Ԍ��所以输入参数是代表颜色的整形的数组。��M��Q�要保证原始颜色与目标颜色一一对应��好了。方法简单实用�?br />
�Ƣ迎大家使用�q�留下宝�늚�意见。当�Ӟ��也可以修改一下这个函敎ͼ�做一些特�D�的效果。这里就不多说了�?br /> 不过�q�个代码用处已经不大�Q�因为现在的手机基本上都支持midp2.0所以可以��用更方便的方法替换颜艌Ӏ?br />
��M��Q�再�ơ感谢这位已�l�被我忘掉名字的大侠�Q�关键代码是他写的，我只是修�Ҏ��理而已�?

LukeW 2008-08-15 14:34 发表评论

Autostarting MIDlets in JP-7 phones using PushRegistry

LukeW — Fri, 15 Aug 2008 02:57:00 GMT

In Sony Ericsson Java Platform 7 (JP-7) phones, such as the K610 or W850, a new push functionality is introduced. PushRegistry Alarm and PushRegistry SMS are supported throughout the Java Platforms, with JP-4 adding PushRegistry CBS and now JP-7 includes PushRegistry auto start.

Note: the auto start functionality is not supported by the first released K610 and K800 JP-7 phones. Please use the phone's upgrade service to ensure that you are using the latest firmware version.

The auto start functionality can be set static by including it in the .jad file or it can be set dynamically in the code. Two example MIDlets are included as an example of this.

Download the source code and examples here>>

To enable the auto start functionality in the .jad file include the line:

//MIDlet-Push-: , ,
MIDlet-Push-1: autostart://:, AutoStartStatic, *

To make the MIDlet auto start from the source code the following methods can be used to register and un-register the MIDlet.

//Registers the pushRegistry
public void Register(){
        // List of registered push connections.
        String connections[];
        // Check to see if the connection has been registered.
        // This is a dynamic connection allocated on first
        // time execution of this MIDlet.
        connections = PushRegistry.listConnections(false);
        if (connections.length == 0) {
                try {
                        //Register so the MIDlet will wake up when phone is started.
                        PushRegistry.registerConnection("autostart://:", "AutoStartDyn", "*");
                        sDisplayString = "MIDlet is registered";
                } catch (Exception ex) {
                        System.out.println("Exception: " + ex);
                        sDisplayString = "Fail: " + ex;
                }
        } else {
                sDisplayString = "Already registered";
        }
        displayForm.deleteAll();
        displayForm.append(sDisplayString);
}

//Unregisters the pushRegistry
public void Unregister(){
        if (PushRegistry.unregisterConnection("autostart://:")){
                System.out.println("The pushRegistry is unregistered");
                sDisplayString = "MIDlet is unregistered.";
        }else{
                System.out.println("There is no pushRegistry to unregister");
                sDisplayString = "No MIDlet to unregister or failed to unregister";
        }
        displayForm.deleteAll();
        displayForm.append(sDisplayString);
}

To find out if the MIDlet is started via pushRegistry or manually you can inspect the connections registered by pushRegistry. Below is a small example of how to handle pushRegistry autostart or manual startup.

    private void handlePushActivation() {
        String[] connections = PushRegistry.listConnections(true);
        if (connections != null && connections.length > 0) {
            for (int i = 0; i < connections.length; i++) {
                if (connections[i].startsWith("autostart")) {
                    alert("Application autostarted");
                }
            }
        } else {
            alert("User started the application");
        }
    }

LukeW 2008-08-15 10:57 发表评论

Using simultaneous sounds

LukeW — Fri, 15 Aug 2008 02:34:00 GMT

The code sample below describes how to play two sounds at the same time. This feature is supported by the Sony Ericsson JP-5 platform and onwards.

Only one wav file can be played simultaniously but several midi files might be played at the same time. The number of existing players is only limited by available memory, and you have the possibility to play more than one player at the same time. The example below shows how to do this using one midi file and one wav file.

The code is straight-forward - just load the resource, create the player and start playing the file.

Here's the sample:

InputStream is = getClass().getResourceAsStream(file);
InputStream is1 = getClass().getResourceAsStream(file1);

player = Manager.createPlayer(is, " audio/midi");
player.setLoopCount(-1);
player.prefetch();
player.realize();

player1 = Manager.createPlayer(is1, "audio/x-wav");
player1.setLoopCount(1);
player1.prefetch();
player1.realize();

player.start();
player1.start();

example code

LukeW 2008-08-15 10:34 发表评论

Serializing an Image

LukeW — Fri, 15 Aug 2008 02:29:00 GMT

Creating an image from an array of data is an easy task, but to create a byte-array of data from an image is a little more complicated. But it's required if you want to send a modified image to a server.

To create a byte-array of data from an image, we can use the getRGB(..) method in the image class in MIDP 2.0. From the getRGB method we get an int-array of data containing all the ARGB values of each pixel in the image. Integers in java are four bytes, so we need to split each int value into four byte values.

To mask out each of the bytes in the int we can use the 'AND &' operator and the 'shift-right >>' operator. Here's an example:

int ARGB = 0xFFFFFFFF; // AARRGGBB
int a = (ARGB & 0xFF000000);
int r = (ARGB & 0x00FF0000);
int g = (ARGB & 0x0000FF00);
int b = (ARGB & 0x000000FF);

// Here we move each bit to the right.
a = (a >> 24); // a = 0x000000FF
r = (r >> 16); // r = 0x000000FF
g = (g >> 8); // g = 0x000000FF
b = b; // b = 0x000000FF

When we convert the integer to a byte, there are some problems since there are only signed bytes in Java. A byte may contain values between -128 and 127 and from our integer we'll have a value between 0 and 255.

Here are some conversion examples:

Integer val 127 = byte val 127.
Integer val 128 = byte val -128.
Integer val 255 = byte val -1.

byte ba = (byte)(a); // if a=0x000000FF (255), then ba = -1
byte br = (byte)(r);
byte bg = (byte)(g);
byte bb = (byte)(b);

We have to loop though each pixel in the image and get each pixel value to our byte-array. When that's done, we can send the image to a server where we can convert the byte-array back to a integer-array and then show our picture.

So, when we convert the byte back to an integer we have to check if the byte value is less than zero.

int a, r, g, b;
if(ba<0)
a = 256 - a;

Now our new integer value will be between 0 and 255 and we just have to use the 'shift-left <<' operator and add the values together to get our new int-array.

a = (a << 24);
r = (r << 16);
g = (g << 8);
b = (b);

0xFF000000 (a)
+ 0x00FF0000 (r)
+ 0x0000FF00 (g)
+ 0x000000FF (b)
= 0xFFFFFFFF (argb)

int ARGB = a+r+g+b;

After the integer-array is recreated we can use the createRGBImage(..) method in the Image class to create our image.

Be aware that the byte-array is quite large as it contains all of the ARGB values for each pixel. If an image is 100*60 px, where each pixel is four bytes, the byte array will be 24kb.
example code

LukeW 2008-08-15 10:29 发表评论

Fade in and out images in MIDP 2.0

LukeW — Fri, 15 Aug 2008 02:18:00 GMT

This tip describes how to change the alpha value of an image to make it appear blended. There's also an example MIDlet with source code.

In MIDP 2.0 there's a new method in the Image class, getRGB(...) that will get all Alpha, Red, Green, Blue (ARGB) values from the image to an int array. We can use this method, and the resulting array, to change the alpha value of the image.

Integers in J2ME are four bytes, each pixel in the image is described by the ARGB values where each color is one byte 0 to 255. If the alpha value is 0, the corresponding pixel will be transparent, if the alpha is 255, the pixel will be opaque.

To get a specific color from the int array, it's possible to use the AND '&' operator and to accomplish the blending effect we need to get the colors, without the alpha value, from the int array.

FF = 11111111 = 255
0xFFFFFFFF - Alpha = 255, Red =255 Green = 255, Blue = 255
(0xFFFFFFFF & 0x00FFFFFF) = 0x00FFFFFF

By doing this we'll only get the RGB colors from the int array, alpha equals zero.

Now when we just have the RGB colors and alpha equals zero, we can just add our new alpha value.

If we have the alpha value 255 and want to add this to our color, we need to use the shift left operator.

To change:

(00000000 00000000 00000000 11111111) to
(11111111 00000000 00000000 00000000)
use the shift left '<<' operator.
(0xFF << 24) = 0xFF000000.

With this knowledge we now can change the alpha value or do masking for specific colors.

Use the getRGB(...) method in the image class to get all the ARGB values from the image to a int array.
Use the blend function below to change the alpha value for each value in the array.
Use the createRGBImage(...) method in the image class to create a new image from the edited int array.

There are examples on how to use the getRGB and createRGBImage methods in the MIDlet below.

public static void blend(int[] raw, int alphaValue){
    int len = raw.length;
    // Start loop through all the pixels in the image.
    for(int i=0; i<len; i++){
        int a = 0;
        int color = (raw[i] & 0x00FFFFFF); // get the color of the pixel.
        a = alphaValue;     // set the alpha value we want to use 0-255.

        a = (a<<24);    // left shift the alpha value 24 bits.
        // if color = 00000000 11111111 11111111 00000000 (0xFFFF00 = Yellow)
        // and alpha= 01111111 00000000 00000000 00000000
        // then c+a = 01111111 11111111 11111111 00000000
        // and the pixel will be blended.
        color += a;
        raw[i] = color;
    }
}

example code

LukeW 2008-08-15 10:18 发表评论

Fast stream reading in Java

LukeW — Fri, 15 Aug 2008 02:11:00 GMT

To increase the performance of your Java™ application when reading from an InputStream, there are a few key areas to look into. If possible, don't make any reallocations of memory. Allocate the input buffer once. Let the Java™ VM do the bulk of the reading, i.e. read data in big chunks. Some coding examples show a loop that reads data a small amount at a time. Using a too small buffer is inefficient. A much better technique is to use a relatively large buffer.

If you know the maximum content length, the most optimal read would be a single line, like this:

len = instream.read( buf, 0, MAX_BUFFER_SIZE );

In this case we assume that the buffer has already been allocated with a size set to MAX_BUFFER_SIZE.

If the content is of varying size, we have to make a tradeoff between performance and memory usage. If you keep your buffer size just above the average content length, then the number of reallocations of the data buffer and the number of reads will be kept at a minimum. There is no defined size of what a large buffer is but a good rule of thumb might be to keep the buffer size at most about 0,5 to 1 MB.

LukeW 2008-08-15 10:11 发表评论