[求助]又一个关于Java的问题。。。

duzhi5368

[s:5] 要是学过PNG图形格式的话，你就清楚90，180，270，360度和其他角度的实现难度差别有多大了。涉及到部分程度的拉伸和缩放的。你可以尝试点象素画，30度和43度，你看看如何表示。^ ^
但是nokia   API支持任意角度的旋转。
自己写类模拟的话，难度实在够大，而且效果不好，建议使用多张美工图来实现这种效果。 [s:5]

若一定要程序实现的话，可以看下面的代码《注：转载，本人不负任何责任》
web2.0package rorate;

import javax.microedition.lcdui.Graphics;

/
* 图片任意角度翻转算法
* 同时实现了Nokia特有API
* author. BB
* Sprite.java
*
*
*/


public class Sprite {

/** SIN TABLE **/

public final static int SIN_TABLE[] =

{

0, 4, 8, 13, 17, 22, 26, 31, 35, 39,

44, 48, 53, 57, 61, 65, 70, 74, 78, 83,

87, 91, 95, 99, 103, 107, 111, 115, 119, 123,

127, 131, 135, 138, 142, 146, 149, 153, 156, 160,

163, 167, 170, 173, 177, 180, 183, 186, 189, 192,

195, 198, 200, 203, 206, 208, 211, 213, 216, 218,

220, 223, 225, 227, 229, 231, 232, 234, 236, 238,

239, 241, 242, 243, 245, 246, 247, 248, 249, 250,

251, 251, 252, 253, 253, 254, 254, 254, 254, 254,

255, 254, 254, 254, 254, 254, 253, 253, 252, 251,

251, 250, 249, 248, 247, 246, 245, 243, 242, 241,

239, 238, 236, 234, 232, 231, 229, 227, 225, 223,

220, 218, 216, 213, 211, 208, 206, 203, 200, 198,

195, 192, 189, 186, 183, 180, 177, 173, 170, 167,

163, 160, 156, 153, 149, 146, 142, 138, 135, 131,

127, 123, 119, 115, 111, 107, 103, 99, 95, 91,

87, 83, 78, 74, 70, 65, 61, 57, 53, 48,

44, 39, 35, 31, 26, 22, 17, 13, 8, 4,

0, -4, -8, -13, -17, -22, -26, -31, -35, -39,

-44, -48, -53, -57, -61, -65, -70, -74, -78, -83,

-87, -91, -95, -99, -103, -107, -111, -115, -119, -123,

-127, -131, -135, -138, -142, -146, -149, -153, -156, -160,

-163, -167, -170, -173, -177, -180, -183, -186, -189, -192,

-195, -198, -200, -203, -206, -208, -211, -213, -216, -218,

-220, -223, -225, -227, -229, -231, -232, -234, -236, -238,

-239, -241, -242, -243, -245, -246, -247, -248, -249, -250,

-251, -251, -252, -253, -253, -254, -254, -254, -254, -254,

-255, -254, -254, -254, -254, -254, -253, -253, -252, -251,

-251, -250, -249, -248, -247, -246, -245, -243, -242, -241,

-239, -238, -236, -234, -232, -231, -229, -227, -225, -223,

-220, -218, -216, -213, -211, -208, -206, -203, -200, -198,

-195, -192, -189, -186, -183, -180, -177, -173, -170, -167,

-163, -160, -156, -153, -149, -146, -142, -138, -135, -131,

-127, -123, -119, -115, -111, -107, -103, -99, -95, -91,

-87, -83, -78, -74, -70, -65, -61, -57, -53, -48,

-44, -39, -35, -31, -26, -22, -17, -13, -8, -4

};

/** 透明 **/
public static int TRANSPARENT = 0;

/** 不透明 **/
public static int OPAQUE = 15;

/**
* 像素值
* 包含动画帧，主下标标识动画帧数，副下标的值代表像素值
*/
public short pixels[][];

/*
* 精灵的宽度
* 该版本默认精灵各个帧的宽度一样大小
*/
public short width;

/*
* 精灵的高度
* 该版本默认精灵各个帧的高度一样大小
*/
public short height;

/*
* 动画帧数
*/
public byte numOfFrame;

/*
* 当前动画帧数
*/
public int curFrame;

/*
* 构造函数
* 确保传入的各个值为合理的值
* 使用的话请自己做判断
* @param _pixels 动画的像素值
* @param _width 宽度
* @param _height 高度
*/
public Sprite(short[][] _pixels, short _width, short _height){

pixels = _pixels;

width = _width;

height = _height;

numOfFrame = (byte)_pixels.length;

curFrame = 0;

}

/*

* 构造函数

*/

public Sprite(){

}

/*
* 投影
* @param cData
* @param l
* @param i1
* @param j1
* @param k1
* @return
*/
private final short project(short cData[], int l, int i1, int j1, int k1) {

int j3 = (j1 & 0xffff) >> 8;

int k3 = (k1 & 0xffff) >> 8;

int l3 = (256 - j3) * (256 - k3);

int i4 = j3 * (256 - k3);

int j4 = (256 - j3) * k3;

int k4 = j3 * k3;

int l2 = j1 >> 16;

int i3 = k1 >> 16;

l2 %= l;

i3 %= i1;

i3 *= l;

short word0 = cData[l2 + i3];

short word2 = cData[(l2 + (i3 + l)) % (cData.length - 1)];

l2 = ++l2 % l;

short word1 = cData[l2 + i3];

short word3 = cData[(l2 + (i3 + l)) % (cData.length - 1)];

int l1 = word0 >> 12 & 0xf;

int i2 = word1 >> 12 & 0xf;

int j2 = word2 >> 12 & 0xf;

int k2 = word3 >> 12 & 0xf;

int k5 = l1 * l3 + i2 * i4 + j2 * j4 + k2 * k4 >> 16;

l1 = word0 >> 8 & 0xf;

i2 = word1 >> 8 & 0xf;

j2 = word2 >> 8 & 0xf;

k2 = word3 >> 8 & 0xf;

int l4 = l1 * l3 + i2 * i4 + j2 * j4 + k2 * k4 >> 16;

l1 = word0 >> 4 & 0xf;

i2 = word1 >> 4 & 0xf;

j2 = word2 >> 4 & 0xf;

k2 = word3 >> 4 & 0xf;

int i5 = l1 * l3 + i2 * i4 + j2 * j4 + k2 * k4 >> 16;

l1 = word0 & 0xf;

i2 = word1 & 0xf;

j2 = word2 & 0xf;

k2 = word3 & 0xf;

int j5 = l1 * l3 + i2 * i4 + j2 * j4 + k2 * k4 >> 16;

return (short) ((k5 << 12) + (l4 << 8) + (i5 << 4) + j5);

}

/*
* 翻转
* @param count 在360度内翻转的个数
* @return 返回一个翻转后的精灵
*/
public Sprite rotate(int count) {

Sprite sprite = new Sprite();

short radius = 30;

if (width == 40 && height == 40)

radius = 57;

else if (width == 8 && height == 8)

radius = 12;

else if (width == 6 && height == 12)

radius = 14;

else if (width == 24 && height == 24)

radius = 34;

else if (width == 26 && height == 26)

radius = 27;

else if (width == 8 && height == 15)

radius = 17;

else if (width == 10 && height == 16)

radius = 19;

else if (width == 11 && height == 15)

radius = 19;

else
System.out.println("Wrong size: " + width + " " + width + " "

+ count);

sprite.width = radius;

sprite.height = radius;

sprite.numOfFrame = (byte)count;

sprite.pixels = new short[count][];

for (int j1 = 0; j1 < count; j1++) {

sprite.pixels[j1] = new short[sprite.width * sprite.height];

int k1 = j1 * (360 / count);

short l1 = width;

short i2 = height;

width = sprite.width;

height = sprite.height;

int j2 = width / 2;

int k2 = l1 / 2;

int l2 = i2 / 2;

int i3 = SIN_TABLE[k1 % 360] << 8;

int j3 = SIN_TABLE[(k1 + 90) % 360] << 8;

int k4 = 0;

int i4 = -j2 * j3;

int j4 = -j2 * i3;

for (int l4 = 0; l4 < sprite.pixels[j1].length; l4++)

sprite.pixels[j1][l4] = (short) (TRANSPARENT << 12);

for (int i5 = 0; i5 < height; i5++) {

int k3 = -j2 * j3;

int l3 = -j2 * i3;

for (int j5 = 0; j5 < width; j5++) {

//x

int k5 = (k3 - j4 >> 16) + k2;

//y

int l5 = (i4 + l3 >> 16) + l2;

if (k5 >= 0 && l5 >= 0 && l5 < i2 && k5 < l1)

sprite.pixels[j1][k4] = project(pixels[0], l1, i2,

(k3 - j4) + (k2 << 16), i4 + l3 + (l2 << 16));

k3 += j3;

l3 += i3;

k4++;

}

i4 += j3;

j4 += i3;

k4 -= width;

k4 += sprite.width;

}

height = i2;

width = l1;

}

return sprite;

}

/*
* 绘制精灵
* 在这里实现了Nokia的特有API
* 大家可以扩展该方法
* @param g
* @param x
* @param y
* @param manipulate
*/
public void draw(Graphics g, int x, int y, int manipulate) {

drawPixels(g, true, 0, width, x, y, width,

height, manipulate, 4444);

}

public void drawPixels(Graphics g, boolean transparency,

int offset, int scanlength, int x, int y, int width, int height,

int manipulation, int format) {

int l1 = map2Manipulation(manipulation);

int j1;

int k1;

if ((l1 & 4) != 0) {

j1 = height;

k1 = width;

} else {

j1 = width;

k1 = height;

}

short newPixels[] = new short [j1 * k1];

if(manipulation == 0){

newPixels = pixels[curFrame];

}

else
for (int i2 = 0; i2 < k1; i2++) {

for (int j2 = 0; j2 < j1; j2++) {

int j = j2;

int k = i2;

if ((l1 & 1) != 0)

j = j1 - 1 - j;

if ((l1 & 2) != 0)

k = k1 - 1 - k;

if ((l1 & 4) != 0) {

int k2 = j;

j = k;

k = k2;

}

newPixels[j1 * i2 + j2] = pixels[curFrame][width * k + j];

}

}

int off = offset;

int vw = x + j1;

int vh = y + k1;

for(int idy = y; idy < vh; idy++){

int voff = off;

for(int idx = x; idx < vw; idx++){

short pixel = newPixels[voff++];

int k3 = idx;

for(; idx < vw - 1 && newPixels[voff] == pixel; voff++)

idx++;

if((pixel >> 12 & 0xff) != 0)

{

int l3 = 0xf0 & pixel << 4;

l3 |= 0xf000 & pixel << 8;

l3 |= 0xf00000 & pixel << 12;

g.setColor(l3);

g.drawLine(k3, idy, idx, idy);

}

}

off += j1;

}

}

private static int map2Manipulation(int i) throws IllegalArgumentException {

int j = 0;

if ((i & 0x2000) != 0)

j ^= 1;

if ((i & 0x4000) != 0)

j ^= 2;

switch (i & 0xffff9fff) {

case 90: // 'Z'

j ^= 6;

break;

case 180:

j ^= 3;

break;

case 270:

j ^= 5;

break;

default:

throw new IllegalArgumentException();

case 0: // '/0'

break;

}

return j;

}

}



虽然我没有测试，但是直觉告诉我，这段代码基本上可以做为测试手机性能的工具了T T [s:5]

friday5xue

引用第2楼好人卡的神话于2007-06-19 20:07发表的  :
我还是多画图好了= =||| [s:5] 。。。

正解~~~机能有限的情况~~~还是,浪费电煤共比较好~~~
而且,手机游戏不用角度那么多的!