我正在尝试旋转位图,其中像素存储在数组int pixels[]
中。我得到了以下方法:
public void rotate(double angle) {
double radians = Math.toRadians(angle);
double cos, sin;
cos = Math.cos(radians);
sin = Math.sin(radians);
int[] pixels2 = pixels;
for (int x = 0; x < width; x++)
for (int y = 0; y < height; y++) {
int centerx = this.width / 2, centery = this.height / 2;
int m = x - centerx;
int n = y - centery;
int j = (int) (m * cos + n * sin);
int k = (int) (n * cos - m * sin);
j += centerx;
k += centery;
if (!((j < 0) || (j > this.width - 1) || (k < 0) || (k > this.height - 1)))
try {
pixels2[(x * this.width + y)] = pixels[(k * this.width + j)];
} catch (Exception e) {
e.printStackTrace();
}
}
pixels = pixels2;
}
但它只是给了我疯狂的结果。有谁知道错误在哪里?
答案 0 :(得分:2)
该行
int[] pixels2 = pixels;
应该复制数组,但您只是复制对它的引用。使用pixels.clone()
。实际上,您只需要一个新的空数组,因此new int[pixels.lenght]
就足够了。最后,您需要System.arraycopy
将新内容复制到旧数组中。
您的代码中还有其他问题 - 您正在混合行和列。一些表达式被编写为好像逐行存储图像,其他表达式就像逐列一样存储。如果逐行(我的假设),那么这没有意义:x*width + y
。它应该显示为y*width + x
- 您正在向下跳过y
行,然后向右移动x
列。总而言之,我有这个代码可以正常工作:
import static java.lang.System.arraycopy;
public class Test
{
private final int width = 5, height = 5;
private int[] pixels = {0,0,1,0,0,
0,0,1,0,0,
0,0,1,0,0,
0,0,1,0,0,
0,0,1,0,0};
public Test rotate(double angle) {
final double radians = Math.toRadians(angle),
cos = Math.cos(radians), sin = Math.sin(radians);
final int[] pixels2 = new int[pixels.length];
for (int x = 0; x < width; x++)
for (int y = 0; y < height; y++) {
final int
centerx = this.width / 2, centery = this.height / 2,
m = x - centerx,
n = y - centery,
j = ((int) (m * cos + n * sin)) + centerx,
k = ((int) (n * cos - m * sin)) + centery;
if (j >= 0 && j < width && k >= 0 && k < this.height)
pixels2[(y * width + x)] = pixels[(k * width + j)];
}
arraycopy(pixels2, 0, pixels, 0, pixels.length);
return this;
}
public Test print() {
for (int y = 0; y < height; y++) {
for (int x = 0; x < width; x++)
System.out.print(pixels[width*y + x]);
System.out.println();
}
System.out.println();
return this;
}
public static void main(String[] args) {
new Test().print().rotate(-45).print();
}
}
答案 1 :(得分:0)
public void render(float nx, float ny, float nz, float size, float rotate) {
int wid = (int) ((width - nz) * size);
int hgt = (int) ((height - nz) * size);
if (wid < 0 || hgt < 0) {
wid = 0;
hgt = 0;
}
for (int x = 0; x < wid; x++) {
for (int y = 0; y < hgt; y++) {
double simple = Math.PI;
int xp = (int) (nx +
Math.cos(rotate) * ((x / simple) - (wid / simple) / 2) + Math
.cos(rotate + Math.PI / 2)
* ((y / simple) - (hgt / simple) / 2));
int yp = (int) (ny + Math.sin(rotate)
* ((x / simple) - (wid / simple) / 2) + Math.sin(rotate
+ Math.PI / 2)
* ((y / simple) - (hgt / simple) / 2));
if (xp + width < 0 || yp + height < 0 || xp >= Main.width
|| yp >= Main.height) {
break;
}
if (xp < 0
|| yp < 0
|| pixels[(width / wid) * x + ((height / hgt) * y)
* width] == 0xFFFF00DC) {
continue;
}
Main.pixels[xp + yp * Main.width] = pixels[(width / wid) * x
+ ((height / hgt) * y) * width];
}
}
}
这对我来说只是一个新的旋转,但这个过程就是正常旋转。它仍然需要很多修复 - 它效率低,速度慢。但是在一个小程序中,这段代码可行。我发布这个,这样你就可以接受它,并使它变得更好。 :)