正确的翻转/镜像的像素的图像？

问题描述

$

这两种类型的镜像的代码：

  void mirrorLeftRight（）
 {
 for（int x = 0; x< width / 2; x ++）{ 
 for（int y = 0; y< height; y ++）{
 int temp = pixelData [x] [y]; 
 pixelData [x] [y] = pixelData [width-x] [y] 
 pixelData [width-x] [y] 
} 
} 
} 
 
 void mirrorUpDown（）
 {
 for（int x = 0; x< width; x ++） {
 for（int y = 0; y< height / 2; y ++）{
 int temp = pixelData [x] [y] 
 pixelData [x] [y] = pixelData [x] [height-y] 
 pixelData [x] [height-y] 
} 
} 
} 
  

？

对于翻转，只需使用 width 和 height w / o除以2？

解决方案

上面的代码更像是正确的Flip， 。

镜像我猜你不能切换像素，而是从一边复制到另一边。

有了镜像我会猜测你需要改变

  int temp = pixelData [x] [y] 
 pixelData [x] [y] = pixelData [width-x] [y] 
 pixelData [width-x] [y] 
  

  pixelData [x] [y] = pixelData [width-x] [y] 
  

http://tinypic.com/r/fwubzc/5

That shows what a flip should be and what a mirror should be.

Code for both types of mirrors:

void mirrorLeftRight()
{
    for (int x = 0; x < width/2; x++) {
            for (int y = 0; y < height; y++) {
                int temp = pixelData[x][y];
                pixelData[x][y]=pixelData[width-x][y]
                pixelData[width-x][y]=temp;
            }
    }
}

void mirrorUpDown()
{
    for (int x = 0; x < width; x++) {
            for (int y = 0; y < height/2; y++) {
                int temp = pixelData[x][y];
                pixelData[x][y]=pixelData[x][height-y]
                pixelData[x][height-y]=temp;
            }
    }
}

Does this seem right for mirrors?

And for flip, just a matter of using width and height w/o dividing by 2?

解决方案

The code above seems more like the correct way to Flip, not mirror.

Mirror I would guess that you not switch the pixels, but rather copy from one side to the other.

With mirror I would guess that you need to change

int temp = pixelData[x][y];
pixelData[x][y]=pixelData[width-x][y]
pixelData[width-x][y]=temp;

to something like this only

pixelData[x][y]=pixelData[width-x][y]

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