我正在尝试替换Scala中已缓冲图像像素中每个红色，绿色和蓝色值中的两个最低有效位。但是，当我尝试更改最低有效位时，更改将不成立。更改两个最高有效位时，我已经完成了这项工作，但使用最低有效位将无法工作。这是完成大部分繁重工作的方法
def hideDataInImage(dataToHide: String, imageToHideIn: BufferedImage): Option[BufferedImage] = {
// Check that the binary data is short enough to encode in the image. Assumeing that we can hide
// one 6-bit element in one pixel the size of the array must be less than the area of the array
if(dataToHide.length > (imageToHideIn.getHeight() * imageToHideIn.getWidth())) {
return None
}
val imageToHideInCopy = deepCopyBufferedImage(imageToHideIn)
// Turn the string of data into a string of binary numbers
val binaryToHide = stringToBinaryArray(dataToHide)
// Loop through the binary data and hide it in each pixel
for(i <- binaryToHide.indices) {
// Get the x and y data for the pixel
val y = i / imageToHideInCopy.getWidth()
val x = i % imageToHideInCopy.getHeight()
val newColor = hideDataInColor(imageToHideInCopy.getRGB(x, y), binaryToHide(i))
imageToHideInCopy.setRGB(x, y, newColor.getRGB)
}
// Return some image to hide in if success
Some(imageToHideInCopy)
}
其余代码可以在我的Github上看到。
编辑：这是我的其余代码
def hideDataInColor(pixelValue: Int, binaryDataToHide: String): Color = {
// Get the red green blue and alpha values from the pixel value
// 3 2 1 0
// bitpos 10987654 32109876 54321098 76543210
// ------ +--------+--------+--------+--------+
// bits |AAAAAAAA|RRRRRRRR|GGGGGGGG|BBBBBBBB|
val blueValue = byteToPaddedBinary(pixelValue & 0xff, 8)
val greenValue = byteToPaddedBinary((pixelValue & 0xff00) >> 8, 8)
val redValue = byteToPaddedBinary((pixelValue & 0xff0000) >> 16, 8)
val alphaValue = byteToPaddedBinary((pixelValue & 0xff000000) >>> 24, 8)
// Split the binarydata into three parts
val splitData = binaryDataToHide.split("(?<=\\G.{" + binaryDataToHide.length / 3 + "})")
// Get the modified red blue green values
val newRed = hideBitsInBinaryString(redValue, splitData(0))
val newGreen = hideBitsInBinaryString(greenValue, splitData(1))
val newBlue = hideBitsInBinaryString(blueValue, splitData(2))
// Convert the binary number to an integer and return it
new Color(Integer.parseInt(newRed, 2), Integer.parseInt(newGreen, 2),
Integer.parseInt(newBlue, 2), Integer.parseInt(alphaValue, 2))
}
def hideBitsInBinaryString(binaryString: String, bitsToHide: String): String = {
// Create a string builder from the binary string
val binaryStringStringBuilder = new StringBuilder(binaryString)
// Loop through the bits to hide
for(i <- 0 until bitsToHide.length) {
// replace to chars at the end of the string builder
binaryStringStringBuilder.setCharAt(binaryString.length - bitsToHide.length + i, bitsToHide(i))
// binaryStringStringBuilder.setCharAt(i, bitsToHide(i))
}
// Return the binary string builder to string
binaryStringStringBuilder.toString()
}
最佳答案
您的代码中的任何地方都不应包含x和y坐标。内存中像素的布局完全无关紧要，无论如何您都不在乎原始图像。您应该直接使用基础线性pixels数组，并按索引处理像素。(x, y)坐标对人体有害，它们会在您的代码中至少引起一个错误。这是错误的：
val y = i / imageToHideInCopy.getWidth()
val x = i % imageToHideInCopy.getHeight()
您应该使用
getWidth() getWidth()或getHeight() getHeight()，但不要同时使用两者。较小的演示，以了解出了什么问题。假设您的图片是3x5：
scala> val w = 5
w: Int = 5
scala> val h = 3
h: Int = 3
这是您在做什么：
scala> val coords = (0 until 15).map{ i => (i / w, i % h) }
coords: scala.collection.immutable.IndexedSeq[(Int, Int)] = Vector(
(0,0), (0,1), (0,2), (0,0), (0,1),
(1,2), (1,0), (1,1), (1,2), (1,0),
(2,1), (2,2), (2,0), (2,1), (2,2)
)
注意例如
(0,0)和(0,1)出现两次。信息被两次写入那些像素。您应该执行以下任一操作：
scala> val coords = (0 until 15).map{ i => (i / w, i % w) }
coords: scala.collection.immutable.IndexedSeq[(Int, Int)] = Vector(
(0,0), (0,1), (0,2), (0,3), (0,4),
(1,0), (1,1), (1,2), (1,3), (1,4),
(2,0), (2,1), (2,2), (2,3), (2,4)
)
或这个：
scala> val coords = (0 until 15).map{ i => (i / h, i % h) }
coords: scala.collection.immutable.IndexedSeq[(Int, Int)] = Vector(
(0,0), (0,1), (0,2),
(1,0), (1,1), (1,2),
(2,0), (2,1), (2,2),
(3,0), (3,1), (3,2),
(4,0), (4,1), (4,2)
)
如您所见，无论是行优先还是列优先完全无关紧要。因此，您实际上可以扔掉容易出错的代码，并直接使用
pixels。那位操作代码...好吧，那不是位操作代码。字符串和正则表达式不应靠近此代码。只需阅读有关位操作和颜色的十个随机stackoverflow答案（例如something like this或this），或查找一些教程，以了解代码的外观。