Question

我使用lodePNG库对png图像进行编码，并使用导入的txt文件更改像素的LSB。我已经编译了程序，但我不确定PNG文件是否实际上是按照我的按位操作进行编码的。

lodePNG库从PNG图像解码/编码，并将像素存储在矢量＆＃34;图像＆＃34;，每像素4个字节，有序RGBARGBA ...，

void decodeOneStep(const char* filename)
{

      unsigned width, height;

      //decode
      unsigned error = lodepng::decode(image, width, height, filename);

      //if there's an error, display it
      if (error) std::cout << "decoder error " << error << ": " << 
          lodepng_error_text(error) << std::endl;
}

程序获取文本文件和PNG文件的命令行参数。我还没有为参数包含错误检查。

int const MAX_SIZE = 100; 
std::vector<unsigned char> image;

int main(int argc, char *argv[])
{
const char* filename;
char* textArray = new char[MAX_SIZE];

std::ifstream textfile;
textfile.open(argv[1]);

int numCount = 0;
while (!textfile.eof() && numCount < MAX_SIZE)
{
    textfile.get(textArray[numCount]); //reading single character from file to array
    numCount++;
}

textfile.close();

    filename = argv[2];
    decodeOneStep(filename);

    unsigned width = 512, height = 512;
    image.resize(width * height * 4);

    int pixCount = 0; 

    for (int i = 0; i < numCount - 1; i++) {

        std::cout << textArray[i]; 

        for (int j = 0; j < 8; j++) {

            std::cout << ((textArray[i]) & 1); //used to see actual bit value.
            image[pixCount++] |= ((textArray[i]) & 1);
            (textArray[i]) >>= 1; 
        }                       
        std::cout << std::endl; 
    } 

encodeOneStep(filename, image, width, height);

在for循环中，我将遍历向量中的每个像素，并用char中的一些位替换LSB。由于char是8个字节，因此for循环循环8次。该程序适用于大多数PNG图像和不超过大小的文本，但我不确定按位操作实际上是在做什么。另外，我如何移位这些位以便将char位从MSB存储到LSB？我觉得我对像素值（位）如何存储在数组中的方法有所了解。

编辑：测试我已经运行了新的位操作：

    for (int j = 7; j >= 0; j--) {

        //These tests were written to see if the 4-bits of the pixels were actually being replaced.
        //The LSB of the pixel bits are replaced with the MSB of the text character. 

        std::cout <<"Initial pixel 4-bits: " << std::bitset <4>(image[pixCount]) << "  "; 
        std::cout << "MSB of char: " << ((textArray[i] >> j) & 0x01) << " ";
        std::cout << "Pixel LSB replaced: " << ((image[pixCount] & mask) | ((textArray[i] >> j) & 0x01)) << " ";

        image[pixCount] = (image[pixCount] & mask) | ((textArray[i] >> j) & 0x01);
        pixCount++;

        std::cout << std::endl;

    }

测试结果：

For char 'a' 
Initial pixel 4-bits : 0000 MSB: 0 Pixel LSB replaced: 0
Initial pixel 4-bits : 0001 MSB: 1 Pixel LSB replaced: 1
Initial pixel 4-bits : 0001 MSB: 1 Pixel LSB replaced: 1
Initial pixel 4-bits : 0000 MSB: 0 Pixel LSB replaced: 0
Initial pixel 4-bits : 0000 MSB: 0 Pixel LSB replaced: 0
Initial pixel 4-bits : 0000 MSB: 0 Pixel LSB replaced: 0
Initial pixel 4-bits : 0000 MSB: 0 Pixel LSB replaced: 0
Initial pixel 4-bits : 0001 MSB: 1 Pixel LSB replaced: 1

Answer 1

首先需要在嵌入秘密位之前清除像素的lsb。

unsigned char mask = 0xfe;  // in binary 11111110

// and in your loop
image[pixCount] = (image[pixCount] & mask) | (textArray[i] & 1);
pixCount++;

如果要将秘密的每个字符的位从最高位嵌入到最低位，则需要倒计时j循环。

for (int j = 7; j >= 0; j--) {
    image[pixCount] = (image[pixCount] & mask) | ((textArray[i] >> j) & 0x01);
    pixCount++;
}

编辑：为了解释上面的代码，image[pixCount] & mask是像素和所选掩码值（二进制1111110）之间的AND运算符，因此结果是lsb被清除

(textArray[i] >> j) & 0x01将您的字符向左移动j并仅保留lsb。如果你绘制出数学数据，这就是你得到的数据

// assume our character has the bits `abcdefgh`

j = 7
(character >> j) & 0x01 = 0000000a & 0x01 = a

j = 6
(character >> j) & 0x01 = 000000ab & 0x01 = b

j = 5
(character >> j) & 0x01 = 00000abc & 0x01 = c

// and so on

j = 0
(character >> j) & 0x01 = abcdefgh & 0x01 = h

隐写术：如何改变像素阵列的LSB

1 个答案: