分离 .raw 图像文件的 4 个通道（R、G、G、B）并将它们保存为 C++ 中的有效图像

Question

我必须获取一个 .raw12 文件，分离 4 个通道（R、G、G、B）并将它们保存为有效图像（内存中的 8 位），而不使用任何外部库（.ppm） C++.

您可以阅读有关 raw12 的内容 here and ppm file format here。

我已经编写了代码，但它给了我这个输出。

我已经尝试了很多东西，但它总是给我类似于上面的输出。我认为数据类型转换有问题，但我不确定。

我尝试从 2 天开始调试它，仍然没有运气。

这是这段代码。

#include <fstream>
#include <iostream>
using namespace std;


const int BUFFERSIZE = 4096;


int main () 
{

    ifstream infile;

    infile.open("portrait.raw12", ios::binary | ios::in);

    ofstream outfile;

    outfile.open("Redtwo.ppm", ios::binary);

    //outfile.write("P6 ", 3);
    //outfile.write("1536 2048 ", 8);
    //outfile.write("2048 ", 4);
    //outfile.write("255 ", 4);


    //outfile << "P6\n" << 1536 << "\n" << 2048 << "\n255\n";
    outfile << "P6"     << "\n"
        << 1536  << " "
        << 2048  << "\n"
        << 255   << "\n"
       ;
    uint8_t * bufferRow = new uint8_t[BUFFERSIZE];

    if(!infile)
    {
        cout<<"Failed to open"<<endl;
    }
    int size=1536*2048*3;
    char * RedChannel=new char[size];
    int GreenChannel_1,GreenChannel_2,BlueChannel;
    int rowNum=0;
    int i=0;
    int j=0;
    int pixel=1;
    while(rowNum<3072)
    {
        infile.read(reinterpret_cast<char*>(bufferRow), BUFFERSIZE);
        if(rowNum%2==0)
        {
            while(i<BUFFERSIZE)
            {
                RedChannel[j]=(uint8_t)bufferRow[i];
                GreenChannel_1=((uint8_t)(bufferRow[i+1] & 0x0F) << 4) | ((uint8_t)(bufferRow[i+2] >> 4) & 0x0F);
                i+=3;
                //Collect s;
                //s.r=(char)RedChannel[j];
                //s.g=(char)0;
                //s.b=(char)0;
                //unsigned char c = (unsigned char)(255.0f * (float)RedChannel[j] + 0.5f); 
                //outfile.write((char*) &c, 3);
                //outfile.write((char*) 255, sizeof(c));
                //outfile.write(reinterpret_cast<char*>(&RedChannel), 4);
                if(pixel<=3 && rowNum<5)
                {
                    cout<<"RedChannel: "<<RedChannel[j]<<endl;
                    if(pixel!=3)
                        cout<<"GreenChannel 1: "<<GreenChannel_1<<endl;
                }
                pixel++;
                j++;

            }
            RedChannel[j]='\n';
            j++;

        }
        else
        {
            while(i<BUFFERSIZE)
            {
                GreenChannel_2=(uint8_t)bufferRow[i];
                BlueChannel=((uint8_t)(bufferRow[i+1] & 0x0F) << 4) | ((uint8_t)(bufferRow[i+2] >> 4) & 0x0F);
                i+=3;
                if(pixel<=3 && rowNum<5)
                {
                    cout<<"GreenChannel 2: "<<GreenChannel_2<<endl;
                    if(pixel!=3)
                        cout<<"BlueChannel: "<<BlueChannel<<endl;
                }
                pixel++;
            }
        }
        rowNum++;
        i=0;
        pixel=1;
        if(rowNum<5)
            cout<<" "<<endl;
    }
    infile.close();
    outfile.write(RedChannel, size);
    outfile.close();
}

Github Link To The Code

Answer 1

我已经大大简化了你的代码，让它只输出一个通道，因为你的问题说你应该为每个通道生成一个图像。

现在您有了一些有用的东西，您可以在其他部分添加回来 - 从一些有用的东西开始更容易！我不会为你完成你所有的挑战……那将毫无乐趣，让你没有任何成就感。剩下的你可以做 - 祝你好运！

#include <fstream>
#include <iostream>
using namespace std;

// Enough for one line of the input image
const int BUFFERSIZE = 4096 * 3;

int main(){

    ifstream infile;
    ofstream outfile;

    infile.open("portrait.raw12", ios::binary | ios::in);
    outfile.open("result.pgm", ios::binary);

    // Write single channel PGM file
    outfile << "P5\n2048 1536\n255\n";

    unsigned char * bufferRow = new unsigned char[BUFFERSIZE];

    if(!infile)
    {
        cout<<"Failed to open"<<endl;
    }
    int size=2048*1536;
    unsigned char * RedChannel=new unsigned char[size];
    unsigned char * Redp = RedChannel;

    for(int rowNum=0;rowNum<1536;rowNum++){
        // Read an entire row
        infile.read(reinterpret_cast<char*>(bufferRow), BUFFERSIZE);
        if(rowNum%2==0){
            for(int i=0;i<BUFFERSIZE;i+=3){
                *Redp++=bufferRow[i];
            }
        }
    }
    infile.close();
    outfile.write(reinterpret_cast<char*>(RedChannel), size);
    outfile.close();
}

分离 .raw 图像文件的 4 个通道（R、G、G、B）并将它们保存为 C++ 中的有效图像

Seperate 4 channels (R,G,G,B) of .raw image file and save them as valid image in c++

c++

rgb

image-processing

ppm

raw-file