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

转载 作者:行者123 更新时间:2023-11-30 03:18:26 28 4
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我必须获取一个 .raw12 文件,分离 4 个 channel (R、G、G、B)并将它们保存为有效图像(内存中的 8 位),而无需使用 C++ 中的任何外部库 (.ppm)。

您可以阅读有关 raw12 here 的信息和 ppm 文件格式 here .

我已经编写了代码,但它给了我这个输出。

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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

最佳答案

我已经大大简化了您的代码并使其仅输出一个 channel ,因为您的问题是您应该为每个 channel 生成一个图像。

现在您有了一些有用的东西,您可以重新添加其他部分 - 从有用的东西开始会更容易!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。您可以完成剩下的工作 - 祝您好运!

#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();
}

enter image description here

关于c++ - 分离 .raw 图像文件的 4 个 channel (R、G、G、B)并将它们保存为 C++ 中的有效图像,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54764740/

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