我必须使用.raw12文件,将4个通道(R,G,G,B)分开并将它们保存为有效图像(内存中为8位),而不使用C ++中的任何外部库(.ppm)。
我已经写了代码,但它给了我这个输出。
我尝试了很多东西,但它总是给我类似于上面的输出。我认为数据类型转换存在问题,但我不确定。
我试图从2天调试它,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();
}
我已经将代码简化了很多,并且只输出了一个通道,因为你的问题是你应该为每个通道生成一个图像。
现在你有一些工作,你可以添加回其他部分 - 从更有效的东西开始更容易!我不会为你做所有的挑战......这将是没有乐趣的,让你没有任何成就感。你可以做其余的事 - 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();
}