CImg<unsigned char> src("image.jpg");
int width = src.width();
int height = src.height();
unsigned char* ptr = src.data(10,10);
如何从rgb
获取ptr
?
答案 0 :(得分:6)
在Ubuntu 10.04上测试,手工制作的3x3 RGB图像保存为test.png
:
sudo apt-get install cimg-dev
源文件cimg_test.cpp
:
#include <iostream>
using namespace std;
#include <CImg.h>
using namespace cimg_library;
int main()
{
CImg<unsigned char> src("test.png");
int width = src.width();
int height = src.height();
cout << width << "x" << height << endl;
for (int r = 0; r < height; r++)
for (int c = 0; c < width; c++)
cout << "(" << r << "," << c << ") ="
<< " R" << (int)src(c,r,0,0)
<< " G" << (int)src(c,r,0,1)
<< " B" << (int)src(c,r,0,2) << endl;
return 0;
}
编译并运行:
g++ cimg_test.cpp -lX11 -lpthread -o cimg_test ./cimg_test 3x3 (0,0) = R0 G0 B0 (0,1) = R255 G0 B0 (0,2) = R0 G255 B0 (1,0) = R0 G0 B255 (1,1) = R128 G128 B128 (1,2) = R0 G0 B128 (2,0) = R128 G0 B0 (2,1) = R0 G128 B0 (2,2) = R255 G255 B255
有效。
答案 1 :(得分:3)
从the CImg documentation - 第34页第6.13节和第120页第8.1.4.16节 - 看起来data
方法可以采用四个参数:x ,y,z和c:
T* data(const unsigned int x, const unsigned int y = 0,
const unsigned int z = 0, const unsigned int c = 0)
...其中c
指的是颜色通道。我猜测如果你的图像确实是一个RGB图像,那么c
使用0,1或2的值会给你在给定x, y
位置的红色,绿色和蓝色分量
例如:
unsigned char *r = src.data(10, 10, 0, 0);
unsigned char *g = src.data(10, 10, 0, 1);
unsigned char *b = src.data(10, 10, 0, 2);
(但这只是猜测!)
修改强>
看起来CImg的操作符()也以类似的方式工作:
unsigned char r = src(10, 10, 0, 0);
答案 2 :(得分:1)
访问数据的最简单方法是使用()
运算符:
unsigned char r = img(10,10,0,0);
unsigned char g = img(10,10,0,1);
unsigned char b = img(10,10,0,2);
你可能会感到困惑,因为CImg存储了非交错的原始数据。即您的原始数据存储在R1, R2, ..., G1, G2, ..., B1, B2, ...
而不是R1, G1, B1, R2, G2, B2, ...
,请参阅:http://cimg.eu/reference/group__cimg__storage.html
.data()
只返回一个指针,所以要像上面那样直接访问数据:
CImg<unsigned char> src("image.jpg");
int width = src.width();
int height = src.height();
unsigned char* ptr = src.data(10,10);
unsigned char r = ptr[0];
unsigned char g = ptr[0+width*height];
unsigned char b = ptr[0+2*width*height];
答案 3 :(得分:-1)