我想读取BMP文件中所有像素的RGB值。 我在C++代码,这看起来像:从位图读取RGB - 错误的值(RGB); C++
#include <iostream>
#include <fstream>
using namespace std;
int main() {
FILE *streamIn;
streamIn = fopen("./Untitled.bmp", "r");
if (streamIn == (FILE *)0) {
printf("File opening error ocurred. Exiting program.\n");
return 0;
}
unsigned char info[54];
fread(info, sizeof(unsigned char), 54, streamIn);
int width = *(int*)&info[18];
int height = *(int*)&info[22];
int image[width*height][3];
int i = 0;
for(i=0;i<width*height;i++) {
image[i][2] = getc(streamIn);
image[i][1] = getc(streamIn);
image[i][0] = getc(streamIn);
printf("pixel %d : [%d,%d,%d]\n",i+1,image[i][0],image[i][1],image[i][2]);
}
fclose(streamIn);
return 0;
}
和像这样的图像(网格重叠):
其是6x12像素与两种颜色文件 - 黑色和白色。
我试图找出为什么在执行上述代码后,将图像作为参数,我没有得到像素与RGB:[0,0,0]和[255,255,255],但也[0,248,0], [7,224,0]等。该文件的
十六进制转储:
0000-0010: 42 4d 9a 01-00 00 00 00-00 00 7a 00-00 00 6c 00 BM...... ..z...l.
0000-0020: 00 00 08 00-00 00 0c 00-00 00 01 00-18 00 00 00 ........ ........
0000-0030: 00 00 20 01-00 00 13 0b-00 00 13 0b-00 00 00 00 ........ ........
0000-0040: 00 00 00 00-00 00 42 47-52 73 00 00-00 00 00 00 ......BG Rs......
0000-0050: 00 00 00 00-00 00 00 00-00 00 00 00-00 00 00 00 ........ ........
0000-0060: 00 00 00 00-00 00 00 00-00 00 00 00-00 00 00 00 ........ ........
0000-0070: 00 00 00 00-00 00 00 00-00 00 02 00-00 00 00 00 ........ ........
0000-0080: 00 00 00 00-00 00 00 00-00 00 ff ff-ff ff ff ff ........ ........
0000-0090: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-00a0: ff ff ff ff-ff 00 00 00-00 00 00 00-00 00 00 00 ........ ........
0000-00b0: 00 00 00 00-00 00 00 ff-ff ff ff ff-ff 00 00 00 ........ ........
0000-00c0: 00 00 00 00-00 00 00 00-00 00 00 00-00 00 00 ff ........ ........
0000-00d0: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-00e0: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-00f0: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-0100: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-0110: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-0120: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-0130: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-0140: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-0150: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-0160: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-0170: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-0180: ff ff ff ff-ff ff ff ff-ff ff ff ff-ff ff ff ff ........ ........
0000-0190: ff ff 00 00-00 ff ff ff-00 00 00 ff-ff ff 00 00 ........ ........
0000-019a: 00 ff ff ff-00 00 00 ff-ff ff ........ ..
文件大小为410个字节。它应该是270(6 * 12 * 3 + 54)。这意味着这个文件中有一些额外的信息。
你确定格式是24位BMP,而不是一个PNG重命名? – 2014-10-29 13:37:05
6像素宽的BMP图像将在每行像素之后填充。 (一行像素必须包含可以被4整除的字节数,所以6x3在每个像素之后有两个填充像素,您还应该查看适当的头信息以确保您的BMP是24位RGB bitmap – 2014-10-29 13:38:54
位图文件格式不是那么容易(它取决于图像格式是如何写在磁盘上的)这就是为什么1)为什么你不使用WinSDK结构的BMP? 2)它非常不可移植(读取一个字节数组并访问未对齐的整数...) – 2014-10-29 13:39:25