Question

我有以下算法将24位位图转换为由它组成的像素的十六进制字符串表示：

// *data = previously returned data from a call to GetDIBits
// width = width of bmp
// height = height of bmp
void BitmapToString(BYTE *data, int width, int height)
{
    int total = 4*width*height;
    int i;
    CHAR buf[3];
    DWORD dwWritten = 0;
    HANDLE hFile = CreateFile(TEXT("out.txt"), GENERIC_READ | GENERIC_WRITE, 
                                0, NULL, CREATE_ALWAYS, FILE_ATTRIBUTE_NORMAL, NULL);
    for(i = 0; i < total; i++)
    {
        SecureZeroMemory(buf, 3);
        wsprintfA(buf, "%.2X", data[i]);

        // only write the 2 characters and not the null terminator:
        WriteFile(hFile, buf, 2, &dwWritten, NULL);

    }
    WriteFile(hFile, "\0", 2, &dwWritten, NULL);
    CloseHandle(hFile);
}

问题是，我希望它忽略每行末尾的填充。例如，对于所有像素值为＃7f7f7f的2x2位图，out.txt的内容也包含填充字节：

7F7F7F7F7F7F00007F7F7F7F7F7F0000

如何调整循环以避免包含填充零？

Answer 1

在写入时将项目设置更改为不填充到16个字节是一个想法。另一个是知道pad设置的字节数（在你的例子中为16），并使用模数（或和）来确定每行需要跳过多少字节：

  int offset = 0;
  for(i = 0; i < height; i++)
  {
    // only read 3 sets of bytes as the 4th is padding.
    for(j = 0; j < width*3; j++)
    {
        SecureZeroMemory(buf, 3);
        wsprintfA(buf, "%.2X", data[offset]);

        // only write the 2 characters and not the null terminator:
        WriteFile(hFile, buf, 2, &dwWritten, NULL);
        offset++;
    }

    // offset past pad bytes
    offset += offset % 8;
  }

这个解决方案应该可行，但我不会在没有进一步了解你的pad字节的情况下保证它。

跳过位图到字符串转换器中的填充？

1 个答案: