写入 BMP 数据出现垃圾
我正在努力理解和绘制我自己的 PDF417（二维条形码）DLL。 无论如何，文件的实际绘制是完美的，并且在 32 位的正确边界内（作为单色结果）。 写入数据时，以下是从 C++ Visual Studio 内存转储复制到 bmp 缓冲区的指针的内存转储。 每行在下一行之前都正确分配为 36 宽。
对于帖子中的自动换行感到抱歉，但我的输出本来是与内存转储相同的 36 字节宽，这样您就可以更好地看到失真情况。
当前绘图为 273 像素宽 x 12 像素高，单色...
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
以下是写出文件的代码 - 在从上面进行内存转储时立即逐字记录
FILE *stream;
if( fopen_s( &stream, cSaveToFile, "w+" ) == 0 )
{
fwrite( &bmfh, 1, (UINT)sizeof(BITMAPFILEHEADER), stream );
fwrite( &bmi, 1, (UINT)sizeof(BITMAPINFO), stream );
fwrite( &RGBWhite, 1, (UINT)sizeof(RGBQUAD), stream );
fwrite( ppvBits, 1, (UINT)bmi.bmiHeader.biSizeImage, stream );
fclose( stream );
}
以下是实际写入文件的内容。
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00
00 00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb
00 00 00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02
eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0d 0a
8e 02 eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c
0d 0a 8e 02 eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79
bc 5c 0d 0a 8e 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00
请注意，从第 4 行读回文件的结果中的“0d”开始失真，大约是第 15 个字节...然后，还有一些交错，总共使图像倾斜了9 个字节的价值...
显然，绘图部分工作正常，因为 12 行的所有内容都在内存中保持正确对齐。
I'm working on understanding and drawing my own DLL for PDF417 (2d barcodes). Anyhow, the actual drawing of the file is perfect, and in correct boundaries of 32 bits (as monochrome result). At the time of writing the data, the following is a memory dump as copied from C++ Visual Studio memory dump of the pointer to the bmp buffer. Each row is properly allocated to 36 wide before the next row.
Sorry about the wordwrap in the post, but my output was intended to be the same 36 bytes wide as the memory dump so you could better see the distortion.
The current drawing is 273 pixels wide by 12 pixels high, monochrome...
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
Here is the code to WRITE the file out -- verbatim immediately at the time of the memory dump from above
FILE *stream;
if( fopen_s( &stream, cSaveToFile, "w+" ) == 0 )
{
fwrite( &bmfh, 1, (UINT)sizeof(BITMAPFILEHEADER), stream );
fwrite( &bmi, 1, (UINT)sizeof(BITMAPINFO), stream );
fwrite( &RGBWhite, 1, (UINT)sizeof(RGBQUAD), stream );
fwrite( ppvBits, 1, (UINT)bmi.bmiHeader.biSizeImage, stream );
fclose( stream );
}
Here's what ACTUALLY Gets written to the file.
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00
00 00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb
00 00 00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02
eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0d 0a
8e 02 eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c
0d 0a 8e 02 eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79
bc 5c 0d 0a 8e 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00
Notice the start of the distortion with the "0d" in the result from reading the file back in the 4th line, about the 15th byte over... Then, there are a few more staggered around which in total, skew the image off by 9 bytes worth...
Obviously, the drawing portion is working ok as everything remains properly aligned in memory for the 12 lines.
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评论（3）
难道你不应该以复合模式（即可写和可写）打开文件吗？ 二进制如
wb+中所示？这是回车符 (CR) 的 ASCII 代码——在某些操作系统上添加了换行符（其中换行符实际上是 CR/LF 的序列）。 一旦您开始以二进制模式写入输出，这种情况就会消失。
否则你的代码看起来很整洁。 干杯!
Shouldn't you open the file in a compound mode i.e. writable & binary as in
wb+?That's ASCII code for Carriage Return (CR) -- added on some OSes with newline (where a newline is actually a sequence of CR/LF). This should go away once you start writing the output in binary mode.
Your code looks neat otherwise. Cheers!
您的
0x0A(\n) 会转换为 DOS 格式0x0D0A(\r\n)，因为您'以文本模式重新写入文件。 切换到二进制模式。Your
0x0A(\n) gets converted to DOS format0x0D0A(\r\n), becouse you're write the file in text mode. Switch to binary mode.实际上我只是在java中做了类似的事情（将bmp数据打印到热敏收据打印机）。 我想与您分享以下几点：
bmp 图像数据！= 微软的图像格式。 MS 位图在任何图像数据之前有大约 54 字节的标头信息。 （我花了一两天的时间才意识到其中的差异）
bmp 图像数据从左到右、从上到下读取，最高有效位位于左侧。
确保条形码图像的位深度为 1。这意味着 1 位 = 1 像素。 十六进制“ab”在二进制中为 10101011，这 8 个像素将被相应填充。
如果您的条形码为 36 字节宽，则条形码分辨率为 288 x 12，而不是 273 x 12。(36 * 8 = 288)。
图像数据的大小应为 432 字节（12 行，每行 36 字节）。
我不知道这意味着什么：
<块引用>
无论如何，文件的实际绘制是完美的，并且在 32 位的正确边界内（作为单色结果）。
单色意味着它是一种颜色或另一种颜色。 像素（认为位）要么被填充，要么未被填充。
希望这可以帮助
I actually just did a similar thing in java (printing bmp data to a thermal receipt printer). There are a couple of things i want to share with you:
bmp image data != an image format from microsoft. the MS bitmap has about 54 bytes of header information before any image data. (i spent a day or two working on this before I realized the difference)
bmp image data reads left to right, top to bottom, with the most significant bit on the left.
make sure the barcode image has a bitdepth of 1. this means 1 bit = 1 pixel. hexidecimal "ab" is 10101011 in binary, those 8 pixels will be filled in accordingly.
if you have a barcode 36 bytes wide, the barcode resolution is 288 x 12, not 273 x 12. (36 * 8 = 288).
the image data should be 432 bytes in size (12 rows of 36 bytes).
i dont know what this means:
monochrome means its either 1 color or another. the pixel (think bit) is either filled in or it isnt.
Hope this helps