So, I'm trying to understand how PNG to BMP conversion actually takes place. I created the following 8x8 pixel PNG using a tool online:
I then performed a conversion using the ImageMagik tool in OSX terminal:
$ convert -monochrome pic.png pic.bmp
Afterwards, I did a hexdump of the image:
$ hexdump -C pic.bmp
00000000 42 4d 4a 01 00 00 00 00 00 00 8a 00 00 00 7c 00 |BMJ...........|.|
00000010 00 00 08 00 00 00 08 00 00 00 01 00 18 00 00 00 |................|
00000020 00 00 c0 00 00 00 00 00 00 00 00 00 00 00 00 00 |................|
00000030 00 00 00 00 00 00 00 00 ff 00 00 ff 00 00 ff 00 |................|
00000040 00 00 00 00 00 ff 42 47 52 73 8f c2 f5 28 51 b8 |......BGRs...(Q.|
00000050 1e 15 1e 85 eb 01 33 33 33 13 66 66 66 26 66 66 |......333.fff&ff|
00000060 66 06 99 99 99 09 3d 0a d7 03 28 5c 8f 32 00 00 |f.....=...(\.2..|
00000070 00 00 00 00 00 00 00 00 00 00 04 00 00 00 00 00 |................|
00000080 00 00 00 00 00 00 00 00 00 00 ff ff ff ff ff ff |................|
00000090 ff ff ff ff ff ff ff ff ff ff ff ff ff ff ff ff |................|
*
00000130 ff ff 00 00 00 ff ff ff ff ff ff ff ff ff ff ff |................|
00000140 ff ff ff ff ff ff ff ff ff ff |..........|
0000014a
So, obviously we have a BMP header from 0x00 - 0x8C. Then we have FF for white pixels and 00 for black. That all makes sense, but the structure of the output doesn't. So far, it seems that there is a 3 to 1 ratio. 3 bytes per pixel. I'm assuming this means black, white, no-color?
I need to fully understand why there are 3 bytes per pixel instead of 2. Ideally, I would like a binary 1 or 0 for each pixel instead. Is there a way to do this? And if not, can someone please explain the layout of bytes? Specifically: Why are the 3 00s at 0x132 - 0x134 and not the very beginning?
Thanks
Imagemagick, in fact, does support 1-bit BMP, if using either BMP3 or BMP2 formats, but not BMP4 which is just BMP. So if you want 1-bit depth per channel, then do
Input:
convert image.png -depth 1 BMP3:result.bmp
If you want a 1-bit depth for the whole image (i.e. binary), then do
convert image.png -depth 1 -type bilevel BMP3:result2.bmp
Or
convert image.png -depth 1 -threshold 50% BMP3:result3.bmp
I'm not sure which question this will answer, but from the Usage docs
IM can not produce BMP's at depth levels other than 8. However you can
use NetPBM image processing set to do the final conversion to other depth
levels (This needs at least a Q16 version of IM)...
So if we leverage NetPBM
$ convert pic.png -depth 1 ppm:- | hexdump
Which gives (Note the values are 0 & 1, but you still have a RGB channels)
0000000 50 36 0a 38 20 38 0a 31 0a 00 00 00 01 01 01 01
0000010 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01
*
00000c0 01 01 01 01 01 01 01 01 01
00000c9
So a workaround my be as simple as
$ convert pic.png -depth 1 ppm:- | pnmdepth 1 | ppm2bmp > pic.bmp
On Fedora 27 that command is:
$ convert pic.png -depth 1 ppm:- | pnmdepth 1 | ppmtobmp > pic.bmp
You need:
$ dnf install netpbm-progs
Related
I want to write to a mifare classic 4k, using the following APDU command (UPDATE BINARY):
APDU = {FF D6 00 20 10 00 01 02 03 04 05 06 07 08 09 0A 0B 0C 0D 0E 0Fh}
It seems fine as a get a 90 00 result...
but when i read the card back I always got the following (even with different data...):
sector: 8 (block 32), auth OK
032: D5 41 00 EA 00 FF 13 3E 86 6A 00 00 00 00 69 FF
033: D5 41 00 EA 00 FF 13 3E 86 6A 00 00 00 00 69 FF
034: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
035: 00 00 00 00 00 00 FF 07 80 69 FF FF FF FF FF FF
where does this string D5 41 00 EA 00 FF 13 3E 86 6A 00 00 00 00 69 FF come from?
Note that i didn't change any setting on the card and was properly authenticated. It was a blank card and i didn't touch the trailer.
I m using a ACR122 reader (this command comes direct from the documentation of the reader...)
Ok i found my problem, i was setting the wrong size for the cbSendLength parameter in SCardTransmit.
Now i set the correct one (the whole size of the APDU command: 21) and it works fine.
Sorry.
I'm implementing a PNG encoder in VHDL for learning purposes. It works with image sizes from 1x1 to 4x4. At the image size of 5x5 there is a behaviour I can't understand:
When encoding raw data with values 0...24, the encoding works. However, when using raw data with values 255...231, it generates a broken image.
Input values 0...24:
> hexdump -C png_encoder/gen/test_img_no_compression_5x5.png
00000000 89 50 4e 47 0d 0a 1a 0a 00 00 00 0d 49 48 44 52 |.PNG........IHDR|
00000010 00 00 00 05 00 00 00 05 08 00 00 00 00 a8 04 79 |...............y|
00000020 39 00 00 00 4c 49 44 41 54 78 01 00 04 00 fb ff |9...LIDATx......|
00000030 00 00 01 02 00 04 00 fb ff 03 04 00 05 00 04 00 |................|
00000040 fb ff 06 07 08 09 00 04 00 fb ff 00 0a 0b 0c 00 |................|
00000050 04 00 fb ff 0d 0e 00 0f 00 04 00 fb ff 10 11 12 |................|
00000060 13 00 04 00 fb ff 00 14 15 16 01 02 00 fd ff 17 |................|
00000070 18 0b a4 01 2d d5 1f a2 6d 00 00 00 00 49 45 4e |....-...m....IEN|
00000080 44 ae 42 60 82 |D.B`.|
00000085
> pngcheck -vv png_encoder/gen/test_img_no_compression_5x5.png
File: png_encoder/gen/test_img_no_compression_5x5.png (133 bytes)
chunk IHDR at offset 0x0000c, length 13
5 x 5 image, 8-bit grayscale, non-interlaced
chunk IDAT at offset 0x00025, length 76
zlib: deflated, 32K window, superfast compression
row filters (0 none, 1 sub, 2 up, 3 avg, 4 paeth):
0 0 0 0 0 (5 out of 5)
chunk IEND at offset 0x0007d, length 0
No errors detected in png_encoder/gen/test_img_no_compression_5x5.png (3 chunks, -432.0% compression).
Input values 255...231:
> hexdump -C png_encoder/gen/test_img_no_compression_5x5.png
00000000 89 50 4e 47 0d 0a 1a 0a 00 00 00 0d 49 48 44 52 |.PNG........IHDR|
00000010 00 00 00 05 00 00 00 05 08 00 00 00 00 a8 04 79 |...............y|
00000020 39 00 00 00 4c 49 44 41 54 78 01 00 04 00 fb ff |9...LIDATx......|
00000030 00 ff fe fd 00 04 00 fb ff fc fb 00 fa 00 04 00 |................|
00000040 fb ff f9 f8 f7 f6 00 04 00 fb ff 00 f5 f4 f3 00 |................|
00000050 04 00 fb ff f2 f1 00 f0 00 04 00 fb ff ef ee ed |................|
00000060 ec 00 04 00 fb ff 00 eb ea e9 01 02 00 fd ff e8 |................|
00000070 e7 6a 21 17 bc 9a 17 87 e7 00 00 00 00 49 45 4e |.j!..........IEN|
00000080 44 ae 42 60 82 |D.B`.|
00000085
> pngcheck -vv png_encoder/gen/test_img_no_compression_5x5.png
File: png_encoder/gen/test_img_no_compression_5x5.png (133 bytes)
chunk IHDR at offset 0x0000c, length 13
5 x 5 image, 8-bit grayscale, non-interlaced
chunk IDAT at offset 0x00025, length 76
zlib: deflated, 32K window, superfast compression
row filters (0 none, 1 sub, 2 up, 3 avg, 4 paeth):
zlib: inflate error = -3 (data error)
(0 out of 5)
ERRORS DETECTED in png_encoder/gen/test_img_no_compression_5x5.png
How to interpret the error message zlib: inflate error = -3 (data error)?
I read https://www.zlib.net/zlib_how.html, but didn't find more specific information. My first guess was the row filters are incorrect, but since both files are structured the same, this is unlikely. Is there something wrong with the ADLER32 calculation in the second case (possibly some overflow)?
It was an overflow in the ADLER32 checksum calculation. Specifically, there were two 16 bit numbers added and truncated before applying the modulo with 65521. Unfortunately my ADLER32 unittest didn't catch it, yet.
However, the error message was shown several times during the implementation and I was always not sure about the cause. If anybody could elaborate the error message or explain how to get a better error message, I would be glad to hear it.
I currently took a photo with a camera which returned to me all its values in decimal and after extracting the values my jpeg file are(i converted them to hex here but they were still saved in decimal)
FF
D8
FF
FE
00
24
30
00
9D
08
00
00
00
00
00
00
00
00
00
00
00
00
00
00
00
F0
00
40
01
0C
00
32
............
00
54
65
6B
D9
9F
FF
FF
D9
i saved theses value into a file called image.jpg (they were saved in decimal)
however the image file was not able to read it and I can not use the linux command convert as it tells me "Bogus marker length". What would i be doing wrong that doesnt allow me to see the image?
I am new to this area; so excuse me if I did not describe the issue well.
I have a hex file which should contain three images (width: 640, height: 333). The hex file is 1.2 MB. So if we do a little calculation, we obtain that each pixel should have 16-bit of data.
Some of hex code of the file is as below:
90 eb 6f 14 02 02 fd fd 4e 01 80 02 00 00 00 00
90 eb 6f 14 82 82 7d 7d 4e 01 80 02 03 00 00 00
90 eb 6f 14 c2 c2 3d 3d 4e 01 80 02 00 00 8e 08
a7 33 0f d4 00 01 00 01 00 01 43 01 f8 03 0e 17
00 01 00 00 00 00 02 00 00 00 00 00 00 00 01 00
00 04 00 00 01 01 00 00 00 00 00 00 00 00 00 00
00 00 01 01 00 00 00 00 00 01 00 00 00 00 00 00
00 00 00 00 00 00 00 00 01 00 00 00 00 00 00 00
00 00 00 00 00 00 00 01 00 00 00 00 00 00 01 01
00 00 00 00 01 00 00 00 00 00 00 00 00 00 01 01
00 00 00 00 00 00 01 00 00 00 00 00 00 00 00 00
00 01 00 00 08 01 00 00 00 00 00 00 00 00 00 00
00 01 00 00 00 00 00 01 00 00 00 00 00 0a 01 00
00 00 01 00 01 00 00 00 01 00 00 00 00 00 00 00
02 00 00 01 00 00 00 01 00 00 00 00 00 00 00 01
As you can see there is a header with 4 row content which 3 of them is similar. the bitmap data begins. The repeating row is repeated another two times in the file. So I assume that the repeating structure is at the beginning of each of 3 images. But the data between these headers are 210 KB, which means 8-bit per pixel. So I read every 8-bit as an tiny integer and set it to rgb of corresponding pixels of images. So I obtained 3 gray images. Also 630 KB of data remains unread in the file.
Here is magnified version of the original image as gray (original version of the picture is colored) and obtained image. As you see, there are some pixels (every other pixel) which are totally different in compare to original pixels, but whole image is almost correct.
So, my questions are as below:
What is the true structure of the hex file? How should I read the hex file?
How can I achieve original colored file?
What is the extra 630 KB of data!? And what is the wrong pixels?!
also here is original image(i.stack.imgur.com/NdBOa.png), original image as gray(i.stack.imgur.com/wDUPB.png) and obtained image (i.stack.imgur.com/lY3ib.png).
Nothing conclusive, but here is what I have found...
If you hex dump the file and look at the start, you find 90eb and if you look for that throughout the file, you get this:
xxd a.raw | egrep "90eb"
0000000: 90eb 6f14 0202 fdfd 4e01 8002 0000 0000 ..o.....N.......
0000010: 90eb 6f14 8282 7d7d 4e01 8002 0300 0000 ..o...}}N.......
0000020: 90eb 6f14 c2c2 3d3d 4e01 8002 0000 8e08 ..o...==N.......
0034340: e773 2bf4 90eb 6f14 c2c2 3d3d 4e01 8002 .s+...o...==N...
0068660: 0301 0100 ca03 0104 90eb 6f14 c2c2 3d3d ..........o...==
The data looks to begin 32 bytes after every 90eb. If the image is 640x333, there will be 213,120 bytes per image. So we can extract the basic planes/channels of the image like this with ImageMagick:
dd if=a.raw bs=1 skip=64 count=213120 | convert -depth 8 -size 640x333 gray:- a.png
dd if=a.raw bs=1 skip=213860 count=213120 | convert -depth 8 -size 640x333 gray:- b.png
dd if=a.raw bs=1 skip=427656 count=213120 | convert -depth 8 -size 640x333 gray:- c.png
Now we have a problem - the individual images are not positioned the same in all three images - you can see that if I animate the 3 frames together like this:
convert -delay 80 a.png b.png c.png -normalize anim.gif
So now I am a bit lost - are there multiple cameras since the viewpoint seems to move?
I don't know - maybe my findings will inspire someone else! Let's see.
Another approach may be to compare the statistics - if you look at the "original" image's statistics, you get this:
identify -verbose original.png | egrep "Red:|Green:|Blue:|mean:|deviation"
Red:
mean: 5.77718 (0.0226556)
standard deviation: 17.0501 (0.066863)
Green:
mean: 13.7015 (0.0537312)
standard deviation: 38.4053 (0.150609)
Blue:
mean: 10.2863 (0.0403386)
standard deviation: 30.1792 (0.11835)
If you now look at the statistics for the a.png, b.png and c.png as extracted above, you get this:
identify -verbose a.png | egrep "Red:|Green:|Blue:|Gray:|mean:|deviation"
Gray:
mean: 2.48532 (0.00974635)
standard deviation: 9.00678 (0.0353207)
identify -verbose b.png | egrep "Red:|Green:|Blue:|Gray:|mean:|deviation"
Gray:
mean: 10.1611 (0.0398473)
standard deviation: 30.2288 (0.118544)
identify -verbose c.png | egrep "Red:|Green:|Blue:|Gray:|mean:|deviation"
Gray:
mean: 2.26135 (0.00886804)
standard deviation: 7.43093 (0.0291409)
And there doesn't seem to be any correlation between the statistics of the "original" image and the putative "channels" of the extracted images... I think there is more going on here than I can guess.
I'd like to screen some jpegs for validity before I send them across the network for more extensive inspection. It is easy enough to check for a valid header and footer, but what is the smallest size (in bytes) a valid jpeg could be?
A 1x1 grey pixel in 125 bytes using arithmetic coding, still in the JPEG standard even if most decoders can't decode it:
ff d8 : SOI
ff e0 ; APP0
00 10
4a 46 49 46 00 01 01 01 00 48 00 48 00 00
ff db ; DQT
00 43
00
03 02 02 02 02 02 03 02
02 02 03 03 03 03 04 06
04 04 04 04 04 08 06 06
05 06 09 08 0a 0a 09 08
09 09 0a 0c 0f 0c 0a 0b
0e 0b 09 09 0d 11 0d 0e
0f 10 10 11 10 0a 0c 12
13 12 10 13 0f 10 10 10
ff c9 ; SOF
00 0b
08 00 01 00 01 01 01 11 00
ff cc ; DAC
00 06 00 10 10 05
ff da ; SOS
00 08
01 01 00 00 3f 00 d2 cf 20
ff d9 ; EOI
I don't think the mentioned 134 byte example is standard, as it is missing an EOI. All decoders will handle this but the standard says it should end with one.
That file can be generated with:
#!/usr/bin/env bash
printf '\xff\xd8' # SOI
printf '\xff\xe0' # APP0
printf '\x00\x10'
printf '\x4a\x46\x49\x46\x00\x01\x01\x01\x00\x48\x00\x48\x00\x00'
printf '\xff\xdb' # DQT
printf '\x00\x43'
printf '\x00'
printf '\x03\x02\x02\x02\x02\x02\x03\x02'
printf '\x02\x02\x03\x03\x03\x03\x04\x06'
printf '\x04\x04\x04\x04\x04\x08\x06\x06'
printf '\x05\x06\x09\x08\x0a\x0a\x09\x08'
printf '\x09\x09\x0a\x0c\x0f\x0c\x0a\x0b'
printf '\x0e\x0b\x09\x09\x0d\x11\x0d\x0e'
printf '\x0f\x10\x10\x11\x10\x0a\x0c\x12'
printf '\x13\x12\x10\x13\x0f\x10\x10\x10'
printf '\xff\xc9' # SOF
printf '\x00\x0b'
printf '\x08\x00\x01\x00\x01\x01\x01\x11\x00'
printf '\xff\xcc' # DAC
printf '\x00\x06\x00\x10\x10\x05'
printf '\xff\xda' # SOS
printf '\x00\x08'
printf '\x01\x01\x00\x00\x3f\x00\xd2\xcf\x20'
printf '\xff\xd9' # EOI
and opened fine with GNOME Image Viewer 3.38.0 and GIMP 2.10.18 on Ubuntu 20.10.
Here's an upload on Imgur. Note that Imgur process the file making it larger however if you download it to check, and as seen below, the width=100 image shows white on Chromium 87:
It occurs to me you could make a progressive jpeg with only the DC coefficients, that a single grey pixel could be encoded in 119 bytes. This reads just fine in a few programs I've tried it in (Photoshop, GNOME Image Viewer 3.38.0, GIMP 2.10.18, and others).
ff d8 : SOI
ff db ; DQT
00 43
00
01 01 01 01 01 01 01 01
01 01 01 01 01 01 01 01
01 01 01 01 01 01 01 01
01 01 01 01 01 01 01 01
01 01 01 01 01 01 01 01
01 01 01 01 01 01 01 01
01 01 01 01 01 01 01 01
01 01 01 01 01 01 01 01
ff c2 ; SOF
00 0b
08 00 01 00 01 01 01 11 00
ff c4 ; DHT
00 14
00
01 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00
03
ff da ; SOS
00 08
01 01 00 00 00 01 3F
ff d9 ; EOI
The main space savings is to only have one Huffman table. Although this is slightly smaller than the 125 byte arithmetic encoding given in another answer, the arithmetic encoding without the JFIF header would be smaller yet (107 bytes), so that should still be considered the smallest known.
The above file can be generated with:
#!/usr/bin/env bash
printf '\xff\xd8' # SOI
printf '\xff\xdb' # DQT
printf '\x00\x43'
printf '\x00'
printf '\x01\x01\x01\x01\x01\x01\x01\x01'
printf '\x01\x01\x01\x01\x01\x01\x01\x01'
printf '\x01\x01\x01\x01\x01\x01\x01\x01'
printf '\x01\x01\x01\x01\x01\x01\x01\x01'
printf '\x01\x01\x01\x01\x01\x01\x01\x01'
printf '\x01\x01\x01\x01\x01\x01\x01\x01'
printf '\x01\x01\x01\x01\x01\x01\x01\x01'
printf '\x01\x01\x01\x01\x01\x01\x01\x01'
printf '\xff\xc2' # SOF
printf '\x00\x0b'
printf '\x08\x00\x01\x00\x01\x01\x01\x11\x00'
printf '\xff\xc4' # DHT
printf '\x00\x14'
printf '\x00'
printf '\x01\x00\x00\x00\x00\x00\x00\x00'
printf '\x00\x00\x00\x00\x00\x00\x00\x00'
printf '\x03'
printf '\xff\xda' # SOS
printf '\x00\x08'
printf '\x01\x01\x00\x00\x00\x01\x3F'
printf '\xff\xd9' # EOI
Try the following (134 bytes):
FF D8 FF E0 00 10 4A 46 49 46 00 01 01 01 00 48 00 48 00 00
FF DB 00 43 00 FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF
FF FF FF FF FF FF FF FF FF FF C2 00 0B 08 00 01 00 01 01 01
11 00 FF C4 00 14 10 01 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 FF DA 00 08 01 01 00 01 3F 10
Source: Worlds Smallest, Valid JPEG? by Jesse_hz
Found "the tiniest GIF ever" with only 26 bytes.
47 49 46 38 39 61 01 00 01 00
00 ff 00 2c 00 00 00 00 01 00
01 00 00 02 00 3b
Python literal:
b'GIF89a\x01\x00\x01\x00\x00\xff\x00,\x00\x00\x00\x00\x01\x00\x01\x00\x00\x02\x00;'
While I realize this is far from the smallest valid jpeg and has little or nothing to do with your actual question, I felt I should share this as I'd been looking for a very small JPEG that actually looked like something to do some testing with when i'd found your question. I'm sharing it here because its valid, its small, and it makes me ROFL.
Here is a 384 byte JPEG image that I made in photoshop. It is the letters ROFL hand drawn by me and then saved with max compression settings while still being sort of readable.
Hex sequences:
my #image_hex = qw{
FF D8 FF E0 00 10 4A 46 49 46 00 01 02 00 00 64
00 64 00 00 FF EC 00 11 44 75 63 6B 79 00 01 00
04 00 00 00 00 00 00 FF EE 00 0E 41 64 6F 62 65
00 64 C0 00 00 00 01 FF DB 00 84 00 1B 1A 1A 29
1D 29 41 26 26 41 42 2F 2F 2F 42 47 3F 3E 3E 3F
47 47 47 47 47 47 47 47 47 47 47 47 47 47 47 47
47 47 47 47 47 47 47 47 47 47 47 47 47 47 47 47
47 47 47 47 47 47 47 47 47 47 47 47 01 1D 29 29
34 26 34 3F 28 28 3F 47 3F 35 3F 47 47 47 47 47
47 47 47 47 47 47 47 47 47 47 47 47 47 47 47 47
47 47 47 47 47 47 47 47 47 47 47 47 47 47 47 47
47 47 47 47 47 47 47 47 47 47 47 47 47 FF C0 00
11 08 00 08 00 19 03 01 22 00 02 11 01 03 11 01
FF C4 00 61 00 01 01 01 01 00 00 00 00 00 00 00
00 00 00 00 00 00 04 02 05 01 01 01 01 00 00 00
00 00 00 00 00 00 00 00 00 00 00 02 04 10 00 02
02 02 02 03 01 00 00 00 00 00 00 00 00 00 01 02
11 03 00 41 21 12 F0 13 04 31 11 00 01 04 03 00
00 00 00 00 00 00 00 00 00 00 00 00 21 31 61 71
B1 12 22 FF DA 00 0C 03 01 00 02 11 03 11 00 3F
00 A1 7E 6B AD 4E B6 4B 30 EA E0 19 82 39 91 3A
6E 63 5F 99 8A 68 B6 E3 EA 70 08 A8 00 55 98 EE
48 22 37 1C 63 19 AF A5 68 B8 05 24 9A 7E 99 F5
B3 22 20 55 EA 27 CD 8C EB 4E 31 91 9D 41 FF D9
}; #this is a very tiny jpeg. it is a image representaion of the letters "ROFL" hand drawn by me in photoshop and then saved at the lowest possible quality settings where the letters could still be made out :)
my $image_data = pack('H2' x scalar(#image_hex), #image_hex);
my $url_escaped_image = uri_escape( $image_data );
URL escaped binary image data (can paste right into a URL)
%FF%D8%FF%E0%00%10JFIF%00%01%02%00%00d%00d%00%00%FF%EC%00%11Ducky%00%01%00%04%00%00%00%00%00%00%FF%EE%00%0EAdobe%00d%C0%00%00%00%01%FF%DB%00%84%00%1B%1A%1A)%1D)A%26%26AB%2F%2F%2FBG%3F%3E%3E%3FGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG%01%1D))4%264%3F((%3FG%3F5%3FGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG%FF%C0%00%11%08%00%08%00%19%03%01%22%00%02%11%01%03%11%01%FF%C4%00a%00%01%01%01%01%00%00%00%00%00%00%00%00%00%00%00%00%00%04%02%05%01%01%01%01%00%00%00%00%00%00%00%00%00%00%00%00%00%00%02%04%10%00%02%02%02%02%03%01%00%00%00%00%00%00%00%00%00%01%02%11%03%00A!%12%F0%13%041%11%00%01%04%03%00%00%00%00%00%00%00%00%00%00%00%00%00!1aq%B1%12%22%FF%DA%00%0C%03%01%00%02%11%03%11%00%3F%00%A1~k%ADN%B6K0%EA%E0%19%829%91%3Anc_%99%8Ah%B6%E3%EAp%08%A8%00U%98%EEH%227%1Cc%19%AF%A5h%B8%05%24%9A~%99%F5%B3%22%20U%EA'%CD%8C%EBN1%91%9DA%FF%D9
Here's the C++ routine I wrote to do this:
bool is_jpeg(const unsigned char* img_data, size_t size)
{
return img_data &&
(size >= 10) &&
(img_data[0] == 0xFF) &&
(img_data[1] == 0xD8) &&
((memcmp(img_data + 6, "JFIF", 4) == 0) ||
(memcmp(img_data + 6, "Exif", 4) == 0));
}
img_data points to a buffer containing the JPEG data.
I'm sure you need more bytes to have a JPEG that will decode to a useful image, but it's a fair bet that if the first 10 bytes pass this test, the buffer probably contains a JPEG.
EDIT: You can, of course, replace the 10 above with a higher value once you decide on one. 134, as suggested in another answer, for example.
It is not a requirement that JPEGs contain either a JFIF or Exif marker. But they must start with FF D8, and they must have a marker following that, so you can check for FF D8 FF.