[OpenJDK 2D-Dev] [9] Review request for 8038000: java.awt.image.RasterFormatException: Incorrect scanline stride

[anton nashatyrev]

Hello,

     could you please review a slightly update fix version (the 
regression test upgraded)
fix: http://cr.openjdk.java.net/%7Eanashaty/8038000/webrev.01/
bug: https://bugs.openjdk.java.net/browse/JDK-8038000

Jim, thanks for your in-depth analysis, the validation indeed doesn't 
look ideal. However my fix addresses the concrete regression which was 
introduced by these validations, so I'm leaving the fix as is.

Thank you!
Anton.

On 01.04.2014 19:39, anton nashatyrev wrote:
> Hello Jim,
>
> On 28.03.2014 3:25, Jim Graham wrote:
>> Hi Anton,
>>
>> A lot of those tests seem out of whack in that they test related 
>> conditions, but not the exact condition itself.  What we really want 
>> is for every index of the form:
>>
>> offset + y * scanlineStride + x + {0 -> numcomponents-1} => [0, 
>> buf.length-1]
>>
>> to be in the array for all valid x,y.  There are a lot of tests there 
>> that are sufficient to prove this, but are overly restrictive in that 
>> they reject a bunch of cases.  The fix you propose only fixes this 
>> for a case where h=1, but what about:
>>
>>     w = 10
>>     h = 2
>>     numcomponents = 1
>>     scanlineStride = 1000
>>     buffer.length = 1010
>>
>> The buffer contains the data for all possible pixel fetches, but 
>> since it isn't 2000 in length, we reject it.
>
> My fix actually relaxes the condition, and the case above is not 
> rejected:
> (height > 1 && scanlineStride > data.length) is FALSE here and hence 
> the exception is not thrown
>
>>
>> Also, we restrict a bunch of parameters to "MAX_INT / some other 
>> factor" because we blindly multiply things out and don't want to deal 
>> with overflow, but a better "pixel array" test would use division (I 
>> think we do this in our native code):
>>
>>     buf.length / w <= h
>>
>> except that you need to deal with offset and scanlineStride for h-1 
>> lines and then w for the last one, so this gets complicate, but you 
>> have:
>>
>>     ((buf.length - offset - w) / (h-1)) < scanlineStride
>>
>> but then you have to special case h=1 to avoid the divide by 0 so you 
>> get:
>>
>>     if (w <= 0 || h <= 0 || scanlineStride < 0 || offset < 0) exception;
>>     if (offset >= buf.length) exception;
>>     int len = buf.length - offset; // known positive
>>     if (len < w) exception;
>>     if (h > 1) {
>>          if (((len - w) / (h - 1)) < scanlineStride) exception;
>>     }
>>
>> Note that the test for (len < w) is done in all cases because it 
>> prevents the calculation in the "h>1" case from having a negative 
>> numerator, which would make the test invalid.  These tests are also 
>> valid for scan=0 for the rare case where someone wants every row of 
>> an image to contain the same data (we may use a case like that in the 
>> GIF loading code that needs to replicate incoming data for interlaced 
>> GIFs).  It doesn't go so far as to allow scan=-1 or similar cases 
>> where the user wants to play games with aliasing parts of rows in a 
>> backwards cascade.
>
> There are a lot of checks in the *Raster.verify() methods. I'm not 
> 100% confident but they look pretty equivalent to the algorithm you 
> have described (BTW the 0 scanline is also acceptable). Anyways that 
> was a security fix some time ago when some of those validations have 
> been added and I'm not sure we would like to perform some major 
> refactorings here unless any incompatibilities are found.
>
> Thank you!
> Anton.
>
>>
>>             ...jim
>>
>> On 3/26/14 10:35 AM, anton nashatyrev wrote:
>>> Hello,
>>>      could you please review the following fix:
>>>
>>> fix: http://cr.openjdk.java.net/~anashaty/8038000/webrev.00/
>>> <http://cr.openjdk.java.net/%7Eanashaty/8038000/webrev.00/>
>>> bug: https://bugs.openjdk.java.net/browse/JDK-8038000
>>>
>>> The last row in the Raster shouldn't be necessary of the scanlineStride
>>> length and thus the Raster with height 1 might have a buffer smaller
>>> than a scanlineStride.
>>>
>>> Thanks!
>>> Anton.
>

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