Bitmap lockbits 16bpp

Question

0.00/5 (No votes)

See more:

Hello everybody. I struggle to understand why this is throwing me an error while saving created bitmap. Could anybody help with the following code?

C#

Bitmap ^a=gcnew Bitmap(width,height,System::Drawing::Imaging::PixelFormat::Format16bppGrayScale);
    int i,j;
    BitmapData ^bitmapData=gcnew BitmapData();
    float value;
    System::Drawing::Rectangle b(0,0,width,height);
    a->LockBits(b, System::Drawing::Imaging::ImageLockMode::WriteOnly, a->PixelFormat, bitmapData);
    unsigned short int *ptr=(unsigned short int*)bitmapData->Scan0.ToPointer();
    int stride=bitmapData->Stride;

    for (i=0;i<height;i++)
    {
        unsigned short int * LinePtr = ptr + (i * height);
        for(j=0;j<width;j++)
        {
            //value=mat[j][i];
            LinePtr[j]=0;//65535*value/255;
        }
    }
    a->UnlockBits(bitmapData);
    return a;

Posted 25-Nov-11 2:17am

Peter Kottas

Updated 25-Nov-11 8:44am

Emilio Garavaglia

v2

Add a Solution

5 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Peter Kottas · Answer 1 · 2011-11-25T02:25:00

Solution 1

Hmm just now a realized something. Image stride for height 600 is 1600 while 1600/600 is not decimal number.And 16bpp should produce stride 600*2 in bytes equals 1200. Am i right ? So i guess bitmap class is putting some "imaginary" pixel data at the end of each scan line maybe for performance reasons? But I still dont know what to do with that ...

Posted 25-Nov-11 2:25am

Peter Kottas

Updated 25-Nov-11 2:26am

v2

Comments

Mark Salsbery 25-Nov-11 18:52pm

Last time I checked, that pixel format isn't supported so to even get that far is lucky. Just always use the stride returned by the system :)

Mark Salsbery 26-Nov-11 11:05am

and this

LinePtr = ptr + (i * height);

should be something like this

LinePtr = ptr + (i * stride);

Peter Kottas · Answer 2 · 2011-11-27T06:49:00

Solution 2

thanks for help however that's not the case , of course my first guess was LinePtr = ptr + (i * stride); however stride is one row in bytes and since my variable is typed short int which have 3 bytes than for that purpose i*height seemed right answer neither is working. What i am doing now is that i have 24 bpp rgb image and i put same number in each "color slot" now thats not very elegant and it takes time. What would be the better way ?

Posted 27-Nov-11 6:49am

Peter Kottas

Updated 27-Nov-11 6:50am

v2

Comments

Mark Salsbery 27-Nov-11 14:40pm

>>"and since my variable is typed short int"

Sorry about that. Regardless, you HAVE to use the data pointer and stride values returned by lockbits. You can't assume anything. The stride can even be negative. So use stride/2 (When using pointer arithmetic I prefer to use Byte* then cast to an Int16* in the inner loop, but that's just me). 16bppgrayscale is not supported anyway (or has very limited support) so you may still struggle with that!

Peter Kottas · Answer 3 · 2011-11-27T08:43:00

Solution 3

Negative ?? Do i understand it correctly? How can amount of information in bytes be negative. So i guess than there is no better way to code grayscale than to write to rgb same value in each Chanel. Pity ... thanks anyway

Posted 27-Nov-11 8:43am

Peter Kottas

Mark Salsbery · Answer 4 · 2011-11-27T14:40:00

Yes the stride can be negative. One problem in your code, however, is you're not using the correct LockBits method. You're using the one marked "This version of the LockBits method is intended to be used with a flags value of ImageLockMode::UserInputBuffer".

Maybe try this, adjusted to use correct pointer arithmetic and LockBits method...

C++

Bitmap ^a = gcnew Bitmap(width,height,System::Drawing::Imaging::PixelFormat::Format16bppGrayScale);
//float value;
System::Drawing::Rectangle b(0,0,width,height);
BitmapData ^bitmapData = a->LockBits(b, System::Drawing::Imaging::ImageLockMode::WriteOnly, a->PixelFormat);
unsigned char *ptr = (unsigned char *)bitmapData->Scan0.ToPointer();

for (int i = 0;i < height; i++)
{
    unsigned short int * LinePtr = (unsigned short int *)(ptr + (i * bitmapData->Stride));
    for(int j = 0; j < width; j++)
    {
        //value=mat[j][i];
        LinePtr[j]=0;//65535*value/255;
    }
}

a->UnlockBits(bitmapData);
return a;

Peter Kottas · Answer 5 · 2011-11-28T09:16:00

Solution 5

well nothing seem to work just fine so i guess i am stuck with 24bpp rgb, thanks for help anyway , much appreciated

Posted 28-Nov-11 9:16am

Peter Kottas

Bitmap lockbits 16bpp

5 solutions

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Add your solution here

Preview 0

Existing Members

...or Join us