CIS 068 / 2: Lab-Assignment #6: Loading an image as a binary file

As you have seen in the last assignment,it is possible to store images by saving each single drawing event. However, this is of course not the way it is usually done, because of many reasons:

to restore the image each single drawing event must be done again. This is time consuming.
To redraw the image, each possible drawing action must be implemented. In your example there was only a circle, but imagine an image consisting of circles, boxes, arbitrarily shaped polygons etc.: the program to restore the image must know about all these objects.
Every information necessary to redraw the image must be stored, e.g. shape (in your example: circle), properties of shape (in your example: radius, center), color, transparency etc. Most of this information needed for reconstruction is usually not of interest in the resulting image.

The minimum of information needed to restore an image without loss of detail is

the size of the image
the color value of every pixel

Most of the image formats out there (*.gif, *.bmp, *.jpg ...) out there contain this information, they differ in possibilities of image compression, animation etc. extending them often to fairly complicated structures.

The simplest format you can think of contains exactly the minimum of information:
sizeX, sizeY first, then the color information of every pixel starting with the first row, split up into rates of red, green and blue.

let's imagine the following image or 'bitmap', consisting only of 2 rows with 6 pixels each:

r w k g g g
b b g g k k

where r stands for red, w for white, k for black, g for green b for blue.
each of the color values is represented by their red/green/blue ratios, which are given in values between 0 (dark) and 255 (bright).

r = red = 255, 0, 0 (meaning: red-ratio = 255, green-ratio = 0, blue-ratio = 0)
w = white = 255,255,255
k = black = 0,0,0
g = green = 0,255,0
b = blue = 0,0,255

We now have to decide for a data-format: let's allow the height and width to be between 0 and 65535, assigning 2 bytes to each one, storing the most significant byte first (e.g. a size of 258 x 13 would be stored as 1 2 0 13, because 258 = 1*256 + 2, 13 = 0*256 + 13).
The color ratios only need one byte, their values will not extend the interval of 0..255.
Given this format, the bitmap shown above would be stored as:
0 6 0 2 255 0 0 255 255 255 0 0 0 0 255 0 ...

Let's see why:
The size was given as an x/y-pair, two bytes each. Hence the x-size is 0 6, which means 256*0 + 6, the y-size results in 0 2, being 256*0 + 2: our image is 6 pixels wide, 2 pixels high.
The rest of the data must be red in triplets of single bytes, defining the color of each pixel row-wise starting at the upper left pixel.
The first triplet is 255 0 0, which is red.
The second triplet is 255 255 255, which is white
The third triplet is 0 0 0, which is black
And so on.

The image-file shall be stored as a binary file, not as a textfile.
Let's call this selfmade bitmape-format 'the c68-format'.

And here's your task:
Write a program that reads and shows the image 'assignmentImage.c68'.
assignmentImage.c68 is -you hopefully guessed it- an image stored in the c68-format.
You can read it HERE (260kB).

Hints:

Only one class is needed, extend it from JFrame. Please name it 'C68ImageReader'
Open a binary file, read the size (2 x 2 bytes), set the size of your frame accordingly.
Create an array (the size of the array is known to you by the size of the image, but notice that you need 3 bytes for every single pixel !) and read the color-data
override the paintComponent-method, the image can be drawn by painting each single pixel rowwise. A single pixel can be drawn using a circle with size 1,1 , defining it's color first by setColor(r,g,b).

Scoring: 5 points, as usual.

Good Luck !