Okay, I’m reading in a data file and the parser only reads in a single float value for specular and diffuse. My instructor says that specular, diffuse, and ambient should sum to one, so the parser doesn’t bother to read in a value for ambient.
Fine, but the data file has only ONE value for each? Each one requires an array with 4 float values, I know the first three on ambient is for an RGB color, but could anyone explain what each of the 4 numbers should be? My instructor said that I could easily find all 4 numbers from the single float.
This is really important to my project as some of my objects are getting EXTREME glare causing exaggerated gradients of light and dark. Thank in advance.
Interesting, packing 4 floats in one… Ye may use float as 4byte array and put your Red Green Blue and Alpha values as fixed-point byte.
What I understood from the given explanations your ambient, diffuse and specular values of a material should sum up to 1.0.
That means you need two values and can get the third calculated from them.
You read one float for diffuse and one float for specular, right?
That means you don’t have any color informations in the data, right?
That would mean you can only display greyscale materials where red, green and blue are all identical taken from the float you read or calculated and alpha is assumed to be 1.0 for opaque materials.
The question is, do you have a material COLOR in some RGB in the file as three floats or not?
If yes, and the diffuse and specular are only given as one float, you need to multiply the ambient color channels with the diffuse and specular scaling factors and put those into the diffuse and specular meterial properties.
Though this is no lighting calculation, yet.
Is the ONE value in the file actually a float? If it could be read in as an int, then it could just be a number in the form
0xRRGGBBAA (or 0xAARRGGBB, or something else like that) To get float values from the first one, you could do something like so…
n = ValueFromFile
r = (float)((n & 0xFF000000) >> 24) / 255.0;
g = (float)((n & 0x00FF0000) >> 16) / 255.0;
b = (float)((n & 0x0000FF00) >> 8) / 255.0;
a = (float)((n & 0x000000FF)) / 255.0;