I converted everything to fixed point, and now I can draw twice the number of pixels and still get a 25% speed gain
Fixed point is amazing; I think my next project will be converting Ncaster to fixed point.

The file changes size when VBA uses it. Maybe that causes troubles?

Have you solved the out-of-memory issues? Last time I asked you, you had something like 800 bytes left.
Sorry if you posted this already, but I haven't been following this thread and I don't feel like browsing through 25 pages to find out

Here's a still.
Really quite boring, but I think there's potential.
I need to figure out how to handle the ugliness that occurs when rendering far-away areas.
The checkerboard is not procedural; I'm actually scaling a 64x64 sprite up to a huge area. Yay giant pixels!
I think I have little enough code here so I can convert everything to fixed point without causing myself a debugging nightmare.

2x2 pixels.

It does the linear algebra for every pixel.
Here's are the loop contents:
float y1 = h - y;
float y2 = y - h/2;
float c = 1 + y1 / y2;
int floorX = (int) ((raydx * c) + px);
int floorY = (int) ((raydy * c) + py);
5 adds, 2 multiplies, and 2 divides.
If I did the pre-calculated method, I think it would be 4 adds and 4 multiplies. Definitely faster
However, that method does not easily allow vertical motion
If I draw every 4th pixel like I do now, I actually get quite a decent framerate. I wonder if this is worth continuing, or should I try to figure out how to do proper mode 7?

Fixed in the first post Thanks for the bugfix!
Can someone verify whether the Linux and Windows versions of the tools can coexist in the same directory? That way I can merge the two skeletons.

@graphmastur: I will try to explain the algorithm:

Shanks-Tonelli modular square root algorithm:
The problem: given n and p, with p prime, find x such that x^2 = n (mod p)
First, a digression. The Legendre symbol (n|p) (actually written in a different manner, but the forum does not support typesetting) is defined as follows: (n|p) = 1 if such an x exists, 0 if p divides n, and -1 otherwise. It can be computed as n^((p-1)/2) mod p.
Now we return to the algorithm:
Step 1: Let p - 1 = 2^s * q , with q odd. This can be done by dividing powers of 2 out of p - 1 until we cannot do so any further.
Step 2: Select z such that (z|p) = -1. We do this simply by randomly selecting z in the set {1, 2, ... p - 1} and computing (z|p) as described above.
Step 3: Let:
                    c = z^q mod p
                    r = n^((q + 1) / 2) mod p
                    t = n^q mod p
                    m = s
Note that all of these values can be computed rapidly using the repeated squaring algorithm.
Step 4: We execute the following loop:
                    While t != 1
                       i = 0
                       tmp = t mod p
                       While tmp != 1
                         tmp = tmp * tmp mod p
                         i = i + 1
                       EndWhile
                       b = c
                       For k = 1 to m - i - 1
                         b  = b * b mod p
                       EndFor
                       r = r * b mod p
                       t = t * b^2 mod p
                       c = b^2 mod p
                       m = i
                    EndWhile
Step 5: Return r