Axe Q&A

[Yeong]

I'm sorry, but I didn't quite get what you're saying.

[Runer112]

Using a constant pointer like Str1 instead of a variable pointer like A is an optimization technique. Both methods will produce the same end results. But if you use a constant, your code will be smaller and faster.

[Yeong]

ah, ok.
I was stuck on where you mentioned something about 2 byte storing.

[Deep Thought]

A static variable is the same as a number to Axe, so 0→{L6}^r is exactly the same as 0→{37696}^r to Axe.

Variables are really shortcuts for {POINTER}^r, so 0→A is exactly the same as 0→{°A}^r is exactly the same as 0→{34852}^r.

So if you were to do 0→{A}^r, what you're really doing is  0→{{34852}^r}^r. That's why it's less optimized.

EDIT: Ninja'd.

[mrmprog]

What is the fastest way to count the black pixels in an 8*8 area?

[leafy]

I suppose you could use bit arithmetic with a pointer to a Pt-Get('ed sprite. I'm not too clear on how that would work though.

[jacobly]

Code: Select All | Copy To Clipboard

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:Lbl Count :0→r3 :pt-Get(r1,r2) :For(8) :{→r1} :While :Select(,r3ʳ++) and (-1) .subtract not negative :End :r1+1 :End :Return r3

[calcdude84se]

I suppose you could use bit arithmetic with a pointer to a Pt-Get('ed sprite. I'm not too clear on how that would work though.

If it's byte-aligned, he could even work with the bytes directly, instead of using Pt-Get(. With that said, the fastest way to find the number of set bits in a byte (which is a necessary part of this) is almost certainly a look-up table. Yes, it's 256 bytes, but it's far faster than any bit arithmetic you can do.

[mrmprog]

Thanks. I have another one:If I have four numbers, and one will be zero, what is the fastest way to find the next smallest number?

[calcdude84se]

If exactly one number will be zero and the numbers aren't sorted in any way, I'm pretty sure the best way is to go through them with a variable containing the minimum value, while ignoring zeroes.
Here, for example, is some code that will give the minimum non-zero value in the first 100 bytes of L₁ in M (And -1 if they're all 0):

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-1->M L1->X For(100 If ({X}->V<M)?V=/=0 V->M End X++ End

There's probably a more optimized way to write that, but it should work.
If there could be more than one zero, the essence of what you have to do is have a flag that records whether you've already encountered one zero.
Edit: Runer112 wins by far. The method I've proposed would work better across lots of values (perhaps at least 10) than across 4.

[Runer112]

If it's only four numbers, the fastest way would be a couple of nested min() commands, like this.

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1 2	min(min(A-1,B-1),min(C-1,D-1))+1

EDIT: Actually, I'm not entirely sure if this will work how you want it to. It depends whether or not you want 0 as a result if two or more of the values are 0. For instance, if the numbers were 0, 0, 1, 2, this routine would return 1.

[parserp]

How do I achieve this? (program image and text)

[Yeong]

I'm pretty sure you can do the image with one of the axe command that I don't remember. (Something like #Image or something?)
but text, I have no clue.

[Runer112]

To quote the documentation:

The next thing you absolutely need is an Axe Header. You must start the first line with a
period, which is a comment in Axe, followed by the name you want for the compiled
program. If you want a program description, you can type a space and then the
description on that same line, but this is optional. In the above example your program
might look like this:

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PROGRAM:MARIOSRC :.MARIO A fun platformer :

And to quote the commands list:

#Icon(HEX) _{Key: identity()}   Tells the parser to replace the default icon with the new icon. The icon must be 64 hex characters long.

[parserp]

Ah. That makes sense. Thanks!