Here's a simple, fairly optimized way to do what you are trying to achieve. I have detailed most of the optimizations in comments in the code, but I did not comment about the biggest optimization. This optimization is to draw sprites from a variable pointer, instead of requiring four different copies of the whole loop just for facing in different directions. For instance, here's the line I used to initialize the player sprite to the right-facing sprite:

:2*8+Pic1→S

If you're familiar with how this works, then go ahead and skip to looking at the code to see the final product with the rest of my suggested simple optimizations. If you're not familiar with how this works, I'll give a quick explanation. Because of how Axe includes data like sprites, the order in which you list data (like sprites) is exactly the order it will assume in the compiled program. Since each sprite is 8 bytes, that means that each successive sprite is found in memory 8 bytes after the previous sprite. Pic1 points to the start of the block of 4 sprites, so it points to the first one, the down-facing sprite. 2*8+Pic1 would point to the data 16 bytes after the down-facing sprite, which we know is the right-facing sprite because it's 2 sprites (remember, a sprite is 8 bytes) after the down-facing sprite. This offset from Pic1 logic is used to draw the proper sprite for facing in all 4 directions.


So here's my suggested fairly optimized and hopefully easy-to-understand solution:

:.AXETEST
:identity(3FFC3FFC30CF30CF300C300C0C300C303FFC3FFC330C330C3FFC3FFCF00FF00F)
:
:[]→Pic1
:[7E665A247EC37EC3] .DOWN
:[7ED242247E4A7EC3] .LEFT
:[7E4B42247E527EC3] .RIGHT
:[7E7E42247EC37EC3] .UP
:
:.START FACING RIGHT AT (0,56):
:2*8+Pic1→S
:0→X
:56→Y
:.THIS DOESN'T NEED TO BE INSIDE THE LOOP:
:StorePic
:
:While 1
: Pt-On(X,Y,S)
: DispGraph
: .ERASING WITH Pt-Change() UNNECESSARY BECAUSE:
: RecallPic
: If getKey(1)
:  Y+2→Y
:  0*8+Pic1→S
: End
: If getKey(2)
:  X-2→X
:  1*8+Pic1→S
: End
: If getKey(3)
:  X+2→X
:  2*8+Pic1→S
: End
: If getKey(4)
:  Y-2→Y
:  3*8+Pic1→S
: End
:.OPTIMIZED POST-CHECK LOOP:
:End!If getKey(15)


---

Now, if you want really optimized code... That can be arranged. I'll leave figuring out how it works as an exercise.

:.AXETEST
:identity(3FFC3FFC30CF30CF300C300C0C300C303FFC3FFC330C330C3FFC3FFCF00FF00F)
:
:[]→Pic1
:[7ED242247E4A7EC3] .LEFT
:[7E7E42247EC37EC3] .UP
:[7E4B42247E527EC3] .RIGHT
:[7E665A247EC37EC3] .DOWN
:
:56→Y
: and 0→X
:→S
:
:While 1
: ClrDraw
: Pt-On(X,Y,S*8+16+Pic1)
: DispGraph
: If getKey(1)-getKey(4)
:  →S
:  *2+Y→Y
: End
: If getKey(3)-getKey(2)
:  -1→S+1
:  *2+X→X
: End
:EndIf getKey(15)


Thank you!

Code:
:.KARRACE
:identity(318C4BD24FF24E724FF2366C0FF0166818181BD87BDE981997E99FF99FF967E6)
:.KAR
:[314B4F4E4F360F16]→Pic1
:[8CD2F272F26CF068]→Pic2
:[181B7B98979F9F67]→Pic3
:[18D8DE19E9F9F9E6]→Pic4
:.DRAW
:Repeat getKey(15)
:Pt-On(0,,Pic1
:Pt-On(0,8,Pic3
:Pt-On(8,0,Pic2
:Pt-On(8,,Pic4
:DispGraph
:End


that's all i see
Edit: It will be best to stick with Pt-On IMO, as you already have the command in the previous and assumed related code

What is this "peephole optimizer" of which you speak? How do I use it?

Even this is unnecessary due to the new peephole optimizer

It would make sense to think this, but this is actually not true. A peephole optimization to do this was added in Axe 1.0.5, but it was disabled in the next version (Axe 1.1.0) and has been ever since. Don't ask me why because I don't know; a quick test with Axe 1.0.5 appears to suggest that the optimization works fine. Perhaps I'll ask Quigibo in a more appropriate thread why this optimization (called o_PushPop5 in Commands.inc) is disabled.

Although I believe (I need Runer to confirm) that a While 1 : EndIf getKey(15) might be better than Repeat getKey(15) : End

This is true.

I was looking through the code of Eat Nethams(it seemed easy) to try to understand it. I'm having trouble figuring out how it is displaying so many lobsters at once. Can someone explain this to me?

Quote from: Spenceboy98 on January 02, 2013, 10:51:41 pm

I was looking through the code of Eat Nethams(it seemed easy) to try to understand it. I'm having trouble figuring out how it is displaying so many lobsters at once. Can someone explain this to me?

*BUMP*
Can no one help me with this^^^^?

They're all sprites, and you can draw as many sprites as you want. I don't quite understand what you mean