Oh crap. And all this time I thought that was 0x203988C + something = position of the next pocket. :( I thought so.
Uhm... Just a tiny bit. And all the routine that I ever did from scratch was a multiply variables and a change gender routine, both of which were based off of FBI's examples. I'd be really glad if you'd walk me through it! :D
I'm not sure. But one thing that I'd point out is that it isn't just about IDA itself. One of the biggest draws is being able to use knizz's idb. The power comes from the documentation in that specific file. So there may be some other good interactive disassembler (no clue), but it would be pointless if it couldn't also read the FireRed idb (or Emerald). Use wine :D
As far as I can see, there are a couple of small problems with this routine, as well as some things that we don't know. First of all, where are you hooking from? We need to know the offset of the hook.
.thumb
start:
mov r0, #0x86
lsl r0, r0, #0x4
add r0, r0, #0x7start:
push
load byte from base_power(0x02024400) into r5
get flag in both r0 and r4
bl to flag_check
check if the flag was set
if it wasn't, b to the_end
add five to r5
add 1 to r4
bl to flag_check
check to see if second flag was set
if it wasn't b to the_end
add five to r5
add 1 to r4, then bl to flag_check
(you can do this until you checked all your flags)
the_end:
store r5 in 0x02024400
pop
flag_check:
move r4 to r0
make r1 0x0809D790 + 1
bx r1(Why was I looking for 867 instead of 0x867? derp) I don't understand the maths here. What does the lsl line do that mul r0, #0x10 wouldn't do?
I didn't know it was possible to write to a byte directly. What would be the point in loading the byte then doing the addition? Wouldn't it be simpler to just write the base power value? Then the summary screen value won't be confusing (Since the bp value will be --- and some npc will just tell the player that it changes per badge)
It would only affect the moves with the move effect that calls the routine so it should be fine. AI will be AI, can't do much with that. I don't mind the AI having the same bonus, the badge thing is just a good measurement of game progression so it makes sense that the AI's attacks would hurt more too.
Ah, ok. I didn't know that.
I looked at it on a bitwise level. 0x86 is 10000110b, and 7 is 0111b. Because mov only accepts one byte, I loaded 0x86 into r0. Then I shifted it to the left by four spaces(note that 7 takes up four spaces). This makes 0x860. Adding 7 is obvious after that. It's shorter, and yeah, pokedude9 is right.
You can store a byte, a half-word, or a word with strb, strh, and str respectively. Sometimes this is difficult becuase of malloc-ing, but the BP is stored in one location.
K, just wanted to make sure I understood.
You're really close now, just a few more edits and you've got it.
Still slightly off I'm afraid. In order to shift left by four to r4, it'd be lsl r4, r4, #0x4. The first r4 is the destination, the second is the register you're shifting, and the third parameter is the amount of shifts you want to make. So if you want to just add to one register, you'd do add r5, r5, #0x5 and stuff like that.
start:
push {r0-r5}
mov r5, #0x3C
mov r4, #0x86
lsl r4, r4, #0x4
add r4, r4, #0x7
bl flag_check
cmp r0, #0x0
beq the_end
add r5, r5, #0x5
add r4, r4, #0x1
bl flag_check
cmp r0, #0x0
beq the_end
add r5, r5, #0x5
add r4, r4, #0x1
bl flag_check
cmp r0, #0x0
beq the_end
add r5, r5, #0x5
add r4, r4, #0x1
bl flag_check
cmp r0, #0x0
beq the_end
add r5, r5, #0x5
add r4, r4, #0x1
bl flag_check
cmp r0, #0x0
beq the_end
add r5, r5, #0x5
add r4, r4, #0x1
bl flag_check
cmp r0, #0x0
beq the_end
add r5, r5, #0x5
add r4, r4, #0x1
bl flag_check
cmp r0, #0x0
beq the_end
add r5, r5, #0x5
add r4, r4, #0x1
bl flag_check
cmp r0, #0x0
beq the_end
add r5, r5, #0x5
add r4, r4, #0x1
b the_end
the_end:
ldrb r0, = 0x02024400
strb r5, [r0]
pop {r0-r5}
flag_check:
mov r0, r4
ldr r1, = 0x809D790 +1
bx r1Oh, I see what you did. The first time you made that routine, you made a small mistake at the very end, which is why it wouldn't assemble.