Help/Support Freeze value instruction?

Joka · July 7, 2017

Is there an arm instruction that freezes the default value?

Like, for example, freezes the default "ammo" value...

Updated July 7, 2017 by Joka

WaveCheck · July 8, 2017

There is nothing that freezes it, I don't think, but what I would do is this:

Default Ammo: 30

sub r1, #0x1 // lets say this is what subtracts your ammo count

mov r1, #0x30 // change it to this

Note: this is just an example

Updated July 8, 2017 by oZarmo

Joka · July 8, 2017

4 minutes ago, oZarmo said:

There is nothing that freezes it, I don't think, but what I would do is this:

Default Ammo: 30

sub r1, #0x1 // lets say this is what subtracts your ammo count

mov r1, #0x30 // change it to this

Note: this is just an example

What's the difference if I do mov r1 or mov r0?

WaveCheck · July 8, 2017

Just now, Joka said:

What's the difference if I do mov r1 or mov r0?

In my example r1 is the register that holds the value for the ammo

Joka · July 8, 2017

Just now, oZarmo said:

In my example r1 is the register that holds the value for the ammo

No I mean the first register, not the second.

Like MOV R0 or MOV R1?

If it was orignally MOV R0, R3 or something and you changed it to LDR R1, [R7], why change the R0 to R1?

WaveCheck · July 8, 2017

5 minutes ago, Joka said:

No I mean the first register, not the second.

Like MOV R0 or MOV R1?

If it was orignally MOV R0, R3 or something and you changed it to LDR R1, [R7], why change the R0 to R1?

I'm saying that if it was originally sub r1, #0x1 and you changed it to mov r1, #0x30 it would set your ammo to 30 everytime, which is the default ammo, instead of subtracting your ammo. Therefore, your ammo wouldn't look like it's change and it looks frozen.

Updated July 8, 2017 by oZarmo