Help/Support lldb breakpoint outcome seems weird

[Ted2 39.2k iPhone 13 Pro 16.0 Donor]

Hi,

 

Since gdb doesnt support arm64, I need to breakpoint on lldb.

 

I get this as outcome & it's totally diffrent as gdb..

(lldb) register read
General Purpose Registers:
        x0 = 0x000000010fcc47e0
        x1 = 0x000000000000000c
        x2 = 0x0000000000000000
        x3 = 0x0000000000000000
        x4 = 0x000000016fda18b8
        x5 = 0x0000000000000000
        x6 = 0x0000000000000000
        x7 = 0x00000001100d7954
        x8 = 0x0000000000000009
        x9 = 0x0000000000000003
       x10 = 0x0000000000000003
       x11 = 0x00000001602e0510
       x12 = 0x0000000160d98000
       x13 = 0x0000000000003fff
       x14 = 0xffffffffffffc000
       x15 = 0x0000000000003fff
       x16 = 0x00000001819ce520  libsystem_platform.dylib`_platform_memset
       x17 = 0x00000000fffffffe
       x18 = 0x0000000000000000
       x19 = 0x0000000115180780
       x20 = 0x000000016fda187f
       x21 = 0x0000000160c85510
       x22 = 0x0000000160320250
       x23 = 0x0000000162440b10
       x24 = 0x00000000000032a3
       x25 = 0x0000000160c84740
       x26 = 0x0000000100cce02c 
       x27 = 0x0000000102c876b8  
       x28 = 0x00000001602e1f80
        fp = 0x000000016fda1690
        lr = 0x00000001000b04f0 
        sp = 0x000000016fda1680
        pc = 0x00000001000a763c 
      cpsr = 0x60000000

 

Is this normal?

 

Do i have to do something extra?

[Ted2 39.2k iPhone 13 Pro 16.0 Donor]

Problem solved, lldb just make it look like sh!t with all those 00000 before the actual number.

Also, the outcome is in hex decimal. 

 

It was X1 what I needed, C is a hex number. Converting that to decimal is 12, which was correct.