ARM，帮助LDR指令

Question

I am studying for an ARM test and I have this code

    AREA datos, DATA, READWRITE
long    EQU 7*4   
serie   DCD 1, 2, 4, 6, 8, 7, 9
resul   DCB 0
    AREA prog, CODE, READONLY

    ENTRY
    mov r0, #0
    eor r1, r1, r1  ;result variable

    ldr r2, =serie **This one**
buc ldr r3, [r2, r0]
    add r1, r1, r3
    add r0, r0, #4
    cmp r0, #long
    bne buc

    ldr r2, =resul **This one**
    str r1, [r2]

fin b fin

    END 

And I'm debugging it with Keil, my problem is that I don't understand very well the instructionts marked.

     8:     mov r0, #0 
0x40000000  E3A00000  MOV       R0,#0x00000000
     9:     eor r1, r1, r1  ;result variable 
    10:      
0x40000004  E0211001  EOR       R1,R1,R1
    11:     ldr r2, =serie 
0x40000008  E59F201C  LDR       R2,[PC,#0x001C]
    12: buc     ldr r3, [r2, r0] 
0x4000000C  E7923000  LDR       R3,[R2,R0]
    13:     add r1, r1, r3 
0x40000010  E0811003  ADD       R1,R1,R3
    14:     add r0, r0, #4 
0x40000014  E2800004  ADD       R0,R0,#0x00000004
    15:     cmp r0, #long 
0x40000018  E350001C  CMP       R0,#0x0000001C
    16:     bne buc 
    17:      
0x4000001C  1AFFFFFA  BNE       0x4000000C
    18:     ldr r2, =resul 
0x40000020  E59F2008  LDR       R2,[PC,#0x0008]
    19:     str r1, [r2] 
    20:         
0x40000024  E5821000  STR       R1,[R2]
    21: fin     b fin 

I have this if I dissasembly it with Keil, then I know that LDR R2, =serie its the same that LDR R2,[PC, #offset] but the value of #offset are placed in the Literal Pool? I don't know why the value is 0x001C .

PD: Sorry for my english, I know its not very good.

Answer 1

Here is an object dump of your program (modified to run on Raspberry Pi).

Disassembly of section .text:

00000000 <main>:
   0:       e3a00000        mov     r0, #0
   4:       e0211001        eor     r1, r1, r1
   8:       e59f201c        ldr     r2, [pc, #28]   ; 2c <buc+0x20>

0000000c <buc>:
   c:       e7923000        ldr     r3, [r2, r0]
  10:       e0811003        add     r1, r1, r3
  14:       e2800004        add     r0, r0, #4
  18:       e350001c        cmp     r0, #28
  1c:       1afffffa        bne     c <buc>
  20:       e59f2008        ldr     r2, [pc, #8]    ; 30 <buc+0x24>
  24:       e5821000        str     r1, [r2]
  28:       e12fff1e        bx      lr
  2c:       00000000        andeq   r0, r0, r0
  30:       0000001c        andeq   r0, r0, ip, lsl r0

Disassembly of section .data:

00000000 <serie>:
   0:       00000001        andeq   r0, r0, r1
   4:       00000002        andeq   r0, r0, r2
   8:       00000004        andeq   r0, r0, r4
   c:       00000006        andeq   r0, r0, r6
  10:       00000008        andeq   r0, r0, r8
  14:       00000007        andeq   r0, r0, r7
  18:       00000009        andeq   r0, r0, r9

0000001c <resul>:
  1c:       00000000        andeq   r0, r0, r0

Disassembly of section .ARM.attributes:

00000000 <.ARM.attributes>:
   0:       00001541        andeq   r1, r0, r1, asr #10
   4:       61656100        cmnvs   r5, r0, lsl #2
   8:       01006962        tsteq   r0, r2, ror #18
   c:       0000000b        andeq   r0, r0, fp
  10:       01080206        tsteq   r8, r6, lsl #4
  14:       Address 0x00000014 is out of bounds.

There is a .text section for the program and .data section for the data (DCD, DCB). At the end of the program, there are two words that will contain the address of the .data section that are defined "serie" and "resul". The address of those address in ldr r2, [pc, #28] is the value of the pc reg + dec 28 = hex 2c. The same is true with the ldr r2, [pc, #8] , value in the pc reg + dec 8 = hex 30.