EASY for most

OP (Company: SMU) #2607600 
; I am horrible with ARM and I am having problems this is the program I 
wrote
; but it isn't working the right way it is supposed to do the fibonacci 
seq
; using indirect addressing it keeps looping back to when i LDR r0
; what do i do to fix this? thanks

  AREA  FIBSEQ,CODE
SRAM_BASE EQU 0x04000000    ;static ram location

  ENTRY

MAIN  MOV r1,#0      ;first number used in sequence
  MOV r2,#1      ;second number used in sequence
  MOV r3,#0      ;third number used in sequence
  MOV r4,#1      ;counter
  LDR r5,=SRAM_BASE
;  STR r3,[r5],#0      ;these are supposed to be the first numbers in 
the           ;fib seq
;  STR r4,[r5],#1


FIBO          ;method that solves for fib up to 18th seq
  ADD r3,r2,r1
  ADD r1,r1,r2
  SUB r2,r2,r2
  ADD r2,r3,r2      ;also having problems here with logic
  ADD r4,r4,#1
  STR r3, [r5]      ;here is where it loops back
  CMP r4,#18
  BNE FIBO


stop  B  stop
  END
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Guest 
Your Code doesn't compute the Fibonacci numbers because there are 
logical errors. Code for first 18 Fibonacci numbers:

1
int SRAM_BASE[18];

2
int main(void)

3
{

4
  int r1 = 0;          // MOV r1,#0

5
  int r2 = 1;          // MOV r2,#1

6
  int r3 = 0;          // MOV r3,#0

7
  int r4 = 0;          // MOV r4,#0

8
  int *r5 = SRAM_BASE; // LDR r5,=SRAM_BASE

9


10
  // Init 1st and 2nd Fibonacci numbers

11
  *r5++ = 0;           // STR r3,[r5]

12
                       // ADD r5,r5,#1

13
  r4++;                // ADD r4,r4,#1

14
  *r5++ = 1;           // STR r3,[r5]

15
                       // ADD r5,r5,#1

16
  r4++;                // ADD r4,r4,#1

17


18
  // Compute following Fibonacci numbers

19
  do {

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  // FIBO:

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    r3 = r2 + r1;      // ADD r3,r2,r1

22
    *r5++ = r3;        // STR r3,[r5]

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                       // ADD r5,r5,#1

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    r4++;              // ADD r4,r4,#1

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    r1 = r2;           // MOV r1,r2

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    r2 = r3;           // MOV r2,r3

27
  } while( r4 != 18 ); // CMP r4,#18

28
                       // BNE FIBO                         

29
// STOP:

30
  while(1){}           // B STOP

31
}
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