CODEWITHSUNDEEP
assemblyarm
Ted West
Ted West

Reputation: 81

Understanding ARM Assembly Code

Can anyone explain to me what this short program does?

ORIGIN 0x1000
one DEFW 13
two DEFW 29
three DEFW 0
ORIGIN 0x1010
ENTRY
ADR R0, one
LDR R1, [R0]
LDR R2, [R0, #4]
ADD R1, R2, R1
STR R1, [R0, #8]
SWI 2

If I'm thinking correctly, it adds 'one' to 'two' and places the result in 'three'. Am I correct?

Upvotes: 2

Views: 3064

Answers (1)

Pete Fordham
Pete Fordham

Reputation: 2343

Yes.

ORIGIN 0x1000          # Start at address 0x1000
one DEFW 13            # Allocate 4-bytes of space for a variable called one and set it to 13
two DEFW 29            # Allocate 4-bytes of space for a variable called two and set it to 29
three DEFW 0           # Allocate 4-bytes of space for a variable called three and set it to 0
ORIGIN 0x1010          # Skip ahead to address 0x1010 (this really leaves a 4-byte gap)
ENTRY                  # Mark next instruction as the begining of program
ADR R0, one            # Load address of one into R0
LDR R1, [R0]           # Load contents of one (pointed to but R0) into R1
LDR R2, [R0, #4]       # Load contents of two (pointed to but R0 + 4) into R2
ADD R1, R2, R1         # R1 = R2 + R1
STR R1, [R0, #8]       # Store R1 into three  (pointed to but R0 + 8)
SWI 2                  # Execute a software interrupt

or 

three = one + two

Not sure about the 'SWI 2' It's probably something specific to your platform. Maybe just a generic end of program call.

Upvotes: 7

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