了解ARM汇编代码

Question

任何人都可以向我解释这个简短的程序是干什么的？

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 

如果我想正确的，它增加了“一”到“二”，并将结果在“三”。我对么？

Answer 1

是的。

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 

或

three = one + two 

不知道的 'SWI 2' 这可能是特定于平台的东西。也许只是程序调用的通用结束。