Return value in R0 in ARM Architecture - Time & Space Complexity
We want to understand how the time taken by ARM code changes when it returns a value in register R0.
Specifically, how does the return operation affect execution time as input changes?
Analyze the time complexity of the following ARM code snippet.
MOV R0, #5 ; Load immediate value 5 into R0
BX LR ; Return from function
This code sets the return value to 5 in R0 and then returns from the function.
Look for any repeated steps or loops in the code.
- Primary operation: Single move instruction and return instruction
- How many times: Each runs once per function call
The instructions run the same number of times regardless of input size.
| Input Size (n) | Approx. Operations |
|---|---|
| 10 | 2 instructions |
| 100 | 2 instructions |
| 1000 | 2 instructions |
Pattern observation: Execution time stays constant no matter the input size.
Time Complexity: O(1)
This means the time to return a value in R0 does not grow with input size; it takes a fixed amount of time.
[X] Wrong: "Returning a value in R0 takes longer if the value is bigger or the input is larger."
[OK] Correct: The instructions to move a value into R0 and return run once and take the same time regardless of the value size or input.
Understanding that returning values in registers is a constant-time operation helps you reason about function call costs in low-level programming.
"What if the function returned a value stored in memory instead of R0? How would the time complexity change?"
Practice
Solution
Step 1: Understand ARM calling convention
In ARM, the first register R0 is used to pass the return value from a function to the caller.Step 2: Identify the return value register
Since R0 holds the return value, the caller reads the result from R0 after the function returns.Final Answer:
R0 -> Option CQuick Check:
Return value register = R0 [OK]
- Confusing LR with return value
- Using SP as return value
- Thinking R1 holds return value
Solution
Step 1: Identify return instruction in ARM
The instructionBX LRbranches to the address stored in the link register (LR), effectively returning from the function.Step 2: Confirm usage after setting R0
After placing the return value in R0,BX LRis used to return control to the caller.Final Answer:
BX LR -> Option DQuick Check:
Return instruction = BX LR [OK]
- Using MOV instead of BX LR to return
- Storing R0 to stack before return
- Using ADD as return instruction
MOV R0, #5 BX LRWhat value will the caller receive after this function returns?
Solution
Step 1: Analyze MOV instruction
The instructionMOV R0, #5sets the R0 register to 5, which is the return value.Step 2: Analyze return instruction
BX LRreturns control to the caller, with R0 holding the return value 5.Final Answer:
5 -> Option AQuick Check:
R0 value after return = 5 [OK]
- Thinking LR value is returned
- Assuming R0 resets to 0
- Ignoring MOV sets return value
ADD R0, R0, #10 BX LR
Solution
Step 1: Check initial value of R0
The instructionADD R0, R0, #10adds 10 to R0, but R0 is not initialized, so the result is unpredictable.Step 2: Confirm correct return method
BX LRis correct to return; the problem is the uninitialized R0.Final Answer:
R0 is not initialized before ADD -> Option AQuick Check:
Uninitialized R0 causes wrong return value [OK]
- Assuming R0 starts at zero
- Confusing BX LR with MOV PC, LR
- Using wrong register for return value
Solution
Step 1: Understand input and output registers
Inputs are in R0 and R1; the sum should be stored in R0 as the return value.Step 2: Check each option
ADD R0, R0, R1 BX LR adds R0 and R1, stores result in R0, then returns with BX LR, which is correct.Final Answer:
ADD R0, R0, R1 BX LR -> Option BQuick Check:
Sum in R0 before BX LR returns correct value [OK]
- Storing sum in R1 instead of R0
- Using SUB instead of ADD
- Not returning with BX LR
Ready to prove it?
Pass the test and this pattern moves unknown
