Question

For both of the instructions shown below, the initial conditions are shown. The result (which the instruction stores in r1) is also shown. For each instruction, answer these questions:

•What is the state of the flags after the instruction execute?
• If the operands were signed, are the results valid? Why or why not?
• If the operands were unsigned, are the results valid? Why or why not?
A) initial conditions:
r0 = 0xAAAA_AAAA
r1 = 0x5555_5555
ADDS r1,r1,r0
result = 0xFFFF_FFFF

B) initial conditions:
r0 = 0x7FFF_FFFD
r1 = 0x0000_000F
ADDS r1,r0,r1
result = 0x8000_000C

Answer 1

After executing the ADDS instructions, the state of flags varies, and the results' validity for signed and unsigned operands differs. Instruction A generates an overflow for signed integers, while Instruction B's result is misleading as it should not be negative given one positive and one small negative input.

Step-by-step explanation:

For the given ARM assembly language instructions, the flag state after execution and the validity of the results for signed and unsigned operands are considered.

A) ADDS r1,r1,r0 result = 0xFFFF_FFFF

Flags: Negative (N) = 1, Zero (Z) = 0, Carry (C) = 1, Overflow (V) = 1.
For signed operands, the result is not valid because an overflow has occurred. The result is too large to be represented in a signed 32-bit integer. For unsigned operands, the result is valid as it represents the correct sum of the inputs.

B) ADDS r1,r0,r1 result = 0x8000_000C

Flags: Negative (N) = 1, Zero (Z) = 0, Carry (C) = 0, Overflow (V) = 0.
For signed operands, the result may appear valid as there is no overflow flag set, but interpreting the operands and result as a 2's complement signed integer, an overflow actually has occurred because adding a positive and negative number should not produce a negative result. For unsigned operands, the result is valid as it represents the correct sum of the inputs.