Suppose the ARM 6 has the following initial register and memory contents

Suppose the ARM 6 has the following initial register and memory contents: r1=00000000; r2-87654321; r3-a05877f9; NZCV 0000.identify the new contents of register or flag that is changed by execution of the following instructions: (a) add r1, r2, r3 (b) adds r1, r3, r3

The Correct Answer and Explanation is:

Let’s examine the execution of the two ARM instructions given the initial register contents:

Initial Register and Flag Values:

• r1 = 0x00000000
• r2 = 0x87654321
• r3 = 0xA05877F9
• NZCV (Negative, Zero, Carry, Overflow) = 0000

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(a) ADD r1, r2, r3

This instruction performs:

r1 = r2 + r3

Calculation:

r2 = 0x87654321
r3 = 0xA05877F9

r2 + r3 = 0x87654321 + 0xA05877F9
         = 0x277DBB1A (with carry-out ignored because it's ADD, not ADDS)

Since we’re using ADD, the condition flags are not affected.

New Register Content:

• r1 = 0x277DBB1A
• Flags unchanged (NZCV = 0000)

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(b) ADDS r1, r3, r3

This instruction performs:

r1 = r3 + r3

and updates the flags (since it is ADDS).

Calculation:

r3 = 0xA05877F9

r3 + r3 = 0xA05877F9 + 0xA05877F9
         = 0x140B0EFF2

This is a 33-bit result. ARM registers are 32-bit, so the result stored in r1 will be:
r1 = 0x40B0EFF2 (lower 32 bits)

Carry-out occurred, so **C flag = 1**

Now let’s evaluate flags:

• N (Negative): 0 (MSB of result is 0)
• Z (Zero): 0 (Result is not zero)
• C (Carry): 1 (because the result overflowed 32 bits)
• V (Overflow): 1 (because adding two negative numbers gave a positive result)

New Register and Flags:

• r1 = 0x40B0EFF2
• NZCV = 0011 → N=0, Z=0, C=1, V=1

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Summary of Changes:

Instruction	New r1 Value	Flags Affected	New NZCV
ADD r1, r2, r3	0x277DBB1A	No	Unchanged (0000)
ADDS r1, r3, r3	0x40B0EFF2	Yes	0011 (C=1, V=1)

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300-Word Explanation:

The ARM instructions ADD and ADDS both perform arithmetic addition, but they differ significantly in how they affect the condition flags (NZCV). Let’s break down the effects step-by-step using the given initial values:

• ADD r1, r2, r3 simply computes the sum of r2 and r3 and stores the result in r1. The binary sum of 0x87654321 and 0xA05877F9 results in 0x127DBB1A2, which is more than 32 bits. Since ADD does not update the condition flags, even though an arithmetic carry occurs, it is ignored for flag purposes. The final result stored in r1 is the lower 32 bits: 0x277DBB1A. The flags remain unchanged.
• ADDS r1, r3, r3 performs an addition of r3 with itself and stores the result in r1, but unlike ADD, it updates the condition flags. The sum of 0xA05877F9 + 0xA05877F9 equals 0x140B0EFF2. Only the lower 32 bits (0x40B0EFF2) are stored in r1. Because the result exceeds 32 bits, a carry occurs, so the C flag is set. Moreover, adding two negative numbers (as the sign bit of r3 is 1) resulting in a positive value causes a signed overflow, so the V flag is also set. The N flag is 0 (most significant bit of the result is 0), and Z is 0 (result is not zero).

In summary, ADD only changes the destination register, while ADDS changes both the destination register and the status flags. This distinction is critical in ARM assembly, especially for conditional execution based on result flags

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