Question

Evaluate the following ARM code line by line and show the state of each register and PC after execution.

Initial state: X1 = 0, X2 = 0, X3 = 0, X4 = 0x7, X5 = 0

If there are multiple values for a given register at a particular instruction list them in order

		X1	X2	X3	X4	X5	PC after execution
0x100	ADDI  X1,X1,#4096
0x104	B #8
0x108	ADDI X3,X3,#1
0x10C	CBNZ X3, #2
0x110	CBZ  X2, #2
0x114	B #-1
0x118	AND X5,X3,X4
0x11C	XOR X5,X3,X4
0x120	ORR X2,X3,X4
0x124	CBZ X2, #7
0x128	BREAK

Answer 1

Below are the details of all the instructions:

Note: The instructions do not seem to form an actual practical program and hence, they are merely judging the understanding of these instructions.

In snapshot:

In text:

		X1	X2	X3	X4	X5	PC after execution
0x100	ADDI  X1,X1,#4096	0, 0x1000	0	0	0x7	0	0x104
Equivalent to X1=X1+1000 (4096 = 0x1000). PC points at next instruction at 0x104
0x104	B #8	0, 0x1000	0	0	0x7	0	0x,108, 0x128
Equivalent to a jump instruction with an offset of 8*4. So PC jumps to 0x108 (pointing to next direction) plus 8*4
0x108	ADDI X3,X3,#1	0x1000	0	0, 0x1	0x7	0	0x10C
Equivalent to X3=X3+1. PC points at next instruction at 0x10C
0x10C	CBNZ X3, #2	0x1000	0	0x1	0x7	0	0x110, 0x118
If X3 not equal to 0, then jump 2 steps ahead. PC = PC+2*4
0x110	CBZ  X2, #2	0x1000	0	0x1	0x7	0	0x114, 0x11C
If X2 equal to 0, then jump 2 steps ahead. PC = PC+2*4
0x114	B #-1	0x1000	0	0x1	0x7	0	0x118, 0x114
Equivalent to a jump instruction with an offset of -1*4. So PC jumps to 0x108 (pointing to next direction) minus 4. This might result in an infinite loop
0x118	AND X5,X3,X4	0x1000	0	0x1	0x7	0, 0x1	0x11C
Equivalent to X5=X3 AND X1. This is a bitwise operation [X3 (0000 0001) AND X4 (0000 0111) = 0000 0001] . PC points at next instruction.
0x11C	XOR X5,X3,X4	0x1000	0	0x1	0x7	0x1,0x6	0x120
Equivalent to X5=X3 XOR X1. This is a bitwise operation [X3 (0000 0001) XOR X4 (0000 0111) = 0000 0110] . PC points at next instruction.
0x120	ORR X2,X3,X4	0x1000	0, 0x7	0x1	0x7	0x6	0x124
Equivalent to X2=X3 XOR X4. This is a bitwise operation [X3 (0000 0001) OR X4 (0000 0111) = 0000 0111] . PC points at next instruction.
0x124	CBZ X2, #7	0x1000	0, 0x7	0x1	0x7	0x6	0x128, 0x144
If X2 equal to 0, then jump 2 steps ahead. PC = PC+7*4
0x128	Break	0x1000	0, 0x7	0x1	0x7	0x6	0x129
End program, size of instruction is 1.