The following assumes our implementation of the RV32I architecture.
Consider following table that represents several of the 32-bit registers in the register file:

Register CONTENTS (binary)
R16 0000 0000 0000 0000 0000 0011 0001 1000
R17 0000 0000 0000 0000 0000 0011 0010 0000
R18 0000 0000 0000 0000 0000 0011 0010 0100
... ...

Also consider the following table that represents part of the memory of a 32-bit address space that has an addressability of 1 byte:

ADDRESS (hex)
CONTENTS (binary)

... ...
0x00000324 0000 0000 0100 0010 1000 0011 1011 0011
0x00000320 0000 0000 0101 0011 0000 0011 1011 0011
0x0000031C 0000 0000 0000 0011 1010 0010 1000 0011
0x00000318 0100 0000 0000 0011 1010 0010 0000 0011
... ...
0x0000021C 0000 0000 0000 1000 0000 0000 0110 0111
0x00000218 0000 0000 0000 1001 0000 0000 0110 0111
0x00000214 0000 0000 0000 1000 1000 0000 0110 0111
... ...

Assume the PC has the address 0x00000214 when FETCH phase begins. After that first instruction executes, which one describes the second instruction to execute?

REFERENCE:

Table of RV32I Instructions:

31 25 24 20 19 15 14 12 11 7 6 0
ADD 0000000 rs2 rs1 000 rd 0110011
JR imm[11:0] rs1 000 00000 1100111
LW imm[11:0] rs1 010 rd 0000011

Question

The following assumes our implementation of the RV32I architecture.
Consider following table that represents several of the 32-bit registers in the register file:

Register	CONTENTS (binary)
R16	0000 0000 0000 0000 0000 0011 0001 1000
R17	0000 0000 0000 0000 0000 0011 0010 0000
R18	0000 0000 0000 0000 0000 0011 0010 0100
...	...

Also consider the following table that represents part of the memory of a 32-bit address space that has an addressability of 1 byte:

ADDRESS (hex)
	CONTENTS (binary)

...	...
0x00000324	0000 0000 0100 0010 1000 0011 1011 0011
0x00000320	0000 0000 0101 0011 0000 0011 1011 0011
0x0000031C	0000 0000 0000 0011 1010 0010 1000 0011
0x00000318	0100 0000 0000 0011 1010 0010 0000 0011
...	...
0x0000021C	0000 0000 0000 1000 0000 0000 0110 0111
0x00000218	0000 0000 0000 1001 0000 0000 0110 0111
0x00000214	0000 0000 0000 1000 1000 0000 0110 0111
...	...

Assume the PC has the address 0x00000214 when FETCH phase begins. After that first instruction executes, which one describes the second instruction to execute?

REFERENCE:

Table of RV32I Instructions:

31	25	24	20	19	15	14	12	11	7	6	0
ADD	0000000	rs2	rs1	000	rd	0110011
JR	imm[11:0]	rs1	000	00000	1100111
LW	imm[11:0]	rs1	010	rd	0000011

BlackTom AI · Accepted Answer

The 32-bit word at 0x00000214 decodes to a JALR (JR) that uses R17 as the target, so the next fetch comes from the address in R17; that target word is an ADD instruction adding R5 and R6 and writing the result to R7.

1) Relevant knowledge:
- RV32I is little-endian: the byte at the lowest address is the least-significant byte of the 32-bit instruction.
- JALR (JR) encoding: imm[11:0] (bits31:20), rs1 (19:15), funct3 (14:12)=000, rd (11:7)=00000 for JR, opcode (6:0)=1100111 (0x67).
- R-type (ADD) encoding: funct7 (31:25), rs2 (24:20), rs1 (19:15), funct3 (14:12)=000, rd (11:7), opcode (6:0)=0110011 (0x33).

2) Step-by-step decoding of instruction at PC = 0x00000214:
Memory word shown at 0x00000214: 0000 0000 0000 1000 1000 0000 0110 0111.
Interpret as little-endian bytes (low addr -> high): b0=01100111 (0x67), b1=10000000 (0x80), b2=00001000 (0x08), b3=00000000 (0x00). Full 32-bit value = 0x00088067.

Extract JALR fields:
- imm[11:0] = bits31:20 = 0000 0000 0000 = 0
- rs1 = bits19:15 = 10001 = 17 → R17
- funct3 = 000, rd = 00000, opcode = 1100111 → JALR (JR)
So the instruction is JR R17 (jump to R17 + 0).

3) Compute the jump target:
R17 contents given: 0000 0000 0000 0000 0000 0011 0010 0000 = 0x00000320.
JALR computes PC := (R17 + imm) & ~1 = 0x00000320.

4) Decode the instruction at 0x00000320:
Memory at 0x00000320: 0000 0000 0101 0011 0000 0011 1011 0011.
Bytes (little-endian): b0=10110011 (0xB3), b1=00000011 (0x03), b2=01010011 (0x53), b3=00000000 (0x00). Full value = 0x005303B3.

Partitioning into R-type fields:
- funct7 (31:25) = 0000000
- rs2 (24:20) = 00101 = 5 → R5
- rs1 (19:15) = 00110 = 6 → R6
- funct3 = 000
- rd (11:7) = 00111 = 7 → R7
- opcode = 0110011 → ADD
Thus the instruction is ADD rd=R7, rs1=R6, rs2=R5 (i.e., R7 := R6 + R5), which matches the phrasing “ADD R5 to R6 and save result in R7.”

5) Why other possibilities are incorrect:
- Fetching sequentially (PC+4 = 0x00000218) would be wrong because the first instruction is a JR that redirects PC to R17. 
- Interpreting the word at 0x00000320 as LW or another opcode is incorrect because the opcode bits (0110011) and funct7/funct3 fields uniquely identify an R-type ADD with the shown rd/rs fields.

Conclusion: After the first instruction (JR R17) executes, the second instruction executed is the ADD that computes R7 := R6 + R5, i.e., “ADD R5 to R6 and save result in R7,” matching the provided answer.

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