Chapter 2

2.4

[10] <§2.2, 2.3> For the RISC-V assembly instructions below, what is the corresponding C statement? Assume that the variables f, g, h, i, and j are assigned to registers x5, x6, x7, x28, and x29, respectively, and the base address of the arrays A and B are in registers x10 and x11, respectively.

slli x30, x5, 3       // x30 = f*8
add  x30, x10, x30    // x30 = &A[f]
slli x31, x6, 3       // x31 = g*8
add  x31, x11, x31    // x31 = &B[g]
ld   x5, 0(x30)       // f = A[f]

addi x12, x30, 8
ld   x30, 0(x12)
add  x30, x30, x5
sd   x30, 0(x31)

slli x30, x5, 3       // x30 = f * 8
add  x30, x10, x30    // x30 = &A[f]
slli x31, x6, 3       // x31 = g * 8
add  x31, x11, x31    // x31 = &B[g]
ld   x5, 0(x30)       // f = A[f]
addi x12, x30, 8      // x12 = &A[f] + 8
ld   x30, 0(x12)      // x30 = A[f+1]
add  x30, x30, x5     // x30 = x30 + x5 = A[f+1] + A[f]
sd   x30, 0(x31)      // B[g] = x30 = A[f+1] + A[f]

故整段汇编可以等效为 C 语句：B[g] = A[f+1] + A[f]。

2.8

[10] <§2.2, 2.3> Translate the following RISC-V code to C. Assume that the variables f, g, h, i, and j are assigned to registers x5, x6, x7, x28, and x29, respectively, and the base address of the arrays A and B are in registers x10 and x11, respectively.

addi x30, x10, 8
addi x31, x10, 0
sd   x31, 0(x30)
ld   x30, 0(x30)
add  x5,  x30, x31

addi x30, x10, 8        // x30 = &A[1]
addi x31, x10, 0        // x31 = &A[0]
sd   x31, 0(x30)        // A[1] = x31 = &A[0]
ld   x30, 0(x30)        // x30 = A[1] = &A[0]
add  x5,  x30, x31      // f = &A[0] + &A[0]

故整段汇编可以等效为 C 语句：f = (&A[0]) * 2。

2.12

[5] <§§2.2, 2.5> Provide the instruction type and assembly language instruction for the following binary value:

0000 0000 0001 0000 1000 0000 1011 0011two

Hint: Figure 2.20 may be helpful.

• opcode: 0110011
• rd: 00001
• funct3: 000
• rs1: 00001
• rs2: 00001
• funct7: 0000000

故这条语句是 R 型指令，对应的汇编语言为 add x1, x1, x1。

2.14

[5] <§3.5> Provide the instruction type, assembly language instruction, and binary representation of instruction described by the following RISC-V fields:

opcode=0x33, funct3=0x0, funct7=0x20, rs2=5, rs1=7, rd=6

R 型指令，sub x6, x7, x5

2.17

[5] <§2.6> Assume the following register contents:

x5 = 0x00000000AAAAAAAA, x6 = 0x1234567812345678

2.17.1

[5] <§2.6> For the register values shown above, what is the value of x7 for the following sequence of instructions?

slli x7, x5, 4
or   x7, x7, x6

slli x7, x5, 4   // x7 = x5 << 4 = 0x0000000AAAAAAAA0
or x7, x7, x6    // x7 = x7 | x6 = 0x1234567ababefef8

故结果为：0x1234567ababefef8

2.17.2

[5] <§2.6> For the register values shown above, what is the value of x7 for the following sequence of instructions?

slli x7, x6, 4

结果为：0x2345678123456780。

2.17.3

[5] <§2.6> For the register values shown above, what is the value of x7 for the following sequence of instructions?

srli x7, x5, 3
andi x7, x7, 0xFEF

结果为：0x545。

2.22

Suppose the program counter (PC) is set to 0x20000000.

2.22.1

[5] <§2.10> What range of addresses can be reached using the RISC-V jump-and-link (jal) instruction? (In other words, what is the set of possible values for the PC after the jump instruction executes?)

2.22.2

[5] <§2.10> What range of addresses can be reached using the RISC-V branch if equal (beq) instruction? (In other words, what is the set of possible values for the PC after the branch instruction executes?)

2.24

Consider the following RISC-V loop:

LOOP: beq x6, x0, DONE
      addi x6, x6, -1
      addi x5, x5, 2
      jal x0, LOOP
DONE:

2.24.1

[5] <§2.7> Assume that the register x6 is initialized to the value $10$. What is the final value in register x5 assuming the x5 is initially zero?

执行 x6 次，每次让 x5 += 2，故结果为 x5 的最终结果为 $20$。

2.24.2

[5] <§2.7> For the loop above, write the equivalent C code. Assume that the registers x5 and x6 are integers acc and i, respectively.

int acc = 0, i = 10;
while (i != 0) {
	i = i - 1
	acc = acc + 2
}

2.24.3

[5] <§2.7> For the loop written in RISC-V assembly above, assume that the register x6 is initialized to the value $N$. How many RISC-V instructions are executed?

共 $4N+1$ 条，其中最后一条是 x6 == x0 时跳出循环。

2.24.4

[5] <§2.7> For the loop written in RISC-V assembly above, replace the instruction “beq x6, x0, DONE” with the instruction “blt x6, x0, DONE” and write the equivalent C code.

while (i >= 0) {
	i = i - 1
	acc = acc + 2
}

2.29

[30] <§2.8> Implement the following C code in RISC-V assembly. Hint: Remember that the stack pointer must remain aligned on a multiple of 16.

int fib(int n) {
	if (n == 0)
		return 0;
	else if (n == 1)
		return 1;
	else
		return fib(n-1) + fib(n-2);
}

函数参数 $n$ 存储在 a0（x10）寄存器中，最后返回值 fib(n) 也会存储在 a0 寄存器中。

FIB:
	beq a0, x0, RETURN     // if (n == 0) return 0
	addi t0, x0, 1
	beq a0, t0, RETURN     // if (n == 1) return 1

	addi sp, sp, -16
	sd a0, 0(sp)
	sd ra, 8(sp)
	addi a0, a0, -1
	jal ra, FIB
	mv t0, a0
	ld a0, 0(sp)
	sd t0, 0(sp)
	addi a0, a0, -2
	jal ra, FIB
	ld t0, 0(sp)
	ld ra, 8(sp)
	add a0, a0, t0         // ret value = fib(n - 1) + fib(n - 2)

RETURN:
	jalr x0, 0(ra)            // return