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1Learning Outcomes

No video. We recommend pulling up the memory section of the RISC-V Green Card.

2Example 1

Consider the assembly code:

li x11 0x93F5 
sw x11 0(x5)
lb x12 1(x5)

Suppose that the memory layout starts as in Figure 1. After executing these instructions, what is in x12?

"TODO"

Figure 1:Starting memory layout for Example 1.

Hint: See the section on pseudoinstructions like li.

Show Answer

R[x12] is 0xFFFF FF93.

Explain li x11 0x93F5

Load Immediate (li rd imm) is addi rd x0 0x93F5. The pseudoinstruction therefore maps to addi x11 x0 0x93F5, so set R[x11] (the value of register x11) to the bits 0x000093F5.

"TODO"

Figure 2:Solution (1/3) for Example 1.

Explain sw x11 0(x5)

Compute memory address as base register + offset, or R[x5] + 0 = 0x100 + 0 = 0x100. Store R[x11] (the value of x11) to memory at address 0x100. On this (assumed) little endian architecture, 0xF5 gets stored at the lowest byte 0x100.

"TODO"

Figure 3:Solution (2/3) for Example 1.

Explain lb x12 1(x5)

Compute memory address as base register + offset, or R[x5] + 1 = 0x100 + 1 = 0x101. Load byte 0x93 from memory at address 0x101 into the lowest byte of register x11.

lb means we must sign-extend. The top bit of 0x93 is 1, so fill top 24 bits with 1s:

0x93 = 0b1001 0011
--> 0b1…1 1001 0011
--> 0xFFFF FF93
"TODO"

Figure 4:Solution (3/3) for Example 1.

Solution: R[x11] (the value in x11) is 0xFFFFFF93.

2.1Example 2

Suppose that x and y are int * pointers whose values are in registers x3 and x5.

How do we translate the statement *x = *y; into assembly?

Consider the following instructions:

  1. add x3 x5 zero

  2. add x5 x3 zero

  3. lw x3 0(x5)

  4. lw x5 0(x3)

  5. lw x8 0(x5)

  6. sw x8 0(x3)

  7. lw x5 0(x8)

  8. sw x3 0(x8)

And consider the following choices:

L10 RISC-V Data Transfer
Data Transfer Instructions
L10 RISC-V Data Transfer
Example: Stack Arrays