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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?

"Initial state: registers x5, x11, and x12 hold 0x100, 0xABCDEFAB, and 0xCDEFABCD beside a little-endian memory grid from 0x100 through 0x10C with associated byte values."

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.

"After load immediate: x11 updates to 0x000093F5 while x5 and x12 are unchanged and memory still shows the original byte grid."

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.

"After store word: the word at 0x100 is highlighted in green as 0x000093F5 in little-endian, matching the updated value of x11, with other addresses of memory unchanged."

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
"After load byte: byte 0x93 at offset plus one from 0x100 is highlighted and x12 holds an updated sign-extended value 0xFFFFFF93 while x5 and x11 retain prior values."

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