Chapter 7, Problem 7.12HW

A.

Explanation of Solution

Computing the value of the relocated reference to “swap” in the “callq” instruction:

From the given question, the function “swap” in object file “m.o” is

9: e8 00 00 00 00callqe<main+oxe>

From the given question,

$\begin{array}{c}\text{r}\text{.offset}\text{=}\text{0xa}\\ \text{r}\text{.symbol}\text{=}\text{swap}\\ \text{r}\text{.type}\text{=}\text{R\_X86\_64\_PC32}\\ \text{r}\text{.addend}\text{=}\text{-4}\end{array}$

And then value of “ADDR(.text)” is “0x4004e0” and value of “ADDR(swap)” is “0x4004f8”.

• $\text{ADDR(s)}=\text{ADDR(}\text{.text)}=\text{0x4004e0}$
  • Here “ADDR” means run-time addresses for each section.
• $\text{ADDR(r}\text{.symbol)}\text{=}\text{ADDR(swap)}\text{=}\text{0x4004f8}$

Compute the run time address of the references using the formula of “refaddr = ADDR(s) + r.offset”

$\begin{array}{c}\text{refaddr}\text{=}\text{ADDR(s)}\text{+}\text{r}\end{array}$

B.

Explanation of Solution

Computing the value of the relocated reference to “swap” in the “callq” instruction:

From the given question, the function “swap” in object file “m.o” is

9: e8 00 00 00 00callqe<main+oxe>

From the given question,

From the given question,

$\begin{array}{c}\text{r}\text{.offset}\text{=}\text{0xa}\\ \text{r}\text{.symbol}\text{=}\text{swap}\\ \text{r}\text{.type}\text{=}\text{R\_X86\_64\_PC32}\\ \text{r}\text{.addend}\text{=}\text{-4}\end{array}$

And then value of “ADDR(.text)” is “0x4004d0” and value of “ADDR(swap)” is “0x400500”.

• $\text{ADDR(s)}\text{=}\text{ADDR(}\text{.text)}\text{=}\text{0x4004d0}$
  • Here “ADDR” means run-time addresses for each section.
• $\text{ADDR(r}\text{.symbol)}\text{=}\text{ADDR(swap)}\text{=}\text{0x400500}$

Compute the run time address of the references using the formula of “refaddr = ADDR(s) + r.offset”

$\begin{array}{c}\text{refaddr}\text{=}\text{ADDR(s)}\text{+}\text{r}\end{array}$