This problem considers a linked list that has been created in memory.

We know the linked list consists of $2$ nodes, we know one of the nodes resides at address $0$x$7$fffffffea$10$ and has the following declaration in C$.$

struct lnode $\{$

struct lnode $*$prev;

struct lnode $*$next;

char first;

char $*$word;

int length;

$\}$;

We decide to look at the node in gdb and see the following:

$($gdb$)$ x$/4$xg $0$x$555555554774$

$0$x$555555554774$: $0$x$6$f$722031363253430$x$6$c$65480021736$b$63$

$0$x$555555554784$: $0$x$3$b$031$b$0100216$f$6$c $0$x$000000060000003$c

$($gdb$)$ x$/14$xg $rsp

$0$x$7$fffffffea$00$: $0$x$00007$fffffffeb$580$x$00000001756$e$6547$

$0$x$7$fffffffea$10$: $0$x$00007$fffffffea$400$x$0000000000000000$

$0$x$7$fffffffea$20$: $0$x$00000000000000430$x$0000555555554774$

$0$x$7$fffffffea$30$: $0$x$000000000000000$c $0$x$000055555555473$d

$0$x$7$fffffffea$40$: $0$x$00000000000000000$x$00007$fffffffea$10$

$0$x$7$fffffffea$50$: $0$x$00000000000000480$x$0000555555554781$

$0$x$7$fffffffea$60$: $0$x$00000000000000060$x$2248$a$3$a$88$ddbb$600$

Fill$-$in the following blanks for the node starting at address $0$x$7$fffffffea$10.$

$-$ The field length is stored at address Blank $1($copy the value and paste it here including $0$x$)$

$-$ The field next is stored at address Blank $2($copy the value and paste it here including $0$x$)$

$-$ The value stored in the field prev is Blank $3($copy the value and paste it here including $0$x$)$

$-$ the second ASCII character referenced by word is Blank $4($you can use an ASCII table$)$

$-$ The value stored in the field length is Blank $5($number in decimal$)$This problem considers a linked list that has been created in memory.

We know the linked list consists of $2$ nodes, we know one of the nodes resides at address $0x7$ ffffffea $10$ and has the following

declaration in C$.$

struct Inode $\{$

struct Inode ${?}^{\text{p}\text{p}\text{r}\text{e}\text{v}\text{;}}$

struct Inode ${?}^{**}$ next;

char first;

char "word;

int length;

l;

We decide to look at the node in gdb and see the following:

$($gdb$)\frac{}{4}g0x555555554774$

$0$x$555555554774$: $0$x$6$f$722031363253430$x$6$c$65480021736$b$63$

$0555555554784$ : $0$x$3$b$031$b$0100216$f$6$c $0$x$000000060000003$c

$(gdb)\frac{}{14}xg$$$rsp$

Ox$7$ffffffea$00$: $0$x$00007$ffffffeb$580$x$00000001756$e$6547$

$0$x$7$ffffffea$10$: $0$x$00007$ffffffea$400$x$0000000000000000$

Ox$7$ffffffea$20$: $0000000000000004300000555555554774$

Ox$7$ffffffea$30$: $0000000000000000c0000055555555473d$

$0$x$7$ffffffea$40$: $0$x$00000000000000000$x$00007$ffffffea$10$

Ox$7$ffffffea$50$: $0000000000000004800000555555554781$

$0$x$7$ffffffea$60$: $0000000000000000602248a3a88ddbb600$

Fill$-$in the following blanks for the node starting at address $0$x$7$ffffffea$10.$

The field length is stored at address Blank $1($copy the value and paste it here including $0x)$

The field next is stored at address Blank $2($copy the value and paste it here including $0$x$)$

The value stored in the field prev is Blank $3($copy the value and paste it here including $0$x$)$

the second ASCll character referenced by word is Blank $4($you can use an ASCll table$)$

The value stored in the field length is Blank $5($number in decimal$)$

Blank $1$

Blank $2$ Add your answer

Blank $3$

Blank $4$

Blank $(\frac{A}{5})$