For this problem we will focus on a hypothetical $16-$bit machine $($instead of $32$ or $64$

as we are accustomed to$).$ This means that, for this problem, an int is $2-$bytes, a long is $4-$bytes,

while a char still is $1-$byte. Also assume that signed values are encoded using a two$$s$-$complement

encoding. Recall that the printf options relevant to this question are as follows:

$\%$p $-$ print pointer$/$address $($in hex$)$

$\%$c $-$ print byte as ascii character

$\%$x $-$ print int$/$unsigned in hex

$\%$u $-$ print unsigned int in decimal

$\%$d $-$ print $($signed$)$ int in decimal

$\%$lx $-$ print long$/$unsigned long in hex

Note: If the variable passed into printf is not the same as what the format string expects, the

input will be converted to the correct type before printing out. As an example, if I try to print a

char using $\%$d $($which is expecting an int$),$ the char will be converted to an int first, and then

the value of that int will be printed out in decimal.

Consider the following portion of memory, displayed as a table as in class and in previous labs.

Each table cell is a memory location, which contains a single byte, with values shown in hex, and

the address of the rightmost byte on each row is shown to the right of the row. Addresses increase

as we move to the left and up in the table.

Address

d$49$c $0$x$7$d$4$e

$4$e $880$x$7$d$4$c

ff $8$d $0$x$7$d$4$a

$1$a $3$c $0$x$7$d$48$