The NOT is an assembly operator that can be used in RISC$-$V assembly programs. However, NOT is not listed as a RV$32$I operator. This means that the NOT operator is translated into a

different statement. What is the translation of 'not rd$,$ rs$1\text{'}$ in RISC$-$V ISA? Justify why the translation works correctly.

write RISC$-$V assembly code for placing the following immediate

constants in register a$0.$ Use a minimum number of instructions. Comment each

line of assembly code.

a$.-2037$

b$.4169$

c$.0$xD$6$A

d$.10101101101100000100112$

Convert the following high$-$level code into RISC$-$V assembly code.

Assume that the signed integer variables a$,$ b$,$ c$,$ d$,$ e and f are in registers t$0,$ t$1,$ t$2,$

t$3,$ t$4$ and t$5.$ Use only RV$32$I operators. Comment each line of assembly code.

if $(($a $>$ b$)$ && $($c $>$ d$))\{$

e $=$ e $/4$;

f $=$ f $*4$;

$\}$

else$\{$

e $=($a $-$ b$)+($c $-$ d$)$;

f $=($a $+$ b$)-($c $+$ d$)$;

$\}$

Write a RISC$-$V assembly function called 'reverse$\_$char$()\text{'}$ to reverse an

array of characters. The function takes the base address of the array and the

number of elements as arguments. It should return the base address of the array

as the return value. Comment each line of assembly code

Convert the following 'fibonacci$()\text{'}$ function, which calculates the

Fibonacci number recursively, into RISC$-$V assembly code. The function takes a

number as an argument. Comment each line of assembly code.

int fibonacci $($int num$)\{$

if $($n $<=1)$

return n;

else

return $($fibonacci$($n$-1)+$ fibonacci$($n$-2))$;

$\}$

Convert the following high$-$level code into RISC$-$V assembly code.

Comment each line of assembly code.

int sum$\_$number$($int n$)\{$

int result $=0$;

$//$ calculate n $+$ n$-1+...+1$

while $($n $>=1)\{$

result $+=$ n;

n$--$;

$\}$

return result;

$\}$

int swap$\_$sum$($int $*$x$,$ int $*$y$)\{$

$//$ swap numbers

int temp;

temp $=*$y;

$*$y $=*$x;

$*$x $=$ temp;

$//$ return the sum of number

int sum$\_$x $=$ sum$\_$number$(*$x$)$;

int sum$\_$y $=$ sum$\_$number$(*$y$)$;

return sum$\_$x $+$ sum$\_$y;

$\}$