here is an assembly code for muliply and division. can someone explain why my debug for dx say $2000,$ ax for $13107,$ eax for $858993459,$ and edx for $2000?$ The code is correct but I don't understand why ax dx eax and edx are in those numbers. please explain by using calculations. Thank you!$($gdb$)$ b $\_$start imull $\%$eax #multipli eax bu itself

#After imull, eax$=1000*1000=1000000$

#edx:eax$=0$:$1000000($since the result fits in eax, edx is zero$)$

movw opWord, sax #load opWord$(100)$ into ax

movw opWord, sdx #load opWord$(100)$ into dx

idivw sourceWord #divide dx:ax by sourceWord$(4000)$

#before idivw, dx:ax$=100*100$

#idivw sourceWord$(4000)$

#Quotient$=($dx:ax$)/$sourceWord$=(100$:$100)/4000$

#Qoutieny$=1($ax$),$ remainder $=100($dx$)$

movl opLong, seax #Load opLong$(1000)$ into eax

movl opLong, sedx #Load opLong$(1000)$ into edx

idivl sourceLong #divide edx:eax by sourceLong$(5000)$

#before idivl, edx:eax$=1000$:$1000$

#idivl sourceLong$(5000)$

#Quotient $=($edx:eax$)/$sourceLong $=(1000$:$1000)/5000$

done:

nop

Function $"\_$start" not defined.

Make breakpoint pending on future shared library load? $($y or $[$n$])$ y

Breakpoint $1(\_$start$)$ pending.

$($gdb$)$ P $dx

$($gdb$)$ b

Breakpoint $2$ at $0$x$8049036$

$($gdb$)$ P $ax

$($gdb$)$ b

Note: breakpoint $2$ also set at pc $0$x$8049036,$

Breakpoint $3$ at $0$x$8049036$

$($gdb$)$ P $dx

$($gdb$)$ p $eax

$($gdb$)$ p $edx

$($gdb$)$

$.$data

opByte: $.$byte $8$

opWord: $.$word $100$

opLong: $.$long $1000$

sourceByte: $.$byte $64$

sourceWord: $.$word $4000$

sourceLong: $.$long $5000$

$.$text

$.$global $\_$main

$\_$main:

movw opWord, $\%$ax #Load opWord$(100)$ into ax

imulw $\%$ax #multiply ax by itself

#After imulw, ax$=100*100=10000$

#dx:ax$=0$:$10000($since the result fits in eax, dx is zero$)$

movl opLong, $\%$eax #Load opLong$(1000)$ into eax

imull $\%$eax #multipli eax bu itself

#After imull, eax$=1000*1000=1000000$

#edx:eax$=0$:$1000000($since the result fits in eax, edx is zero$)$

movw opWord, $\%$ax #load opWord$(100)$ into ax

movw opWord, $\%$dx #load opWord$(100)$ into dx

idivw sourceWord #divide dx:ax by sourceWord$(4000)$

#before idivw, dx:ax$=100*100$

#idivw sourceWord$(4000)$

#Quotient$=($dx:ax$)/$sourceWord$=(100$:$100)/4000$

#Qoutieny$=1($ax$),$ remainder $=100($dx$)$

movl opLong, $\%$eax #Load opLong$(1000)$ into eax

movl opLong, $\%$edx #Load opLong$(1000)$ into edx

idivl sourceLong #divide edx:eax by sourceLong$(5000)$

#before idivl, edx:eax$=1000$:$1000$

#idivl sourceLong$(5000)$

#Quotient $=($edx:eax$)/$sourceLong $=(1000$:$1000)/5000$

#Quotient $=0,$ remainder $=1000($edx$)$

done:

nop