using this code, answer the question in the image

render$\_$hw$01($pFrame$\_$Buffer& fb$,$ HW$01\_$Data& hw$01\_$data$)$

$\{$

$//$ Window Size, in Pixels

$//$

const int win$\_$width $=$ fb$.$width$\_$get$()$;

const int win$\_$height $=$ fb$.$height$\_$get$()$;

const auto& tryout$1[[$maybe$\_$unused$]]=$ hw$01\_$data.tryout$1$;

const auto& tryout$2[[$maybe$\_$unused$]]=$ hw$01\_$data.tryout$2$;

kbd$\_$input$\_$handle$($fb$,$hw$01\_$data$)$;

image$\_$box$\_$drag$($fb$,$hw$01\_$data$)$; $//$ Update image box corners.

$///$ Coordinates Used for Placement of Other Objects

$//$

const int sq$\_$slen $=$ min$($win$\_$width,win$\_$height$)*\mathrm{.}45$;

const int sq$\_$x$0=$ sq$\_$slen $*0\mathrm{.}1,$ sq$\_$y$0=$ win$\_$height $-150$;

const int sq$\_$x$1=$ sq$\_$x$0+$ sq$\_$slen, sq$\_$y$1=$ sq$\_$y$0-$ sq$\_$slen;

$//$

$//$ One vertex of the square is at $($sq$\_$x$0,$sq$\_$y$0)$ and the opposite

$//$ vertex is at $($sq$\_$x$1,$sq$\_$y$1).$

$//$ RRGGBB

const uint$32\_$t color$\_$red $=0$xff$0000$;

const uint$32\_$t color$\_$blue $=0$xff;

const uint$32\_$t color$\_$green $=0$xff$00$;

$///$ Demo Code: Draw Sine Waves $--$ No need to modify this.

$//$

int num$\_$waves $=4$;

float plot$\_$ht $=$ sq$\_$slen$/$float$($num$\_$waves$)$;

for $($ float

yb $=$ sq$\_$y$0-$ plot$\_$ht$/2,$ freq $=4*$ M$\_$PI $/($ win$\_$width $-40-$ sq$\_$x$1)$;

yb $>$ sq$\_$y$1$;

yb $-=$ plot$\_$ht$,$ freq $*=2)$

for $($ int x $=20$; x $$ win$\_$width $-20$; x$++)$

$\{$

int y $=$ yb $+$ sin$($ x $*$ freq $)*$ plot$\_$ht $*\mathrm{.}45$;

fb$[$ y $*$ win$\_$width $+$ x $]=0$xff$00$;

$\}$

$///$ Demo Code: Draw Lines Radiating From a Point $--$ No need to modify.

$//$

$//$ This code calls the line$\_$draw routine.

$//$

int n$\_$rects $=24$;

int n$\_$pts $=$ n$\_$rects $*3$;

float c$\_$r$1=$ sq$\_$slen$/6$; $//$ Inner radius

float c$\_$r$2=$ sq$\_$slen$/2$; $//$ Outer radius.

pCoor c$\_$ctr$($ c$\_$r$2+20,$ c$\_$r$2+20)$; $//$ Center of "circle"

float delta$\_$theta $=2*$ M$\_$PI $/$ n$\_$pts;

vector vecs;

for $($ int i$=0$; i$=$ img$\_$pixels.size$()$;

$//$

$//..$ and read pixel if they are valid, otherwise use color red.

$//$

uint$32\_$t img$\_$pixel $=$ out$\_$of$\_$range $?$ color$\_$red : img$\_$pixels$[$ img$\_$idx $]$;

$//$

$//$ Finally, write image pixel into frame buffer.

$//$

fb$[$ fb$\_$y $*$ win$\_$width $+$ fb$\_$x $]=$ img$\_$pixel;

$\}$Modify the code in render$\_$hw$01$ so that the aspect ratio is preserved. When zoom is $1($the

default$)$ one image should be across with width of the box, regardless of how the image box is

resized. For this problem don't worry about unfilled parts of the image box $($shown in red in the

screenshot$).$

Hint: This is easy, just one line needs to be changed.