Consider the following C code

for$($int i$=0$; i$<5$; i$++)$

printf$("\%$d

$",$ i$)$;

How would you write this in Haskell?

doPrint $=$ doPrint' $0$

where doPrint' n

$|$ n $<5=$ print n $>>$ doPrint' $($n$-1)$

$|$ otherwise $=$ return $()$

doPrint $=$ let n $=0$ in

do print n

n $<-$ n$+1$

if n $==5$ then break

for $05($

$->$ print $5)$