sum 1 = 0 sum2 = 0 cnt = 0 while read W do echo $W | grep A >/dev/null && cnt= 'echo -n $W | wc -c' && let sum1=sum1+cnt cnt=0 echo $W | grep i >/dev/null | cnt='echo -n $W | wc -c && let sum2=sum2+cnt done echo $sum1 echo $sum2
Adams Aspirin Asimov Brunner Clarke
Dick Gibson Grimwood Haldemann Herbert
Lem Niven Simmons Strugatzki Verne
The solution for this is according to the tutor
cnt’echo -n $W | wc -c’ 3 times
Line 8 should be 7 times
sum 1 = 17 times
sum 2 = 42 times
My problem is I dont understand how he comes to this count,
I understand that the first echo has something to do with the letter ‘A’
and the 2nd echo is looking for ‘i’.
sum1=sum1+cnt means that the value of cnt gets added to sum1. Last lines are just printing out the value of sum1 and sum2.
Can someone explain me the middle part please?