shell script related to square of number question
 

Hi,
Im trying to write a shell script to input few numbers and print every number that is a square of atleast one number given as input.The code i have written is as follows,

echo "Enter total no. of numbers"
read tot
no=0
i=0
num=0
echo "Enter $tot Numbers"
while test $i -lt $tot
do
read no
a[i]=$no
i=`expr $i + 1`
done

echo "DISPLAY NOS"
i=0
while test $i -lt $tot
do
echo ${a[$i]}
i=`expr $i + 1`
done

echo "SQUARE"
i=0
while test $i -lt $tot
do
num=${a[$i]}
for (( j = 0 ; j < $tot ; j++ ))
do
x=${a[$j]}
if test `expr $num \* $num` -eq $x
echo "$x is the square of $num"
fi
j=`expr $j + 1`
done
i=`expr $i + 1`
done

It accepts the nos and displays them too.But later on this error appears,
SQUARE
./m10: line 33: syntax error near unexpected token `fi'
./m10: line 33: ` fi'

What is my mistake?

"if" needs a "then" for the "fi" to work

Once you put the 'then' back, you will discover you have incremented j twice in the for loop, which means it skips every second number.

Thanks for your reply.I have added the then and removed the extra incrementation of j.But now it gives me the error ,
'integer expression expected' when i multiply num using expr.
Is my way of storing the array value in the variable wrong??Should it be,
num={${a[$i]}} instead??
Please reply..

I suspect it is a problem with the input data. Are you entering each number on a new line?

Yes, i am entering each number on a new line..

studywell09 -

To understand what a variable really looks like (even though you *think* it it should be an integer, it might be a letter like "a", or a blank space)...

... then "echo" is your friend:

http://dollhousewiki.fox.com/page/Echo