CODEWITHSUNDEEP
bashshell
Shesh Narayan Deshmukh
Shesh Narayan Deshmukh

Reputation: 333

Why bash consider every character as a number

This is the code:

sum=0
i=1

while [ $i -le $# ]
do
if [[ $i =~ ^[0-9]+$ ]];
then
        sum=$(($sum+$(eval echo '$'$(eval echo ${i}))))
        echo "$(eval echo '$'$(eval echo $i)) is number"
        i=$((i+1))
else
        echo "$(eval echo '$'$(eval echo $i)) is not a number"
        i=$((i+1))
fi
done
echo "sum is $sum"

and this is the output when I try to distinguish b/w numbers and characters:

$ bash sumOfGivenNums.sh 3 4 3 a
3 is number
4 is number
3 is number
a is number
sum is 10

Upvotes: 0

Views: 58

Answers (1)

larsks
larsks

Reputation: 311665

You've got a lot of really unnecessary code in your script; you almost never want to use the eval function. I would probably write something like this:

sum=0

for arg in "$@"; do
  if [[ $arg =~ ^[0-9]+$ ]]; then
    echo "$arg is a number"
    (( sum += arg ))
  else
    echo "$arg is not a number."
  fi
done

echo "sum is $sum"

Upvotes: 4

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