Floating point in UNIX using bash Shell

Question

Please help me to find out this, I am very new to UNIX, My code is really can't get the right answer...

#! /usr/bin/env bash

echo -n "How many number : "; read num
for ((i=0; i<$num; i++));
do 
    echo -n "Enter your number : "; read number 
total+=$(echo "${number}" | bc)
echo "${total}"

done

#! /usr/bin/env bash echo -n "How many number : "; read num for ((i=0; i<$num; i++)); do echo -n "Enter your number : "; read number total+=$(echo "${number}" | bc) echo "${total}" done — user3183246
– user3183246, Commented Jan 10, 2014 at 19:49
Please edit your question instead of putting what is missing in a comment. — Jonathan Drapeau
– Jonathan Drapeau, Commented Jan 10, 2014 at 20:08

John1024 · Accepted Answer · 2014-01-10 20:50:58Z

2

echo -n "Enter your numbers: "
read numbers
total=$(echo "$numbers" | sed 's/ \+/ + /g' | bc)
echo "The total is $total"

Sample use:

Enter your numbers: 4 6 2.47
The total is 12.47

MORE: In your script, the problem was the line total+=$(echo "${number}" | bc). This fixes it:

#! /usr/bin/env bash
total=0
echo -n "How many number : "; read num
for ((i=0; i<$num; i++));
do 
    echo -n "Enter your number : "; read number 
    total=$(echo "$total + ${number}" | bc)
    echo "${total}"
done

edited Jan 10, 2014 at 20:50

answered Jan 10, 2014 at 19:54

John1024

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