Reading user-input in the same line as the script execution

Question

I have a BASH script named fib.sh. The script reads a user input (number) and performs a calculation. I want to be able to type

$ ./fib.sh 8

where 8 is the input

Currently, I have to wait for the next line to enter the input.

$ ./fib.sh 
$ 8

Script

#!/bin/bash

read n

a=0
b=1
count=1
fib=$a

while [ $count -lt $n ]; 
do
    fib=$[$a+$b]
    a=$b
    b=$fib
    count=$[$count+1]
done

echo "fib $n = $fib"

exit 0

$[…] is deprecated; use $((…)) (part of the POSIX standard) instead. Example fib=$(($a+$b)). — chepner
– chepner, Commented Oct 28, 2013 at 15:45
True; I was just trying to move the code off the deprecated syntax first; using it to its full capabilities can come later. — chepner
– chepner, Commented Oct 28, 2013 at 16:21

fedorqui · Accepted Answer · 2013-10-28 15:42:04Z

2

So you want to pass a parameter to the script instead of reading it. In this case, use $1 as shown here:

#!/bin/bash

n=$1 <---- this will take from the call of the script
echo "I have been given the parameter $n"

a=0
b=1
count=1
fib=$a

while [ $count -lt $n ]; 
do
    fib=$[$a+$b]
    a=$b
    b=$fib
    count=$[$count+1]
done

echo "fib $n = $fib"

exit 0

answered Oct 28, 2013 at 15:42

fedorqui

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