arithmatic in shell script

[blackbox111]

I am writing a script that outputs the fibanocci sequence and gives as many terms as specified by the argument.

#!/bin/sh
clear
i=0
a=0
b=0
echo "this is i $1"
while [ `expr $i` -lt `expr $1` ]
do
c=`expr $a+$b`
echo " $c"
a=$b
b=$c
i=`expr $i+1`
done

The error I get is this is i 6
0+0
./Fibanocci.sh: line 7: [: 0+1: integer expression expected

I'm guessing the problem is that arguments are automatically a string type so that's why I have the `expr`s inside the while loop.

[dxqcanada]

Try this syntax:

#!/bin/sh
clear
i=0
a=0
b=0
echo "this is i $1"
while [ $i -lt $1 ]
do
c=$(($a+$b))
echo " $c"
a=$b
b=$c
i=$(($i+1))
done

[secondmouse]

You can't start a fibanocci with two 0's.

#!/bin/bash
[ $# -lt 1 ] && echo Usage: $0 count && exit 0
a=0
b=1
echo -n $a $b
for i in `seq 1 $1`; do
    c=$((a+b))
    echo -n " "$c
    a=$b
    b=$c
done
echo

[blackbox111]

thanks that did it.
why was using expr wrong and what happens when you put the expression in $(...) ?

[secondmouse]

For now you don't have to worry about type conversion in bash, types are automatically decided unless you specify (declare) otherwise. For example:

$ a=3
$ b="4"
$ [ $a -lt $b ] && echo yes
yes

Regarding $((..)), please read Arithmetic Expansion

[lomcevak]

thanks that did it.
why was using expr wrong and what happens when you put the expression in $(...) ?

Your expr's weren't working because you didn't have spaces around the '+' sign.

c=`expr $a+$b`
should be:
c=`expr $a + $b`

i=`expr $i+1`
should be:
i=`expr $i + 1`

[clickalot]

you can also try to have a look at the bc "command"

an example usage from a script:

percent=`echo "100 * $counter / $total_lines" | bc -l` # calculate percent