arithmatic in shell script

2 Weeks Ago	#1 (permalink)
blackbox111 Just Joined! Join Date: Oct 2009 Posts: 8	arithmatic in shell script 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.

2 Weeks Ago	#2 (permalink)
dxqcanada Linux User Join Date: Sep 2006 Location: Canada Posts: 258	Try this syntax: Code: #!/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 __________________ Men occasionally stumble over the truth, but most of them pick themselves up and hurry off as if nothing had happened. Winston Churchill ... then the Unix-Gods created "man" ...

2 Weeks Ago	#3 (permalink)
secondmouse Linux Newbie Join Date: May 2008 Location: NYC, moved from KS & MO Posts: 247	You can't start a fibanocci with two 0's. Code: #!/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

2 Weeks Ago	#4 (permalink)
blackbox111 Just Joined! Join Date: Oct 2009 Posts: 8	thanks that did it. why was using expr wrong and what happens when you put the expression in $(...) ?

2 Weeks Ago	#5 (permalink)
secondmouse Linux Newbie Join Date: May 2008 Location: NYC, moved from KS & MO Posts: 247	For now you don't have to worry about type conversion in bash, types are automatically decided unless you specify (declare) otherwise. For example: Code: $ a=3 $ b="4" $ [ $a -lt $b ] && echo yes yes Regarding $((..)), please read Arithmetic Expansion

2 Weeks Ago

#6 (permalink)

Linux Newbie

Join Date: Mar 2009

Posts: 136

Quote:

Originally Posted by blackbox111

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.

Code:

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

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

2 Weeks Ago	#7 (permalink)
clickalot Just Joined! Join Date: Nov 2009 Location: Erlangen, Germany Posts: 31	bc - An arbitrary precision calculator language you can also try to have a look at the bc "command" an example usage from a script: Code: percent=`echo "100 * $counter / $total_lines" \| bc -l` # calculate percent