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Whats the error in this script ?

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Top Forums Shell Programming and Scripting Whats the error in this script ?
# 1  
Old 08-11-2010
Whats the error in this script ?
Can someone help me figure out the error with this simple script:


Code:
#!/bin/sh

fact()
{


        if [ $1 -gt 1 ] ; then

                p=`fact expr $1 - 1`
        else
                echo $1
        fi
        echo `expr p \* $1`

}

echo "Enter a number you wish to calculate factorial of"

read NUM

echo "Factorial of $NUM is : " `fact $NUM`

Thanks for help!

---------- Post updated at 07:12 PM ---------- Previous update was at 05:46 PM ----------

Code:
$ ./factorial.sh 
Enter a number you wish to calculate factorial of
3
./factorial.sh: line 7: [: expr: integer expression expected
expr: non-numeric argument
expr: non-numeric argument
Factorial of 3 is : 
$

Last edited by pludi; 08-12-2010 at 02:01 AM.. Reason: code tags, please...
# 2  
Old 08-11-2010
Two things. First, you need a return after your 'echo 1' or you need to move the last echo into the true branch of the if.

Secondly, you need to first do the math before passing the value to the recursive call to fact.

Code:
#!/bin/sh

fact()
{
   if [ $1 -gt 1 ] ; then
        # eval first to do the expr and pass value to fact.  
        eval p=\`fact `expr $1 - 1`\`   
   else
        echo $1
        return      # needed so that you dont echo two things
   fi
   echo `expr $p \* $1`
}

echo "Enter a number you wish to calculate factorial of"
read NUM

echo "Factorial of $NUM is : " `fact $NUM`

If you were using Ksh or bash, the eval statement becomes a bit less muddled:

Code:
p=$( fact $(( $1 - 1)) )

# 3  
Old 08-12-2010
There's really no need for that eval. A simple command substitution will do (with perhaps a variable to hold the result of expr for clarity's sake).

Code:
arg=`expr $1 - 1`
p=`fact $arg`

Or with any posix-compliant sh:
Code:
p=$(fact $(expr $1 - 1))

Or your example:
Quote:
Originally Posted by agama
If you were using Ksh or bash, the eval statement becomes a bit less muddled:

Code:
p=$( fact $(( $1 - 1)) )

which is also posix-compliant.

Regards,
Alister
Last edited by alister; 08-12-2010 at 12:35 AM..
# 4  
Old 08-12-2010
Still getting an error
I modified the script to:


Code:
#!/bin/sh

fact()
{

        if [ $1 -gt 1 ] ; then

                res=`expr $1 - 1`
                p=`fact $res`
        else
                echo $1
        fi
        echo `expr $p \* $1`

}

echo "Enter a number you wish to calculate factorial of"

read NUM

echo "Factorial of $NUM is : " `fact $NUM`


~~~~~~~~~~~~~~~~~~~~~~~~
Code:
$ ./factorial.sh 
Enter a number you wish to calculate factorial of
4
expr: syntax error 
Factorial of 4 is :  24

WHy am I still getting this : "expr: syntax error"
Last edited by Franklin52; 08-13-2010 at 04:19 AM.. Reason: Please use code tags!
# 5  
Old 08-12-2010
You're getting the error because you forgot to add the return after your first echo.

Code:
if [ $1 -gt 1 ] ; then
   res=`expr $1 - 1`
   p=`fact $res`
else
   echo $1
   return        # missing return needed to avoid second echo
fi
echo `expr $p \* $1`

If you don't add the return, the second echo will execute and p will not be set. An empty variable will result in the error from expr. Another solution would be to move your 'echo `expr $p \* $1`' statement into the true branch of the if (right after the p= statement). If you do that you don't need the return.
# 6  
Old 08-13-2010
Thanks that worked. However I thought "echo" also did the return and an explicit return was not needed, as in this script:
Code:
#!/bin/sh

func()
{

        echo 1

}

echo " The result is :" `func`

~~~~~~~~~
Code:
$ ./echotest.sh 
 The result is : 1

Last edited by Franklin52; 08-13-2010 at 04:21 AM.. Reason: Please use code tags!
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