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Shell script to find the sum of argument passed to the script


 
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# 1  
Old 07-15-2012
Shell script to find the sum of argument passed to the script

I want to make a script which takes the number of argument, add those argument and gives output to the user, but I am not getting through...

Script that i am using is below :
Code:
#!/bin/bash
sum=0
for i in $@
do
    sum=$sum+$1
    echo $sum
    shift
done

I am executing the script as below
Code:
sh script 4 5 6 7

the output that I am getting is
Code:
0+4
0+4+5
0+4+5+6
0+4+5+6+7

# 2  
Old 07-15-2012
Code:
#!/bin/bash
sum=0
for i in $@; do sum=$((sum+i)); done
echo $sum
exit 0

Code:
~/unix.com$ bash script 4 5 6 7
22

See man bash for more infos
Quote:
((expression))
The expression is evaluated according to the rules described
below under ARITHMETIC EVALUATION. If the value of the expres-
sion is non-zero, the return status is 0; otherwise the return
status is 1. This is exactly equivalent to let "expression".
This User Gave Thanks to tukuyomi For This Post:
# 3  
Old 07-15-2012
may be this ?

Code:
$ cat script.sh
#!/bin/bash
sum=0
for i in $@
do
    sum=`expr $sum + $i`
    echo $sum
    shift
done

output:

Code:
$ ./script.sh 4 5 6 7

4
9
15
22

# 4  
Old 07-15-2012
Code:
printf '%s\n' "$@" | paste -sd+ - | bc

Regards,
Alister

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