bash scripting taking integers as args and adding them

Question

so i have this code here:

    #!/bin/bash
sum=0
for i;
do 
   if [ "$i" -ge 0 ]
   then
       sum=$((sum + i))
   else
       sum=$((sum + (i * -1)))
   fi
done
echo $sum

so the user types blin.sh 1 3 2 5 -5 -8, and the program is supposed to take all of the integers, and make them positive, if it is a negative, and then echo the sum. but my bash code only echo's 0 for some reason unknown to science.

And why do you talk about command line arguments when your script looks at filenames instead? — Shawn
– Shawn, Commented May 5, 2020 at 2:31
file names?, how can i make it to where i takes in a integer?? — this is me
– this is me, Commented May 5, 2020 at 2:32
Because the file names are included in the loop instead of the arguments..., stdin_data and stdout_data are both files that are in the same directory where you execute that script..., Do you know how to access the arguments/positional parameter in a bash shell script? The glob * expands to all files/directories in the current directory... — Jetchisel
– Jetchisel, Commented May 5, 2020 at 2:43
execute as sh -x blin.sh 1 3 2 5 -5 -8 This will print all command execution on stdout — Digvijay S
– Digvijay S, Commented May 5, 2020 at 3:45

Jetchisel · Accepted Answer · 2020-05-05 03:46:15Z

1

Here is an example of your script.

#!/bin/bash

sum=0
for i ; do
   if [ "$i" -ge 0 ]
   then
      sum=$((sum + i))
   else
      sum=$((sum + (i * -1)))
   fi
done
echo $sum

The name of the script is foo.sh, Now execute that script with the arguments.

./foo.sh  1 3 2 5 -5 -8

Out put

Using sh

sh ./foo.sh  1 3 2 5 -5 -8

Out put

Using dash

dash ./foo.sh  1 3 2 5 -5 -8

Output

Using zsh

zsh ./foo.sh  1 3 2 5 -5 -8

Output

I don't see any errors there.

answered May 5, 2020 at 3:46

Jetchisel

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Comments

Vunq1601 · Accepted Answer · 2020-05-05 03:44:27Z

1

use $* for all arguments (without the semicolon ;)

for i in $*

your code works for me.

#!/bin/bash
sum=0
for i in $*
do
        if [ "$i" -ge 0 ] 
        then
                sum=$((sum + i))
        else
                sum=$((sum + (i * -1)))
        fi
done
echo $sum

edited May 5, 2020 at 3:44

answered May 5, 2020 at 3:25

Vunq1601

463 bronze badges

2 Comments

this is me Over a year ago

still nothing, it only returns zero, can you try it??

this is me Over a year ago

can you help me please?

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