How do I create a bash script that sums any number of command line arguments?

Question

How would I create a bash script that the user can use to sum any amount of command line arguments? For example, say my script is called sum:

sum 3

3

sum 3 5

8

sum 9 8 21

38

And so on.. I realize that I will have to loop through the command line arguments and increment them to the total, but I am not really sure how to do that. I am not sure how to reference say the 2nd command line argument to a specific variable or how to get the number of total command line arguments.

Solution 1

No need for bash, plain sh will do as well:

#! /bin/sh - 
IFS=+; echo "$(($*))"

$* in POSIX shells, expands to the list of positional parameters (in this case, the arguments to the script) separated by the first character of $IFS (or space if $IFS is unset or nothing if $IFS is empty). $((...)) is the shell internal arithmetic expansion operator (note that it supports decimal, octal and hexadecimal numbers)

If you need floating point support, that's where you'll need a different shell like ksh93 or zsh (not bash as bash only supports integer arithmetic), though you could also use awk:

#! /usr/bin/awk -f
BEGIN {t=0; for (i in ARGV) t+=ARGV[i]; print t}

That will use long (for integer) and double (for floating point) type numbers as implemented by your system. The input numbers must be decimal floating point or engineering notation in the English style (floating point delimiter is the period character regardless of the locale). With some awk implementations, it will fail if the first number is negative as awk would try to interpret it as an option.

Some awk implementations like GNU awk when POSIXLY_CORRECT is in the environment also support hexadecimals including with binary exponent notations. Or with --non-decimal-data, it understands octals and hexadecimals:

$ POSIXLY_CORRECT=1 ./sum  0xap3 0xa
90 # (0xa * 2^3) + 0xa
$ awk --non-decimal-data -f ./sum  010
8

Solution 2

You can use the following bash function:

sum() {
    local sum=0
    for arg in "$@"; do
        (( sum += arg ))
    done   
    echo $sum
}

Solution 3

A non-looping variant:

{ printf %d+ "$@"; echo 0; } | bc

Example

Put the above in a script file, sum.

#!/bin/bash

{ printf %d+ "$@"; echo 0; } | bc

Run it like so:

$ ./sum 4
4
$ ./sum 4 4 5
13

Comments

[user4591106]

The native bash feature won't handle floating point values (eg: 3.5). For that you'd need to use a program such as awk, python, dc, bc or perl. See @rici's answer.

[John]

As a side question, is there a way to determine how many args are in the command line without looping?

[rici]

$# is the number of args

[John]

Thank you, perfect. One last question I forgot to clarify.. is there any way I can detect if it is like the very first argument? So say it is the very first argument, I can do something special with it. For example, as a random example, if the very first argument is a 7, I echo that the very first argument is a 7.

[Radu Rădeanu]

@John $1 is the first argument, $2 is the second argument, and so on. The variable $0 is the script's name. The total number of arguments is stored in $#. The variables $@ and $* return all the arguments. See also: how-to.wikia.com/wiki/…

[Olivier Dulac]

clever. But I really hope you are not confusing him ^^ It seems he's learning shell scripting basics, and you throw him IFS manipulation ^^ It's always good to know, but probably a bit too early for him... Or provide explanations such as IFS=+ means afterward the "$*" will expand to "all the arguments separated by a '+'", hence the $((computation)) will be replaced with the sum of all arguments

[icarus]

@John typically if you want to process the first argument differently you first copy it to something else and then shift all the rest of the arguments with first=$1;shift