Shell script variable & loop

Question

I want to print the largest number passed in on the command line but I am not sure how to assign a variable correctly.

What is the correct way to solve this issue?

#!/bin/sh
x=0
for var in "$@";
do 
if [ $var -gt $x ]; then
$x=$var       #this does not work
fi

done
echo "Largest number: $x"

To assign a value to a variable you do var=$something. Change $x=$var to x=$var. — fedorqui
– fedorqui, Commented Oct 16, 2013 at 10:24

Chris Seymour · Accepted Answer · 2013-10-16 10:31:29Z

3

Your mistake is $x=$var. It should be x=$var. You can think of it as $x is the value contained the in variable x so you don't want to assign to value of x but to x itself.

#!/bin/sh
x=0
for var in "$@"; do
    if [ "$var" -gt "$x" ]; then
        x="$var"
    fi
done
echo "Largest number: $x"

Here is an alternative script using some core-utils that will work with negative numbers:

#!/bin/bash
echo "Largest: $(printf '%i\n' $@ | sort -n | tail -n 1)"

edited Oct 16, 2013 at 10:31

answered Oct 16, 2013 at 10:24

Chris Seymour

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1 Comment

ojhawkins Over a year ago

Brilliant, thought I was close @sudo_O

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