I'm trying to write a script that adds three specified arguments together, and if there are no arguments outputs: "No arguments".
The trouble is that "No arguments" is always output even when there are three arguments.
I am very new to shell script.
Here is my script:
#!/bin/sh
if [[("$#"==0)]]; then
echo "No arguments specified"
exit 1
fi
sum=0
sum=$(expr $1 + $2 + $3)
echo "$sum"
exit 0
Either change your shebang to #!/bin/bash and use
if (( $# == 0 )); then
or use the POSIX-compatible [:
if [ $# -eq 0 ]; then
Don't forget that [ and [[ are both commands, not syntax, so as with any other command, you need to separate the arguments you pass to the command with spaces.
If you are using bash features, such as [[, you should always use the #!/bin/bash shebang, as otherwise you will run into problems.
As pointed out in the comments below the other answer, it is possibly a better idea to check that you have been passed three arguments:
#!/bin/bash
if (( $# < 3 )); then
echo "Insufficient number of arguments specified"
exit 1
fi
sum=$(( $1 + $2 + $3 ))
echo "$sum"
I have made a couple of other changes to your script, such as not initialising sum to 0 and using the more modern $(( )) to evaluate the sum of the variables.
I guess what you ask for is:
if [ $# -eq 0 ];
