CODEWITHSUNDEEP
bashif-statementlazy-evaluation
Dori
Dori

Reputation: 1055

Bash - test existence and value of argument in one if statement

I'm writing a script that needs two input files. I wrote the following code:

if (( $# < 3 || ! ( -f $1 && -f $2 ) )); then
    echo Cannot open input files
    exit 1
fi

But if I supply only one input file, I get this error:

line 4: ((: 1 < 3 || ! ( -f tests/example && -f  ) : missing `)' (error token is "tests/example && -f  ) ")

I guess the OR evaluation isn't "lazy", so the -f $1 works because it's replaced by -f tests/example, but -f $2 causes an error.

Is there some way to "force lazy evaluation" (if $# < 3 don't try to evaluate the rest) in the if statement? Or should I be doing things differently?

Upvotes: 1

Views: 393

Answers (2)

devnull
devnull

Reputation: 123608

I'm writing a script that needs two input files.

Why are you checking if the number of arguments is less than 3 in order to exit? You need to ensure that there are no fewer than 2 arguments. Use short-circuting for all the checks. Say:

if [[ $# < 2 || ! -f "$1" || ! -f "$2" ]]; then
    echo Cannot open input files
    exit 1
fi

Actually, even the following would suffice for your question:

if [[ ! -f "$1" || ! -f "$2" ]]; then
    echo Cannot open input files
    exit 1
fi

Upvotes: 3

anubhava
anubhava

Reputation: 785856

(( ... )) is for maths operations in shell.

Rewrite your if condition like this:

[[ $# < 3 || ! ( -f "$1" && -f "$2" ) ]]

Upvotes: 2

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