Testing presence of multiple files within loop

Question

Assume a bash script to which four commandline arguments are passed. Each of these arguments represents a path to, and a name of, an input file (i.e., INF1, INF2, INF3, INF4). I would like to evaluate for each of these four input files, if it exists and - if a file does not exist as specified - exit the script.

#!/bin/bash

# Assigning commandline arguments
INF1=$1
INF2=$2
INF3=$3
INF4=$4

# Check if input files exist
if [ ! -f $1 ]; then
    echo -ne " ERROR | File not found: $1\n"
    exit 1
fi
if [ ! -f $2 ]; then
    echo -ne " ERROR | File not found: $2\n"
    exit 1
fi
if [ ! -f $3 ]; then
    echo -ne " ERROR | File not found: $3\n"
    exit 1
fi
if [ ! -f $4 ]; then
    echo -ne " ERROR | File not found: $4\n"
    exit 1
fi

I would think that having four individual if-statements here is unnecessary, and that the same functionality can be achieved with a single if-statement, wrapped into a loop. How can I reduce the code in this aspect?

Answer 1

in a script

#!/bin/bash

for v; do
if [ ! -f "$v" ]; then
    echo "ERROR | File not found: $v"
    exit 1
fi
done