I have this code here:
try {
System.out.print("There are args!: " + args[0]);
} catch (Exception e) {
System.out.print("No args");
}
I am checking if arguments were passed when I executed the program. If no arguments are passed, and I try to print the first index of arguments, an Exception is thrown. I can use a Try-Catch block to catch the Exception, and declare there is no arguments.
How could I do this with an 'If Statement'?
I tried:
if (args[0] == null)
//And
if (args[0] == '')
And still got an Exception. What is "args[0]" equal to if no arguments are passed?
....................
args
is a normal array. If you try to access an item beyond the last index (args[0]
when args
is empty, for example), then you'll always get an exception.
To avoid it, you can test its length before accessing the element:
if (args.length > 0) {
System.out.println("We have one or more arguments.");
// You can safely use `args[0]` here.
}