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Programming-Idioms

  • C
  • Java
  • Php

Idiom #60 Read command line argument

Assign to x the string value of the first command line parameter, after the program name.

$x = $argv[1];
void main(int argc, char *argv[])
{
    char *x = argv[1];
}

argv[0] would be the program name. See §5.1.2.2.1 Program startup in linked doc n1570.pdf .
String x = args[0];

This works only in method main(String[] args) .
public static void main(String[] args) {
    if (args.length != 0) {
        String x = args[0];
    }
}

In most JVM implementations, the `args` parameter will be non-null.
int main(int argc, char *argv[])
{
 vector<string> args(1 + argv, argc + argv);

 string x = args.at(0);
}

Notice that this version checks for errors (and throws if args[0] does not exist).
New implementation...
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programming-idioms.org 
$x = $argv[1];
void main(int argc, char *argv[])
{
    char *x = argv[1];
}
String x = args[0];
public static void main(String[] args) {
    if (args.length != 0) {
        String x = args[0];
    }
}
int main(int argc, char *argv[])
{
 vector<string> args(1 + argv, argc + argv);

 string x = args.at(0);
}
static void Main(string[] args)
{
    string x = args[0];
}
auto x = args[1];
main(args) {
  var x = args[0];
}
  character(len=:), allocatable :: x
  integer :: n
  call get_command_argument (1, length=n)
  allocate (character(n):: x)
  call get_command_argument (1, x)

x := os.Args[1]
x <- return.head =<< System.Environment.getArgs

const x = process.argv[2]
x = arg[1]
var
  _x: String;
begin
  _x := ParamString(1);
end.
$x = $ARGV[0];
x = sys.argv[1]
x = ARGV.first
let x = env::args().nth(1).unwrap_or("".to_string());
let first_arg = env::args().skip(1).next();

let fallback = "".to_owned();
let x = first_arg.unwrap_or(fallback);