问将Bash脚本参数$@存储在变量中

EN

args=$@
args2="$@"        # the arguments are combined (see the last output paragraph)

s.sh A B="       " C

#!/bin/bash

args=$@
args2="$@"
#args3 = ? - TODO

echo $args
echo $args2
echo $@
#echo $args3   

echo

echo "$args"
echo "$args2"
echo "$@"
#echo "$args3"

echo

java A $args
java A $args2
java A $@
#java A $args3

echo

java A "$args"
java A "$args2"
java A "$@"
#java A "$args3"

import java.util.Arrays;

public class A {
    public static void main(String args[]){
        System.out.println(args.length + ": " + Arrays.asList(args));
    }
}

A B= C
A B= C
A B= C

A B= C
A B=        C
A B=        C

3: [A, B=, C]
3: [A, B=, C]
3: [A, B=, C]

1: [A B= C]
1: [A B=        C]
3: [A, B=       , C]

A B= C
A B= C
A B= C
A B= C

A B= C
A B=        C
A B=        C
A B=        C

3: [A, B=, C]
3: [A, B=, C]
3: [A, B=, C]
3: [A, B=, C]

1: [A B= C]
1: [A B=        C]
3: [A, B=       , C]
3: [A, B=       , C]