Home  >  Article  >  Never-ending loop for user-populated array list

Never-ending loop for user-populated array list

WBOY
WBOYforward
2024-02-05 22:00:051275browse
Question content

I need to write a method to get the car specifications entered by the user and add them to an arraylist. This needs to accept any number of specifications the user wants to enter, including none. This is my first post here, apologies for any bad grammar.

public static arraylist gettrim() {

scanner t = new scanner(system.in);
system.out.println("enter car trim");
arraylist<string> trim = new arraylist<>();

while (t.hasnext()) {
trim.add(t.next());
}

return trim;

}

I think this condition will return false if a space is entered. Such continuous iteration can only be exited manually.

I also tried it

if (t.hasNext()) {
trim.add(t.next());
}

else {
t.close();
}

This iterates once and returns the arraylist, but I need to be able to enter more. Changing the if or while condition to hasnextline() gives the same result, here I am using hasnext() because the car trim level has specific formatting expectations. I don't understand why hasnext() doesn't return false when no input is given.


Correct answer


It..No.

Think about it. How does a computer differentiate between a user who is thinking about what to type and may leave it until after lunch, and a user who is "done"? Fire up the webcam and do some AI analysis and see if the user looks like they've finished typing?

system.in is not the keyboard. It is the "standard input of the jvm process", default, and at least, if you start your java application from the command line, set it to read from the keyboard. It doesn’t have to be:

java -jar myapp.jar </some/path/to/some/file.txt

Now system.in is read from this file. The process cannot be read from the keyboard.

java -jar myapp.jar </dev/barcodescanner1

Now, presumably whenever you scan a barcode, the java application will receive the barcode just as if you entered the digits of the barcode and pressed "Enter" 1.

This depends on the "source of this input" to the "end". The file ends when the end of the file is reached.

It is impossible for the keyboard to "end" - therefore, .hasnext() cannot return false.

blank

You talked about "entering spaces" in your comments.

Then you don’t understand scanner. You are in good company; this is probably the most misunderstood thing. If we look at the number of questions flooding stackoverflow, this is indeed the case.

scanner has nothing to do with the keyboard.

The scanner simply takes any text input source and splits it into chunks. Blocks are called "tokens", and "tokens" are defined by "all text between delimiters". Delimiters are in turn defined by regular expressions, the regular expression used by default is

\s . For example, any amount of white space.

therefore:

public static void main(string[] args) {
  scanner s = new scanner(system.in);
  while (true) {
    system.out.println("token: ≥" + s.next()) + "≤");
  }
}

If you run this command and type in the command line:

hello world!, then press enter and type my name is lajos, you will see:

token: ≥hello≤
token: ≥world!≤
token: ≥my≤
token: ≥name≤
token: ≥is≤
token: ≥lajos≤

Things about the scanner:

It is not possible to register or otherwise obtain the content of anything under "separator". You can't ask the scanner: How many blanks are there?

You also can't ask: "stop" when you encounter a space. No, spaces just separate one token from the next. For scanners, there is no

any difference between pressing the enter key and pressing the space bar and then the enter key. This is all "1 or more whitespace characters" so are interchangeable and cannot be detected since this is about delimiters.

You may not want to use a scanner at all. system.in itself can certainly differentiate it.

Solution 1

A common strategy is to mention some magic word that means "done" in the prompt. For example:

static void main(string[] args) throws exception {
  var s = new scanner(system.in);
  system.out.println("welcome to the fruit stand! enter the fruit you'd like to buy, one at a time:");

  var basket = new arraylist<string>();
  while (true) {
    system.out.print("fruit (type 'done' when done): ");
    string fruit = s.next();
    if (fruit.equalsignorecase("done")) break;
    basket.add(fruit);
  }

  system.out.println("here's your basket: " + basket);
}

Solution 2

Trench Scanner. Or at least, drop everything and just use

nextline, which interacts very with all the other methods it has - choose one and only one (nextline, or Everything except nextline):

static void main(string[] args) throws exception {
  var s = new scanner(system.in);
  system.out.println("welcome to the fruit stand! enter the fruit you'd like to buy, one at a time:");

  var basket = new arraylist<string>();
  while (true) {
    system.out.print("fruit (enter when done): ");
    string fruit = s.nextline();
    if (fruit.isempty()) break;
    basket.add(fruit);
  }

  system.out.println("here's your basket: " + basket);
}

[1] In fact, most barcode scanners look and act like keyboards, and have no device because it does not exist in

/dev/ and cannot be piped into such a process. But, as an example, it works.

You can try this:

public static ArrayList getTrim() {

Scanner t = new Scanner(System.in);
System.out.println("Enter car trim");
ArrayList<String> trim = new ArrayList<>();

while (t.hasNext()) {

if (!t.next().trim().equals("") {

trim.add(t.next());
      }

else {
t.close();
      }
   }

return trim;

}

The above is the detailed content of Never-ending loop for user-populated array list. For more information, please follow other related articles on the PHP Chinese website!

Statement:
This article is reproduced at:stackoverflow.com. If there is any infringement, please contact admin@php.cn delete