How does this snippet work with the buffer? Java - java

So basically this code is supposed to keep asking "Please insert a valid number" until an integer is introduced through the keyboard. It works just fine, I'm just curious as to how it actually does what it does.
Scanner input = new Scanner(System.in);
System.out.println("Please insert the number of lives.");
while (!input.hasNextInt()) {
input.next();
System.out.println("Please insert a valid number.");
}
numberOfLives = input.nextInt();
In my head it would make more sense to put the numberOfLives = input.nextInt(); line inside of the while loop. This is basically what I'm understanding from this snippet once it reaches the while loop:
Check condition in while, if input scanner variable doesn't have an int value, do the following
Discard what is in the input scanner variable
Print out the valid number message
Obviously this isn't how it goes, else this snippet wouldn't work. I'm thinking maybe each time it checks the condition, the actual !input.hasNextInt() opens up the buffer again for input and THEN checks what's inside, but I have no clue.
Sorry if I'm not using the terms correctly, pretty now to all of this stuff.
Thanks!

Like an iterator, scanner will always point to the element before the stream starts. i.e. if the list starts with position 0, scanner will point to -1st position. next() will advance the scanner and return the element present at location.
In this code, we are checking if the next element is integer. If it is not, why to advance the scanner?

Related

Don't understand how Java Scanner object is working in this example (hasNextInt, next...?)

In my program I need to take an input from a user, but it must be an integer. If the user doesn't input an integer, I need the program to reprompt them to enter an integer, and keep doing so until they enter one. I found an example while loop code snippet online which works perfectly, but I am having problems understanding why and how it works, and I'd really like to understand better:
Scanner reader = new Scanner(System.in);
int guess=0;
System.out.println("Guess the number");
while (!reader.hasNextInt()) { //I get that this is a "not have" boolean
System.out.println("That's not a number. Please enter a valid number");
reader.next();
}
guess= reader.nextInt();
System.out.println(guess);
My questions are:
Does the !hasNext condition in the while loop actually trigger the scanner object to ask the user for input and then check that input, or does it only check anything that has already been inputted?
Why is it necessary to have the reader.next(); line after the while loop's main statement and what is this doing exactly? I know it's necessary as the program doesn't work when I take it out. But is it prompting for input? If yes what happens to this input?
When guess= reader.nextInt(); runs, why doesn't it re-prompt the user for input at this point?
Sorry that these are probably really basic questions, I'm new to coding and Java and just can't get my head around what's happening internally in this particular example, though have no problem doing other basic stuff with the Scanner.
It's not particularly intuitive, for sure, so don't feel bad for not getting it immediately.
It seems to me that the problem you're having with your understanding is that that methods such as nextInt may or may not prompt for user input, depending if anything is already in the Scanner.
Here's the sequence of events:
All your code executes until you hit !reader.hasNextInt(). There's no input so it "blocks" (waits) until there is some input from the user.
If the user enters 'A', that's not an integer so we enter the body of the while loop. We then print the error message.
Now, hasNextInt doesn't "consume" (process) the user input when it's checking whether or not it's an integer, so we still have that invalid user input of 'A' sitting in our scanner. We call reader.next() to effectively discard that value.
Now we're back to !reader.hasNextInt(). The scanner is empty once again, so we prompt for user input. If they enter another non-integer, that process will simply keep repeating.
Say this time we do have a valid user input - they've entered the number 2. This passes the check so our while loop ends and we continue along.
We've now got some input in our scanner, but we've not consumed it. We're sure it's an integer because of the while loop condition. We can now consume the input with reader.nextInt() and assign it to our variable.
hasNextInt() only checks whether the next input is an integer. It won't consume any input at all. The ! is just negate theu outcome of this function.
As hasNextInt() won't consume, then we need to use next() to consumeo the user input to let hasNextInt() to check with user's next input value. As you won't need it at all, then no need to assign it to any variables.
Scanner won't display the prompt at all. The prompt is printed by System.out.println().
For the line nextInt(), it is used to consume the next user input. Since hasNextInt() must be true when it execute this line, there must be one integer input waiting for consumption, so this method can return immediately with that user input.
Try it :)
public static void main(String[] args) {
int option;
if(args!=null&&args.length>0){
option = Integer.parseInt(args[0]);
}
else{
System.out.println("Enter:\n 1 to run x \n 2 to run y \n");
Scanner keyboard = new Scanner(System.in);
option = keyboard.nextInt();
}
switch (option) {
case 1:
xx
break;
case 2:
yy
break;
default:
break;
}
}

Java "While" Loop -- Counting Matches Algo - Assigning value must occur inside the while loop

I apologize in advance; I couldn't find the answer to this simple question in Search.
I'm a newbie to coding. Currently on Udacity working my way through the Intro to Java Programming course.
There's something that I'm not understanding as it relates certain Algorithms.
Take the "counting matches" algo for example.
The assignment of double input = in.nextDouble(); needs to occur inside the while-loop.
If I place it just above the while-loop, it breaks the program. Why?
It seems to me that Java shouldn't care when the value is stored in the variable.
import java.util.Scanner;
public class CountingMatches
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
int counter = 0;
System.out.print("Enter a value, Q to quit: ");
// double input = in.nextDouble(); // remember to place the assignment of this variable inside the while loop.
// I tend to want to place this outside the while loop because I still don't
// understand why it necessarily must occur inside the while loop.
while (in.hasNextDouble())
{
double input = in.nextDouble(); // this assignment is properly located here as opposed to just above the while-loop
if (input < 0)
{
counter++;
}
System.out.print("Enter a value, Q to quit: ");
}
System.out.println("The water line fell on " + counter + " years.");
}
}
Because in.nextDouble() can only be used if Scanner has already confirmed the next token can be parsed as a double. In addition to waiting for user input, this is what hasNextDouble() guarantees for you. If you take it out of the loop, not only are you skipping that hasNextDouble() guarantee (and not giving the user a chance to input anything), you are also only running nextDouble() one time, so you wouldn't have the newest value anyway.
Returns true if the next token in this scanner's input can be interpreted as a double value using the nextDouble() method. The scanner does not advance past any input.
-- https://docs.oracle.com/javase/8/docs/api/java/util/Scanner.html#hasNextDouble--
since you can´t do any further input when you move it outside the loop the condition for the loop in.hasNextDouble() will allways be true. Afterwards you will be stuck in an infinite loop, which makes the programm like like it´s brocken, but actually it´s just repetitively looping and printing until you stop the java process by yourself.
You are right that java shouldn't (and doesn't) care when the value is being stored in a variable, but there is more to it.
in.hasNextDouble() doesn't return until the line inside the wrapped input stream has ended.
It does not alter the stream in any way, but it guarantees there is a character sequence waiting in the stream and if true can be parsed as a double. The buffer is not altered until nextDouble() is called.
This is what breaks your code. When you remove the line from the buffer without it being parseable to a double an exception is thrown, because no characters are waiting in System.in

Continue reading console input after Ctrl+Z in Java

For a university assignment, from the console I need to be able to read in multiple lines until the user enters Ctrl+Z. That I have no issue with, my problem lies with me being unable to read anything in from System.in after that because it always throws a NoSuchElementException.
The following is the code that I am using for reading in the multiple lines, it was provided by the instructor and so I don't want to change it.
System.out.println("To terminate input, type the correct end-of-file indicator ");
System.out.println("when you are prompted to enter input.");
System.out.println("On UNIX/Linux/Mac OS X type <ctrl> d");
System.out.println("On Windows type <ctrl> z");
Scanner input = new Scanner(System.in);
while (input.hasNext()) {
//Processes input
addFileIntoDirectory(input.nextLine());
}
I understand that this is caused by the Ctrl+Z which equates to an EOF marker, but I don't know how to move past it. No matter how many reads after it I do, regardless of if I have typed more to the console, I just instantly get back another NoSuchElementException.
I have tried having a separate scanner for the menu, closing the scanner above and opening a new one for the menu but neither works.
Is there a way to flush/purge System.in or to reset it?
Please let me know if you would like more details. I have kept the rest of the program vague as it is homework.
EDIT 1: Assignment says "The system provides a textual selection menu as follows, and
runs in loop." Which means the program is not to terminate on Ctrl+Z.
add files from user inputs
display the whole directory
display the size of directory
exit
Please give a selection [0-4]:
The short answer is:
if (!input.hasNext())
input = new Scanner(System.in);
The long answer is:
Below is code, with comments made for removing later, to demonstrate the problem as I understand it and solution.
Note this demo is intended to only work with integer and ^Z inputs.
Copy out the code and save into file HN.java to compile.
Run the program and it will prompt One. Enter integers to your heart's content and exit the loop by entering ^Z.
Two will be displayed followed by the NoSuchElementFound exception in the question. This is because the ^Z remains in the Scanner object input and thereby fails on the nextInt() method.
The first inclination might be to use hasNext() to account for this. So go ahead and uncomment /*Comment1 and recompile and run again.
Now you avoid the exception, but the program blazes through to the end, skipping over the nextInt() originally intended.
So now, uncomment /*Comment2, recompile, and run again. Now the program will wait at the Two prompt as intended. If you enter an integer here, you will proceed to the Three prompt for another entry.
However, if you enter ^Z at Two, again, the program skips the next input. To correct this, uncomment /*Comment3, recompile, and run, and you will see the program works for all combos of integers and ^Z at the various inputs.
Now, you might be wondering why I didn't just reuse the !input.hasNext() solution in /*Comment3. Here's a demo why:
Put back the /*Comment3 and uncomment the /*Comment4, compile and run again. The program works fine and dandy with a ^Z input at the Two prompt, but if you enter an integer there, you'll see the system waiting for input, but there is no prompt Three!
This is because the integer you entered was used in the preceding nextInt() so when the program gets to the hasNext(), it stops and waits for input.
The lesson here is you use the !input.hasNext() when you KNOW you have a ^Z in queue such as when it was used to escape the while loop in this program and the original poster's question. Otherwise the else structure is better suited.
hasNext() is confusing to work with. You have to keep in mind that if nothing is in queue the program will stop and wait for input at that point. This may mess up your prompting if you're not careful.
import java.util.Scanner;
public class HN
{
public static void main(String args[])
{
Scanner input = new Scanner(System.in);
int temp = 0;
System.out.println("One");
while (input.hasNext())
{
temp = input.nextInt();
}
System.out.println("Two");
/*Comment2
if(!input.hasNext()) input = new Scanner(System.in);
Comment2*/
/*Comment1
if (input.hasNext())
Comment1*/
{
temp = input.nextInt();
System.out.printf("%d\n", temp);
}
/*Comment3
else input = new Scanner(System.in);
Comment3*/
/*Comment4
if(!input.hasNext()) input = new Scanner(System.in);
Comment4*/
System.out.println("Three");
if (input.hasNext())
{
temp = input.nextInt();
System.out.printf("%d\n", temp);
}
System.out.println("Four");
}
}

Why doesn't Scanner seem to look for a new int every time nextInt is called

Why does it continue in the while loop repeating "Please enter a valid number" and it keeps repeating without stopping to let the user input something.
while (true) { //This will continually run until something is returned, AkA the number 1 or 2
Scanner s = new Scanner(System.in);
try {
input = s.nextInt();
} catch (Exception e) {
System.out.println("Please enter a valid number!");
}
s.close();
}
When I use the debugger nothing seems to be the cause of this problem and no errors are thrown. What is causing this problem and how can it be fixed?
Closing a scanner also closes the underlying stream if it implements the Closable interface.
That's System.in in this case and, once closed, you won't be able to create a scanner using it again.
See http://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#close() for the gritty detail.
But, even if you fix that problem, an exception from nextInt will not advance the stream pointer so, the next time you call it, it will find exactly the same data in the input stream again.
You need to clear out the erroneous data before trying again. Since you're accepting user input, one solution is to call nextLine and throw that away;
string junk = s.nextLine();
Remove
s.close();
And Define your Scanner object outside the while loop as it not an Issue. Btw, I've tried your code, it worked fine until I entered a input which is not an int. When I entered a string, the exception sysout statement keeps printing infinitely
So you've to add
string junk = s.nextLine();
in your Exception block like paxdiablo said so that it wont repeat continuously.

Curious why this recursive code won't work, and exhibits really anamolous behavior in another method

New programmer here, having some problems with some code about what I would think is recursion.
public static int sum (int a) {
int input = goodInput(); //get input from below method without having to put its code in this one
if (input==-1)//so user has the ability to exit at any time
return a; //when user has finally entered -1, the final sum is sent out
else; //for scenarios before last input
int b = a + input; //adding the newest input to the sum
int c = sum(b); //throw the total into the method again until -1 is read
return c; //once -1 is read, a in that iteration is sent up through every iteration of the method until the original method gets its return
}
public static int goodInput () { //code to get input of correct type
Scanner input = new Scanner (System.in);
while (!input.hasNextInt()) { //if I put in integer input, the loop should not be entered, but this isn't happening
System.out.println("Integers only please");
input.next();
}
int finalInput = input.nextInt(); //Finally set good input
input.close(); //close scanner
return finalInput;
}
First method here is clearly just a way to get a sum. I know there are multitudes of other ways to just sum some numbers together, I've done a few myself, but when my class had its lesson on methods and wrote it up having something like the code I listed was the first thing I could think of as a solution, rather than what the teacher ended up recommending. Thought it would be a good learning exercise, in any case.
This code doesn't show any errors in eclipse, so I am stumped as to why it refuses to work. Indeed, it produces results I am really curious of; it naturally asks input at the beginning of the program, but when I enter 0, 1, or any other int, despite the scanner actually having an integer, "Integers only please" is printed, followed by Java announcing exceptions at the end of the while loop, at goodInput's calling in sum, at the return of c, and at the execution of sum in the main method, as well as at java.util.NoSuchElementException, java.util.Scanner.throwFor, and at java.util.Scanner.next.
I have very little idea what is happening here; my best guess would be memory issues, but bugs start occuring at the very first occasion! And the code in goodInput works perfectly well when just used as the main method; not sure why it being called by sum would cause problems.
Again, I don't just want some sum method. I just want to know why the code is behaving in this manner, and how an implementation of sum with my approach would actually work.
Not recursion is the problem here, but your Scanner. I have to admit that I am not too familiar with the Scanner class, but it seems that if you call input.close() and then reenter goodInput later, your System.in is closed. At least, stepping through with the debugger, I found that the line "Integers only please", is printed in the second invocation of goodInput. Deleting the line input.close(); did the trick for me, and your method worked as intended.
I'd suggest you initialize your scanner in the main method, pass it as an argument, and close it in the main method afterwards.
Edit:
The close method of the Scanner states the following:
If this
scanner has not yet been closed then if its underlying
java.lang.Readable readable also implements the java.io.Closeable
interface then the readable's close method will be invoked.
So, the underlying reader, i.e., System.in, was closed when you called close on the Scanner.

Categories

Resources