I am playing with Java and want to do a simple while loop that keeps going until the user presses ctrl+z.
I have something like this:
public static void main(String[] args) {
//declare vars
boolean isEvenResult;
int num;
//create objects
Scanner input = new Scanner(System.in);
EvenTester app = new EvenTester();
//input user number
System.out.print("Please enter a number: ");
num = input.nextInt();
while() {
//call methods
isEvenResult = app.isEven(num);
if(isEvenResult) {
System.out.printf("%d is even", num);
} else {
System.out.printf("%d is odd", num);
}
}//end while loop
}//end main
I tried while( input.hasNext() ) { ... but the code inside the while loop wouldn't execute.
//input user number
System.out.print("Please enter a number: ");
do {
try {
num = input.nextInt();
} catch (Exception e) {
break;
}
// call methods
isEvenResult = app.isEven(num);
if (isEvenResult) {
System.out.printf("%d is even", num);
} else {
System.out.printf("%d is odd", num);
}
} while (true);
When the user writes something non-numeric, the loop breaks.
while (num != 'z')
Although if you are expecting a 'z' why are doing input.getInt()?
You may want to check out the Console class too.
If you want to loop until the user has to force break via Ctrl+Z, then just do while(true). But you want your nextInt() to be inside the loop, and maybe also your prompting statement.
TrueSoft's solution is dead on. The reasons it may not be working for the asker is are kinda outside the scope of the program.
The program works for me: I'm running it under Linux and enter Ctrl-D as the first thing on a line. Ctrl-D is end-of-file for Linux the same way that Ctrl-Z is for Windows. Program stops dead in its tracks, perfectly.
The Windows console (the black DOS box, whatever you want to call it) has a wrinkle: It reads input line-by-line. It won't see the Ctrl-Z until it's read the line, so it needs an Enter keyin before it will see the Ctrl-Z.
I'm unwilling to fire up Windows just to try this, but my guess is that CTRL-Z followed by the Enter key (just like after the number entries) should cause the program to stop cleanly.
There are system-y ways to make a Java program work on a character-by-character basis so you can handle any characters directly and respond immediately to Ctrl-Z. But that's advanced stuff and doesn't belong in a simple programming exercise like this. I think Ctrl-Z / Enter is an acceptable way to have the program end.
You need to implement KeyBindings. Then you can make the determination to exit based off of what keys were pressed.
you are doing the input outside the loop,and it will run for only once.
System.out.print("Please enter a number: ");
num = input.nextInt();
Put your above code inside the loop.
Since you are having a system out inside the loop you will also come to know whether the control went inside the loop, obviously it should.
Also, try
while(true)
I wonder if while() alone is working.
This looks like Exercise 6.16 out of Deitel's book Java How to Program, 9th Edition.
The CTRL-Z charcter does, indeed, end input on a Windows platform just as CTRL-D ends input on most any UNIX or Linux platform.
Also, there are logic errors in the construction of the program that indicate Scanner methods and the System.in byte stream (i.e. the standard input from the console) are not well understood.
In your posted program, the statement:
num = input.nextInt();
executes unconditionally. It will block execution until some kind of input is received. If the input is not an integer, it will throw an exception. If the input received is an integer, then num will be assigned the integer value and the integer in the input stream (input) will be discarded from the input stream. There may be remainaing stuff on the input line up to the end of line, depending on what the user typed in before hitting the enter key that ended the input line and placed it into the System.in byte stream that Scanner is scanning.
If you were to leave your program as written except for putting input.hasNext() into the while statement's test condition, it would block until more input was in the input stream after the integer that nextInt() processed.
Some answer(s) suggest using KeyBindings as a solution. Whilst that may work, it gets into waiting for keypress events at nearly the hardware level and is NOT friendly to platform independence. It is a potential rabbit-hole into Alice's Wonderland for having to figure out all kinds of event processing and the code having to know what platform it is running on. Using the hasNext() boolean false return to indicate the end of the input stream should work on any platform and will avoid potentially non-portable gee-whiz code for processing the keyboard and key presses at nearly the hardware event level.
The following program is one that does what you (and the exercise) intended and will end the input if the user presses CTRL-Z on a Windows platform or a CTRL-D on a UNIX/Linux platform without you having to determine the platform on which the code is executing.
// Exercise 6.16: EvenOrOddTest.java
// Write a method isEven that uses the remainder operator (%)
// to determine whether an integer is even. The method should
// take an integer argument and return true if the integer is
// even and false otherwise. Incorporate this method into an
// application that inputs a sequence of integers (one at a time)
// and determines whether each is even or odd.
import java.util.Scanner;
public class EvenOrOddTest {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int integer;
System.out.println("Odd even integer test.");
System.out.printf("Input CTRL-Z on Windows or CTRL-D on UNIX/Linux to end input\n"
+ "or an integer between values\n"
+ "%d and %d\n"
+ "to test whether it is odd or even: ",
Integer.MIN_VALUE, Integer.MAX_VALUE);
// the input.hasNext() will block until
// some kind of input, even a CTRL-Z,
// arrives in the stream
// the body of the while loop will execute
// every time input appears for as long as the input
// is not a CTRL-Z
while (input.hasNext()) { // repeat until end of input
// prompt user
// now see if the input we did get is an integer
if (input.hasNextInt()) { // we got an integer...
integer = input.nextInt();
System.out.printf("\n%d is "
+ (EvenOrOdd.isEven(integer) ? "even.\n\n" : "odd.\n\n"), integer);
} else { // we got a non-integer one too large for int
System.out.printf("\nInput %s invalid! Try again...\n\n", input.next());
} // end if...else
// white space (i.e. spaces and tabs) are separators
// next and nextInt get only to the first separator
// so it is possible for the user to enter an integer
// followed by tabs and/or spaces followed by more
// input, integer or not up to the end of the input line
// input.nextLine() flushes everything not processed
// by the nextInt() or next() to the input line end
// won't block execution waiting for input
// if there is nothing left on the input line
input.nextLine();
// prompt for user input again
System.out.printf("Input CTRL-Z to end input\n"
+ "or an integer between values\n"
+ "%d and %d\n"
+ "to test whether it is odd or even: ",
Integer.MIN_VALUE, Integer.MAX_VALUE);
} // end while
} // end main
static boolean isEven(int integer) {
// integer modulus 2 is zero when integer is even
return ((integer % 2) == 0);
} // end isEven
} // end class EvenOrOddTest
Related
The code below is for a simple calculator with the four basic mathematical operators. It is a working program, it works as expected. However, I have a few questions to understand and improve both my program as well as my understanding of Java. (I have used google but the amount of redundant info confuses me and haven't found any perfect answers on StackOverflow too, though there are dozens of related questions. Believe me, I did tried before posting here).
How can I make sure that the user input is exactly and strictly one char?
here in my program, it accepts more than one character (+-*) but operates on the first char (+) only. I want to make sure more than one character is not accepted as input.
After successful execution of the program, how can I somehow let the user repeat the main method? I mean, a user adds two numbers, gets his answer and he wants to do another calculation, maybe multiply two numbers this time. I can ask the user for yes or no to continue but how do I take him/her back to the beginning? (will a loop work? how?)
A the end of the program I used two methods to output a message. The system.out.print works fine but the JOptionPane method doesn't display the message and the program doesn't terminate (I have commented it out). I would like to understand why?
Is the default case required in the switch? And Am I following the correct code structure? (the arrangements and uses of curly braces)
NB: As I said this calculator works fine and can be used by newbies like myself to better understand the concept as I have commented on every detail. Please understand that I couldn't add everything in the question title due to limits...
package mycalculator;
import javax.swing.JOptionPane;
import java.util.*;
public class MyCalculator{
public static void main (String [] args){
// Let us code a simple calculator//
// Variable type declarations//
char OP;
int firstNum;
int secNum;
// Display an explanation of what this program does//
System.out.println("This is a simple calculator that will do basic
calculations such as :"
+ "\nAddition, Multiplication, Substraction and Division.");
// Create a scanner object to Read user Input.//
Scanner input = new Scanner(System.in);
// Ask user to input any positive number and read it//
System.out.println("Enter Any positive number followed by pressing
ENTER.");
firstNum = input.nextInt();
// Ask user to input/decide his choice operator and read it //
System.out.println("Enter a valid OPERATOR sign followed by pressing
ENTER.");
OP = input.next().charAt(0);
// Loop the below statement till one of the four (+,-,*,/) is entered.//
while(OP != '+' && OP != '-' && OP != '*' && OP != '/'){
System.out.println("Please Re-enter a valid Operator (+,-*,/)");
OP = input.next().charAt(0);}
// Ask user for any second number and read it//
System.out.println("Enter your Second number followed by an ENTER
stroke.");
secNum = input.nextInt();
// Various possible Resolution based on OP value.//
int RSum = firstNum+secNum;
int RSubs= firstNum-secNum;
int RPro = firstNum*secNum;
double DPro = firstNum/secNum;
// Conditional statements for Processing Results based on OP and display.//
switch(OP){
case '+': System.out.println("The Resulting sum is "+ RSum);
break;
case '-': System.out.println("The Resulting sum is "+ RSubs);
break;
case '*': System.out.println("The Resulting Product is "+ RPro);
break;
case '/': System.out.println("The Resulting Divisional product is "+
DPro);
break;
//Maybe the default case is actually not needed but is added for totality//
default : System.out.println("Try Again");
break;
}
// The following code maybe deleted, it is for experimental purpose//
// Just checking if additional statements executes after a program
completes//
System.out.println("Test Message ");
// JOptionPane.showMessageDialog(null, "The Process Ends Here!");
//The test message works fine//
//The JOptionPane statement don't work and program doesn't end. WHY?//
}
}
How can I make sure that the user input is exactly and strictly one
char? here in my program, it accepts more than one character (+-*) but
operates on the first char (+) only. I want to make sure more than one
character is not accepted as input.
If you use console application and Scanner, only thing that you can do is read a String and check its length. In case you use Swing, you could implement KeyPressListener and proceed exactly after user press a button (but not for console application).
After successful execution of the program, how can I somehow let the
user repeat the main method? I mean, a user adds two numbers, gets his
answer and he wants to do another calculation, maybe multiply two
numbers this time. I can ask the user for yes or no to continue but
how do I take him/her back to the beginning? (will a loop work? how?)
You can't repeat main method. In Java main method is been executing only once. To repeate your code, you could wrap whole main method content to the infinite loop or move the content to the separate method and call it from the loop in the main method.
A the end of the program I used two methods to output a message. The
system.out.print works fine but the JOptionPane method doesn't display
the message and the program doesn't terminate (I have commented it
out). I would like to understand why?
JOptionPane works only for graphic application (Swing/AWT). This is not available in console. You have only standard input and output there.
Is the default case required in the switch? And Am I following the
correct code structure? (the arrangements and uses of curly braces)
No, default case is optional by JVM syntax. I remember, that e.g. in C++ there was reccomendation to place it (event empty), to exclude side effects of compilators. I do not know, is there such reccomendation in Java, but when I use switch, I prefer to always add it to exclude logical problem (but this is definetly optional according to syntax case). You use switch correctly.
public static void main(String[] args) {
System.out.println("This is a simple calculator that will do basic calculations such as :"
+ "\nAddition (+)"
+ "\nMultiplication (*)"
+ "\nSubtraction (-)"
+ "\nDivision (/)");
System.out.println();
try (Scanner scan = new Scanner(System.in)) {
while (true) {
System.out.println("Enter Any positive number followed by pressing ENTER.");
int first = scan.nextInt();
System.out.println("Enter a valid OPERATOR (+,*,-,/) sign followed by pressing ENTER.");
String operator = scan.next();
while (operator.length() != 1 || !"+*-/".contains(operator)) {
System.out.println("Please Re-enter a valid Operator (+,-*,/)");
operator = scan.next();
}
scan.nextLine();
System.out.println("Enter your Second number followed by an ENTER stroke.");
int second = scan.nextInt();
if ("+".equals(operator))
System.out.println("The Resulting sum is " + (first + second));
else if ("*".equals(operator))
System.out.println("The Resulting mul is " + (first * second));
else if ("-".equals(operator))
System.out.println("The Resulting sub is " + (first - second));
else if ("/".equals(operator))
System.out.println("The Resulting div is " + ((double)first / second));
System.out.println();
System.out.println("Do you want to exit ('y' to exit)?");
if ("y".equals(scan.next()))
return;
System.out.println();
}
}
}
1) you can check size of string input.next() .If it is one then continue else again prompt for operator choice .
2)I would suggest better create a different method and put all logic in it and call it the number of time you want or call infinite number of times.
4)Should switch statements always contain a default clause?
How does this conditional statement (from Udacity's Intro to Java Programming | Problem Set 4 | Question #20) work?
import java.util.Scanner;
public class MonthPrinter {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.print("Enter a month number (1 through 12) ");
if (!in.hasNextInt()) {
System.out.println("Not an integer. Terminating");
} else {
int theMonthNumber = in.nextInt();
if (!(theMonthNumber >= 1 && theMonthNumber <= 12)) {
System.out.println("Number must be 1 through 12");
} else {
Month test = new Month(theMonthNumber);
System.out.print(test.getMonthName() + " " + test.getNumberOfDays());
}
}
}
}
The first if (!in.hasNextInt()) checks to see if the user input is an integer. If it's NOT an integer, the main method prints Not an integer. Terminating. That makes complete sense.
However, in the event that the user inputs an integer, the code proceeds to the else statement where the next line of code is int theMonthNumber = in.nextInt();
When the program runs and I provide an integer as an input, I'm NOT prompted for another input. I'm thinking that both the hasNextInt() method and nextInt() method should request an input from the user. Therefore, I should be prompted for a total of two inputs (assuming I provide an integer). When I dry-run this scenario, I input an integer 3. This passes the if(!hasNextInt()) check.
What am I missing in the logical flow of the statement(s)?
I'm thinking in my mind that both the hasNextInt() method and nextInt() method should request an input from the user.
No. The Scanner.hasNextInt() Javadoc says (in part)
Returns true if the next token in this scanner's input can be interpreted as an int value in the default radix using the nextInt() method. The scanner does not advance past any input.
The last sentence is telling you that it does not consume the int.
Function hasNextInt() is just for checking the input value. It will not consume the input. What happens is, it will advance the Scanner and check the data but after that it will go back to it's previous position. So technically, the position of Scanner has not changed. That's why it's not prompting for next input.
Only nextInt() will consume the data. Scanner will advance to the next position after calling this function.
I want to read user input like: 11 12 13 14 15 16
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
System.out.println(sc.next());
}
System.out.println("Test");
but it newer goes out of while loop and prints "Test".
How could i read that input?
The method hasNext() works like this:
If it sees the end of the file, it returns false;
If it sees another valid, non-whitespace input, it returns true;
If neither of the above is true, it waits for the next input the user is going to enter, and doesn't return until he does.
Usually, if you use Scanner for files, such a loop will work correctly, because a file has a definite end, and it usually doesn't get stuck waiting for more input.
But when you are working with console input (System.in, not redirected), then usually the user does not send the end-of-file signal. He just presses Return, and so, hasNext() sits and waits to see if the user will enter more input on the next line and so on.
There are two general ways to deal with this:
The user has to actually terminate the input. After you finish entering all your numbers and press Return, you also need to send the end-of-file sequence, which is usually ctrlD or ctrlZ.
If you do that, you will not be able to enter any more input to that program.
The program tells the user to enter some particular value that will tell it that the input is over. For example, the string "DONE". When you do that, you have to change the loop to something like:
String nextInput;
while( sc.hasNext() && ! (nextInput = sc.next()).equals( "DONE" ) ){
System.out.println(nextInput);
}
You can break the loop depending whether you want to quit or not E.g.
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
String next = sc.next();
if (next.equals("q")) { //if user press q then break the loop
break;
}
System.out.println(next);
}
System.out.println("Test");
}
Use api like:
while(sc.hasNextInt()){
int aba= sc.nextInt();
if (aba == 0) {//or even non numeric value here would let this loop exit
break;
}
}
So you need to enter 0 or even in other way enter non numeric value inorder to come out of loop. nextLine method will read whole line just once and then you will need to parse it and then convert to integer so it's good to use sc.nextInt which will do the work for you.
New programmer here. This is probably a really basic question, but it's stumping me nevertheless.
What I'm trying to do is write a method that supplies only one integer input so I can use that input in my main program without having to mess around with non-integer inputs. However, even writing the method to do that in its own method seems to be problematic.
public static int goodInput () {
Scanner input = new Scanner (System.in); //construct scanner
boolean test = input.hasNextInt(); //set a sentinel value
while (test == false) { //enter a loop until I actually get an integer
System.out.println("Integers only please"); //tell user to give me an integer
test = input.hasNextInt(); //get new input, see if it's an integer
}
int finalInput = input.nextInt(); //once i have an integer, set it to a variable
input.close(); //closing scanner
return finalInput; //return my integer so I don't have to mess around with hasNextInt over there
}
This seems to be broken in multiple levels, but I'm not really sure why.
If I enter an integer value like 0 or 1 when I'm first asked for input, it should skip the loop entirely. But, instead, it enters the loop, and prints "Integers only please". Even worse, it doesn't actually ask for input while I'm in there, and just prints that line repeatedly.
I understand the latter problem is probably due to token issues, but I'm not necessarily sure how to solve them; closing and then reopening the scanner gets Eclipse to bug me over "duplicate objects", simply assigning the old input to a garbage String variable that is never used tells me that "No line was found" at runtime, and I'm not experienced enough to think of other ways to get new input.
Even once that's solved, I need to find some way to avoid entering the loop in the case of having an integer. I don't really understand why integer inputs inter the loop to begin with, so I'm not sure how this would be possible.
Please help? Sorry if this is an old question; tried looking at past questions but none of them seem to have the same problem that I have.
You were close: this works fine for me:
Scanner input = new Scanner(System.in); //construct scanner
while(!input.hasNextInt()) {
input.next(); // next input is not an int, so consume it and move on
}
int finalInput = input.nextInt();
input.close(); //closing scanner
System.out.println("finalInput: " + finalInput);
By calling input.next() in your while loop, you consume the non-integer content and try again, and again, until the next input is an int.
//while (test == false) { // Line #1
while (!test) { /* Better notation */ // Line #2
System.out.println("Integers only please"); // Line #3
test = input.hasNextInt(); // Line #4
} // Line #5
The problem is that in line #4 above, input.hasNextInt() only tests if an integer is inputted, and does not ask for a new integer. If the user inputs something other than an integer, hasNextInt() returns false and you cannot ask for nextInt(), because then an InputMismatchException is thrown, since the Scanner is still expecting an integer.
You must use next() instead of nextInt():
while (!input.hasNextInt()) {
input.next();
// That will 'consume' the result, but doesn't use it.
}
int result = input.nextInt();
input.close();
return result;
Sorry if this sounds too simple. I'm very new to Java.
Here is some simple code I was using to examine hasNextLine(). When I run it, I can't make it stop. I thought if you didn't write any input and pressed Enter, you would escape the while loop.
Can someone explain to me how hasNextLine() works in this situation?
import java.util.*;
public class StringRaw {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNextLine()) {
String str = sc.nextLine();
}
System.out.print("YOU'VE GOT THROUGH");
}
}
When reading from System.in, you are reading from the keyboard, by default, and that is an infinite input stream... it has as many lines as the user cares to type. I think sending the control sequence for EOF might work, such as CTL-Z (or is it CTL-D?).
Looking at my good-ol' ASCII chart... CTL-C is an ETX and CTL-D is an EOT; either of those should work to terminate a text stream. CTL-Z is a SUB which should not work (but it might, since controls are historically interpreted highly subjectively).
CTRL-D is the end of character or byte stream for UNIX/Linux and CTRL-Z is the end of character or byte stream for Windows (a historical artifact from the earliest days of Microsoft DOS).
With the question code as written, an empty line won't exit the loop because hasNextLine() won't evaluate to false. It will have a line terminator in the input byte stream.
System.in is a byte stream from standard input, normally the console. Ending the byte stream will therefore stop the loop. Although nextLine() doesn't block waiting for input, hasNextLine() does. The only way the code terminates, as designed, is with CTRL-Z in Windows or CTRL-D in UNIX/Linux, which ends the byte stream, causes hasNextLine() not to block waiting for input and to return a boolean false which terminates the while loop.
If you want it to terminate with an empty line input you can check for non-empty lines as part of the loop continuation condition. The following code demonstrates how to change the basic question design that uses hasNextLine() and nextLine() to one that terminates if it gets an empty line or an end of input character (i.e. CTRL-Z in Windows or CTRL-D in UNIX/Linux). The additional code in the while condition uses a feature of assignment operators wherein they can be evaluated like an expression to return the value that was assigned. Since it is a String object, the String.equals() method can be used with the evaluation.
Other additional code just adds some printed output to make what is going on obvious.
// HasNextLineEndDemo.java
import java.util.*;
public class HasNextLineEndDemo {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
// this code is a bit gee-whiz
// the assignment expression gets assigned sc.nextLine()
// only if there is one because of the &&
// if hasNextLine() is false, everything after the &&
// gets ignored
// in addition, the assignment operator itself, if
// executed, returns, just like a method return,
// whatever was assigned to str which,
// as a String object, can be tested to see if it is empty
// using the String.equals() method
int i = 1; // input line counter
String str = " "; // have to seed this to other than ""
System.out.printf("Input line %d: ", i); // prompt user
while (sc.hasNextLine() && !(str = sc.nextLine()).equals("")) {
System.out.printf("Line %d: ", i);
System.out.println("'" + str + "'");
System.out.printf("Input line %d: ", ++i);
} // end while
System.out.println("\nYOU'VE GOT THROUGH");
} // end main
} // end class HasNextLineEndDemo
Hit Ctrl + D to terminate input from stdin. (Windows: Ctrl + Z) or provide input from a command:
echo -e "abc\ndef" | java Program
I had a similar problem with a socket input stream. Most solutions I found would still block the execution. It turns out there is a not-blocking check you can do with InputStream.available().
So in this case the following should work:
int x = System.in.available();
if (x!=0) {
//Your code
}
As per my understanding , if you take an example of result set object from JDBC or any iterator then in these cases you have a finite set of things and the iterators each time check whether end of the set has been reached.
However in the above case , their is no way of knowing the end of user input i.e. hasNextLine() has no way of knowing when user wants to terminate, and hence it goes on infinitely.
Best way is to put additional condition on the for loop that checks for some condition inside for loop that fails in the future.
In the above post #Jim 's answer illustrates this.
In fact using hasNextLine() as loop terminator for console input should be discouraged because it will never return false.