I am writing a program in which a user inputs an unknown amount of numbers on a single line, like: 2 6 3 9 12.
I have to insert those numbers in a queue. However the loop doesn't end until I enter a non-integer value. I found one solution online which was to use .useDelimiter(" *"). This works except for when I enter a two digit integer: it splits it into two separate numbers. Is there a way to end this loop without having to enter a non integer value?
Scanner in = new Scanner(System.in)
while(in.hasNextInt())
{
myQueue.insert(in.nextInt());
}
Since the numbers are on one line, you could read the line and construct a Scanner on that text. Something like,
System.out.println("Please enter a line of integer values: ");
Scanner in = new Scanner(System.in);
if (in.hasNextLine())
{
String line = in.nextLine();
Scanner scan = new Scanner(line);
while (scan.hasNextInt()) {
myQueue.insert(scan.nextInt());
}
}
Here's one way:
Scanner in = new Scanner(System.in);
while(in.hasNextLine())
{
try {
myQueue.insert(Integer.parseInt(in.nextLine()));
}
catch (NumberFormatException ex) {
break;
}
}
I think you are on the right track. There are a number of ways to do this. You could modify your current code by adding the line:
in.useDelimiter(System.getProperty("line.separator"));
This will delimit the iterator to a new line correctly on any platform.
Related
This is the program
public class bInputMismathcExceptionDemo {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean continueInput = true;
do {
try {
System.out.println("Enter an integer:");
int num = input.nextInt();
System.out.println("the number is " + num);
continueInput = false;
}
catch (InputMismatchException ex) {
System.out.println("Try again. (Incorrect input: an integer is required)");
}
input.nextLine();
}
while (continueInput);
}
}
I know nextInt() only read the integer not the "\n", but why should we need the input.nextLine() to read the "\n"? is it necessary?? because I think even without input.nextLine(), after it goes back to try {}, the input.nextInt() can still read the next integer I type, but in fact it is a infinite loop.
I still don't know the logic behind it, hope someone can help me.
The reason it is necessary here is because of what happens when the input fails.
For example, try removing the input.nextLine() part, run the program again, and when it asks for input, enter abc and press Return
The result will be an infinite loop. Why?
Because nextInt() will try to read the incoming input. It will see that this input is not an integer, and will throw the exception. However, the input is not cleared. It will still be abc in the buffer. So going back to the loop will cause it to try parsing the same abc over and over.
Using nextLine() will clear the buffer, so that the next input you read after an error is going to be the fresh input that's after the bad line you have entered.
but why should we need the input.nextLine() to read the "\n"? is it necessary??
Yes (actually it's very common to do that), otherwise how will you consume the remaining \n? If you don't want to use nextLine to consume the left \n, use a different scanner object (I don't recommend this):
Scanner input1 = new Scanner(System.in);
Scanner input2 = new Scanner(System.in);
input1.nextInt();
input2.nextLine();
or use nextLine to read the integer value and convert it to int later so you won't have to consume the new line character later.
Also you can use:
input.nextInt();
input.skip("\\W*").nextLine();
or
input.skip("\n").nextLine();
if you need whitespaces before line
I have been using the following code in c and c++ for looping till user feeds the correct value till the program comes out of it:
while((scanf("%d",&num)==1)//same way in for loop
{
//some code
}
Can i some how use the same way to accept and loop the program till i keep entering let's say an integer and floating or a char or a special character breaks it.
Use :
Scanner sc = new Scanner(System.in); // OR replace System.in with file to read
while(sc.hasNext()){
//code here
int x = sc.nextInt();
//...
}
There are different variants of hasNext() for specific expected input types: hasNextFloat(), hasNextInt()..
Same goes for next() method so you can find nextInt(), nextFloat() or even nextLine()
You can go to Java doc for more info.
As proposed in comments, you can use the Scanner class.
Note you need to read the in buffer with a nextLine() when it is not an int.
public static void main(String[] args) {
try (Scanner in = new Scanner(System.in)) {
System.out.println("Enter an int: ");
while (!in.hasNextInt()) {
System.out.println("That's not an int! try again...");
in.nextLine();
}
int myInt = in.nextInt();
System.out.println("You entered "+myInt);
}
}
Ive got a problem.I 'm new to Java,I've started today:D) ..I've programmed before so I know it little bit,but I am new to Java. Here is my code: `
public class Tutorial {
public static void main(String[] args) {
double num1,num2;
String operacia;
Scanner in=new Scanner (System.in);
System.out.println("Write 2 numbers");
num1=in.nextDouble();
num2=in.nextDouble();
System.out.println("Choose the operation");
operacia=in.nextLine();
if (operacia.equals("+")){
System.out.println("Your result is "+(num1+num2)) ;
}
else if (operacia.equals("-")){
System.out.println("Your result is "+(num1-num2)) ;
}
else if (operacia.equals("/")){
System.out.println("Your result is "+(num1/num2)) ;
}
else if (operacia.equals("*")){
System.out.println("Your result is "+(num1*num2)) ;
}
}
}`
It wants from me 2 numbers,I write them and them it writes "Choose the operation" and its over.No more inputs.Thank you very much :)
Your problem is simple.
Just replace the code with next() instead of nextLine().Effectively, the line your code is returning is receiving is a blank line. Hence when it reaches the conditional statement it has an empty string and terminates.
next()
Finds and returns the next complete token from this scanner.
nextLine()
Advances this scanner past the current line and returns the input that was skipped.
Your code should be fixed by a simple change.
public static void main(String[] args) {
double num1,num2;
String operacia;
Scanner in=new Scanner (System.in);
System.out.println("Write 2 numbers");
num1=in.nextDouble();
num2=in.nextDouble();
System.out.println("Choose the operation");
operacia=in.next();
if (operacia.equals("+")){
System.out.println("Your result is "+(num1+num2)) ;
}
else if (operacia.equals("-")){
System.out.println("Your result is "+(num1-num2)) ;
}
else if (operacia.equals("/")){
System.out.println("Your result is "+(num1/num2)) ;
}
else if (operacia.equals("*")){
System.out.println("Your result is "+(num1*num2)) ;
}
}
Scanner#nextDouble() consumes only the next token as a double from the input. It does not consume the new line you typed using the Enter on the keyboard while entering the two numbers. When the execution reaches operacia=in.nextLine();, this new line is consumed, never allowing the user a chance to type the operating string.
To solve this, you need to read the whole line using Scanner#nextLine() and convert it to a double:
String input = in.nextLine();
num1 = Double.parseDouble(input);
input = in.nextLine();
num2 = Double.parseDouble(input);
I believe the in.nextLine(); operation is reading only to the end of the line where you input 2 numbers. If you want your program to only consider the next line, you have to clear the current one first.
Try this, it should work:
System.out.println("Choose the operation");
in.nextLine(); //clear the current line
operacia=in.nextLine();
This is the program
public class bInputMismathcExceptionDemo {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean continueInput = true;
do {
try {
System.out.println("Enter an integer:");
int num = input.nextInt();
System.out.println("the number is " + num);
continueInput = false;
}
catch (InputMismatchException ex) {
System.out.println("Try again. (Incorrect input: an integer is required)");
}
input.nextLine();
}
while (continueInput);
}
}
I know nextInt() only read the integer not the "\n", but why should we need the input.nextLine() to read the "\n"? is it necessary?? because I think even without input.nextLine(), after it goes back to try {}, the input.nextInt() can still read the next integer I type, but in fact it is a infinite loop.
I still don't know the logic behind it, hope someone can help me.
The reason it is necessary here is because of what happens when the input fails.
For example, try removing the input.nextLine() part, run the program again, and when it asks for input, enter abc and press Return
The result will be an infinite loop. Why?
Because nextInt() will try to read the incoming input. It will see that this input is not an integer, and will throw the exception. However, the input is not cleared. It will still be abc in the buffer. So going back to the loop will cause it to try parsing the same abc over and over.
Using nextLine() will clear the buffer, so that the next input you read after an error is going to be the fresh input that's after the bad line you have entered.
but why should we need the input.nextLine() to read the "\n"? is it necessary??
Yes (actually it's very common to do that), otherwise how will you consume the remaining \n? If you don't want to use nextLine to consume the left \n, use a different scanner object (I don't recommend this):
Scanner input1 = new Scanner(System.in);
Scanner input2 = new Scanner(System.in);
input1.nextInt();
input2.nextLine();
or use nextLine to read the integer value and convert it to int later so you won't have to consume the new line character later.
Also you can use:
input.nextInt();
input.skip("\\W*").nextLine();
or
input.skip("\n").nextLine();
if you need whitespaces before line
I have the following code:
Scanner inputSide = new Scanner(System.in);
double side[] = new double[3];
int i = 0;
do{
try{
System.out.println("Enter three side lengths for a triangle (each followed by pressing enter):");
side[i] = inputSide.nextDouble();
i++;
}
catch(Exception wrongType){
System.err.println(wrongType);
System.out.println("Please enter a number. Start again!!");
i=0;
}
}
while(i<3);
It works fine and does what it's meant to if I don't enter a wrong data type but if I enter something other than a double then it loops over and over, printing everything in both try and catch blocks instead of waiting for me to enter another double.
Any help as to why it's doing this - as I can't seem to understand why - would be appreciated.
Thank you :)
The problem is that, you have used input.nextDouble method, which reads only the next token in the input, thus skipping the newline at the end. See Scanner.nextDouble
Now, if you enter wrong value first time, then it will consider the newline as the next input. Which will also be invalid.
You can add an empty input.nextLine in the catch block.
catch(Exception wrongType){
System.err.println(wrongType);
System.out.println("Please enter a number. Start again!!");
i=0;
input.nextLine(); // So that it consumes the newline left over
}
Now, your nextLine() will read the linefeed left over, and linefeed will not be taken as input to your nextDouble next time. In which case, it will fail, even before you giving any input.