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.
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
So I am trying to make a code that will prompt the user to either use a basic calculator, or a word counter that displays how many words are in a given sentence entered by the user, this is done using methods. I have figured out how to properly set up the calculator, but the word counter is giving me some issues:
public static int wordCounter(String str){
String words[]=str.split(" ");
int count=words.length;
return count;
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("What do you want to do( calculator(0)/word counter(1) )? ");
//This runs and I select '1' for word counter
int choice = input.nextInt(); //Input the choice here
if (choice == 0) {
// It runs this selection statment, and since zero is not selected,
//it runs the word Counter branch
calculator();
}else{
System.out.println("Please enter a sentence:"); // Tells me to enter a sentence
String sentence=input.nextLine();
//^ This input is completely skipped and goes
//right to the 'System.out.print(); Statement.
System.out.print("There are "+ wordCounter(sentence) + " words in the sentence.");
//^ This prints a 1 immediately after the branch is selected with '1'
}
}
I'm not sure where it is going wrong since this only happens while it is in the if/else statement. Doing some testing also showed me that it seems that the first scanner "int choice=input.nextInt()" Is somehow interfering with the second scanner for the string. Any ideas keeping a similar formatting would be greatly appreciated.
Please forgive my formatting, it may not look great.
nextLine() will only return the remainder of the current line being scanned. Since you would have pressed enter after selecting the number, all it will capture is an empty string.
To fix it, just add a nextLine() directly after you get the integer.
public String nextLine()
Advances this scanner past the current line and returns the input that was skipped. This method returns the rest of the current line, excluding any line separator at the end. The position is set to the beginning of the next line.
Since this method continues to search through the input looking for a line separator, it may buffer all of the input searching for the line to skip if no line separators are present.
https://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html#nextLine()
The problem is when you enter the number int choice = input.nextInt() it's only scanning the integer, not the newline. So when you call input.nextLine() it instantly returns an empty string. One way to fix this would be to replace that line with
int choice = Integer.parseInt(input.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've been working on a programming assignment that acts as a Scrabble dictionary for a while now. The program takes input from the user and outputs a file with a list of words, depending on what the user requests from a menu. The problem I've been having has to do with Scanner.nextLine().
I'm not aexactly sure why, but for some reason I have to press enter once sometimes before my code will take my input and store it as the variable. Essentially, I end up entering the input twice. I tried inserting Scanner.nextLine() around the code to "take up" the empty enter/spaces but it doesnt work, and I have to press enter multiple times to get it to process what I want.
Does anybody have any suggestions? I'd appreciate any and all help.
Here is a bit of the code:
System.out.println("Enter the length of the word you are" + " searching for.");
int n = -1;
while(!(n >=0)) {
if(in.hasNextInt())
n = in.nextInt();
else {
System.out.println("You have not entered a valid number.
Please enter a real number this time.");
in.nextLine();
}
}
in.nextLine();
System.out.println("Enter the first letter of the words" + " you are searching for.");
String firstLetter = "";
while(!(firstLetter.length() == 1)) {
if(in.nextLine().length() > 1) {
System.out.println("You have not entered a valid letter.
Please press enter and enter only one real letter.");
}
else if(in.hasNextInt()) {
System.out.println("Do not enter a number. Please enter one real letter.");
}
else {
in.nextLine();
firstLetter = in.nextLine();
break;
}
}
At the end of this, I have to press enter once and then input to get it to store anything in the variable firstLetter. I assume it has something to do with the nature of nextLine(), as the conditions using nextInt() give no issues.
It's because you're using both nextLine() and nextInt(), what's going on is that nextLine() is searching for a new line (enter) and nextInt will automatically stop the search if any integer is typed through System.in.
Rule of thumb: Just use Scanner.nextLine() for your input, then convert your string from Scanner.nextLine() accordingly through Integer.parseInt(string), etc.
I think you're overcompensating with too many nextLines. You may want to do that once to clear the line after the int is inputted, for example, to clear the newline, but the second time here just absorbs an extra line of input:
System.out.println("You have not entered a valid number. Please enter a real number this time.");
in.nextLine();//first time
}
}
in.nextLine();//this second time is unnecessary.
The same thing happens with your duplicate uses here:
in.nextLine();
firstLetter = in.nextLine();
break;
You should only add an extra in.nextLine() immediately between inputting nextSOMETHINGELSE() and another nextLine().
EDIT:
Additionally, note that whenever you call in.nextLine(), you are absorbing a line of input. For example, this line should be fixed:
if(in.nextLine().length() > 1){
because it reads in a line, using it up, and then checks whether that (now used-up) line is long enough.
I will admit, this is a school assignment... But I simply cannot figure out what I am doing wrong.
I have a hash table with an insert function. The following code is supposed to take a line of data from System.in in the format "Long String" (i.e. "32452 John"). The first token must be a Long for the ID number, and it must be followed by a String token for the name. When I run the program and I get to the portion where this must be executed (It is in a switch statement), I entered 'a' and hit enter. The command line immediately reads "Invalid value." (note: not VALUES, as that would mean it hit the nested if statement. It won't let me type in any data. Thank you in advance!
System.out.println("Enter ID and Name.");
//temp to take in the next line entered by the user
//inScan is the Scanner for System.in
temp = inScan.nextLine();
//Create Scanner for the line
Scanner tempScan = new Scanner(temp);
if(tempScan.hasNextLong()){
thisID = tempScan.nextLong();
if((tempScan.hasNext()) && (thisID>0)){
thisName = tempScan.next();
//The data will only be inserted if both segments of data are entered
myTable.insert(new Student(thisID, thisName));
}else{
System.out.println("Invalid values.");
}
}else{
System.out.println("Invalid value.");
}
Why do you need the second Scanner?
Example
String input = scanner.nextLine();
String[] tokens = input.split(" ");
Long id = Long.parseLong(tokens[0]);
String name = tokens[1];
And if you wanted to add your validation:
String input = scanner.nextLine();
if(input.contains(" ")) {
// You know there's a space in it.
String[] tokens = input.split(" ");
if(tokens.length == 2) {
// You know it's a value, followed by a space, followed by a value.
if(tokens[0].matches("[0-9]+")) {
// You know it only contains numbers.
Long id = Long.parseLong(tokens[0]);
}
}
}
I've not run it, but i guess your problem is that when you enter the text 'a' and hit enter, this line is false:
if(tempScan.hasNextLong()){
as you haven't entered a number. hence why it drops to the next block. If you enter something numerical first, i suspect your code with work. you probably need to add a 'while' loop around it, to run until it gets a number.
You already have a Scanner which reads from System.in, there's no need for another one. The second one you've made is a scanner for a String, which will never have a nextLong as it has nothing to scan after your String.
I won't write any code for you as this is homework, but stick to your original scanner when checking for user input instead.