Did some metrics around these in a single threaded model and here are the results I got. ~~~~~~~~~~~~~~~~~~Time Metrics~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~ Tokenizer | String.Split() | while+SubString | Scanner | ScannerWithCompiledPattern ~ ~ 4.0 ms | 5.1 ms | 1.2 ms | 0.5 ms | 0.1 ms ~ ~ 4.4 ms | 4.8 ms | 1.1 ms … Read more
I wonder if you’re not handling an end of line token appropriately. Often if you use Scanner#next###() (except for nextLine), and you reach an end of line token as when the user presses enter, if you don’t handle the end of line token, it will prevent the Scanner object from working appropriately. To solve this, … Read more
The class java.io.Console may be useful: System.console().readPassword(); This reads a sequence of chars from the console, without echoing anything. Note that it only works when you launch your java application with a real console. Otherwise, System.console() returns null.
You could try something like this: public static void main (String[] args) { Scanner input = new Scanner(System.in); double[] numbers = new double[5]; for (int i = 0; i < numbers.length; i++) { System.out.println(“Please enter number”); numbers[i] = input.nextDouble(); } } It seems pretty basic stuff unless I am misunderstanding you
The reason for the error is that the nextInt only pulls the integer, not the newline. If you add a in.nextLine() before your for loop, it will eat the empty new line and allow you to enter 3 names. int nnames; String names[]; System.out.print(“How many names are you going to save: “); Scanner in = … Read more
.nextInt() gets the next int, but doesn’t read the new line character. This means that when you ask it to read the “next line”, you read til the end of the new line character from the first time. You can insert another .nextLine() after you get the int to fix this. Or (I prefer this … Read more
There are 2 observations : Using myScannerInstance.nextInt() leaves behind a new line character. So, if you call nextLine() after nextInt(), the nextLine() will read the new line character instead of the actual data. Consequently, you will have to add another nextLine() after the nextInt() to gobble up that dangling new-line character. nextLine() doesn’t leave behind … Read more
Here’s a way: Scanner keyboard = new Scanner(System.in); String line = null; while(!(line = keyboard.nextLine()).isEmpty()) { String[] values = line.split(“\\s+”); System.out.print(“entered: ” + Arrays.toString(values) + “\n”); } System.out.print(“Bye!”);
Your file should directly be under the project folder, and not inside any other sub-folder. So, if your project folder is MyProject, it’s folder structure(not complete though) should be like: – MyProject +- src + | | | +– Your source file +- file.txt It should not be under src folder. Or, you can give … Read more
I am assuming you are using java 7, thus you get a compiler warning, when you don’t close the resource you should close your scanner usually in a finally block. Scanner scanner = null; try { scanner = new Scanner(System.in); //rest of the code } finally { if(scanner!=null) scanner.close(); } Or even better: use the … Read more