package mygradeloops;
import java.io.IOException;
public class MyGradeLoops {
public static void main(String[] args) throws IOException {
char x = 'A';
for (x='0';x<'9';x++){
System.out.println("Please enter in one of your grades.");
System.in.read();
System.out.println("Keep going!");
}
}
}
This code keeps double printing after the first "grade". Does anyone know why it double prints? Have I done the "For Loop" wrong?
It's "double printing" because when you enter a character by pressing return, you're actually writing two characters: the character you typed, and \n (newline character).
Add a second System.in.read(); call to read the newline character:
for (x='0';x<'9';x++){
System.out.println("Please enter in one of your grades.");
System.in.read(); // your character
System.in.read(); // newline
System.out.println("Keep going!");
}
Also, initializing x to 'A' in not needed, char x; is fine. In fact, it doesn't make sense to use a char in this loop, using an int would be preferred.
The read method for System.in (an InputStream) only reads one byte of data from the input stream. Because you must hit "Enter" to send your input to the stream, you have two characters on the stream - the character you typed plus a newline.
The for loop loops twice, and "double prints" Keep going! and Please enter in one of your grades. because each iteration reads one of the two characters on the stream.
It would be easier to wrap System.in in a InputStreamReader then a BufferedReader or just initialize a Scanner with System.in. With a BufferedReader you can just call readLine(), and with a Scanner you can call nextLine().
Also, it's unclear why you are looping from '0' to '9' with chars. It would be clearer to use an int for a and loop from 0 until 9.
Related
In the following code why does the string inside println method shows twice.What should I do to show the message once per iteration
package practicejava;
public class Query {
public static void main(String[] args) throws java.io.IOException {
System.out.println("Guess a capital letter Character");
while ((char) System.in.read() != 'S') {
System.out.println("wrong.guess again to finish the program");
}
}
}
When a user writes in console characters, to confirm fact that his input is ready to be passed to application he presses enter key. But console doesn't pass only provided characters, it also adds to input stream (System.in) OS dependent line separator character(s) after it. Some OS use \r or \n (those are single characters, \x is just notation to represent them) others like Windows use \r\n (two characters) sequence as line separator.
Now those additional characters are also read by System.in.read() and since they are not equal to S System.out.println("wrong.guess again to finish the program"); is executed additional time.
To avoid such problems instead of working with raw data via System.in.read() consider using classes meant to make our life easier like java.util.Scanner
Scanner sc = new Scanner(System.in);
System.out.println("Guess a capital letter Character");
String response = sc.nextLine();
while(!response.equals("S")){
System.out.print("incorrect data, please try again: ");
response = sc.nextLine();
}
Its because the first char that you read is the letter you typed, and then there is a second loop where the character is the line return.
For example on my linux machine, if I input "E" and then press enter, the first loop processes the char 69 'E', and then there is a second loop to process the carriage return (char 10).
What you can do is use a Scanner to get the user's input:
package practicejava;
import java.util.Scanner;
public class Query {
public static void main(String[] args) throws java.io.IOException {
Scanner s = new Scanner(System.in);
char c;
do {
System.out.println("Guess a capital letter Character");
c = s.next().charAt(0);
if (c != 's') {
System.out.println("Wrong! Guess again to finish the program.");
}
} while(c != 's');
}
}
s.next() will get the input from the user as a string and s.next().charAt(0) will return the first character in that string.
I have started to learn Java, wrote couple of very easy things, but there is a thing that I don't understand:
public static void main(String[] args) throws java.io.IOException
{
char ch;
do
{
System.out.println("Quess the letter");
ch = (char) System.in.read();
}
while (ch != 'q');
}
Why does the System.out.println prints "Quess the letter" three times after giving a wrong answer. Before giving any answer string is printed only once.
Thanks in advance
Because when you print char and press Enter you produce 3 symbols (on Windows): character, carriage return and line feed:
q\r\n
You can find more details here: http://en.wikipedia.org/wiki/Newline
For your task you may want to use higher level API, e.g. Scanner:
Scanner scanner = new Scanner(System.in);
do {
System.out.println("Guess the letter");
ch = scanner.nextLine().charAt(0);
} while (ch != 'q');
Using System.in directly is probably the wrong thing to do. You'll see that if your character is changed from q to something in Russian, Arabic or Chinese. Reading just one byte is never going to match it. You are just lucky that the bytes read from console in UTF-8 match the character codes for the plain English characters.
The way you are doing it, you are looking at the input as a stream of bytes. And then, as #Sergey Grinev said, you get three characters - the actual character you entered, and the carriage return and line feed that were produce by pressing Enter.
If you want to treat your input as characters, rather than bytes, you should create a BufferedReader or a Scanner backed by System.in. Then you can read a whole line, and it will dispose of the carriage return and linefeed characters for you.
To use a BufferedReader you do something like:
BufferedReader reader = new BufferedReader( InputStreamReader( System.in ) );
And then you can use:
String userInput = reader.readLine();
To use a Scanner, you do something like:
Scanner scanner = new Scanner( System.in );
And then you can use:
String userInput = scanner.nextLine();
In both cases, the result is a String, not a char, so you should be careful - don't compare it using == but using equals(). Or make sure its length is greater than 1 and take its first character using charAt(0).
As has been mentioned, the initial read command takes in 3 characters and holds them in the buffer.
The next time a read command comes around, it first checks the buffer before waiting for a keyboard input. Try entering more than one letter before hitting enter- your method should get called however many characters you entered + 2.
For an even simpler fix:
//add char 'ignore' variable to the char declaration
char ch ignore;
//add this do while loop after the "ch = (char) System.in.read();" line
do{
ignore = (char) System.in.read();
} while (ignore != '\n');
this way 'ignore' will cycle through the buffer until it hits the newline character in the buffer (the last one entered via pressing enter in Windows) leaving you with an fresh buffer when the method is called again.
My professor tends to do the following to get a number from the user:
Scanner scanner = new Scanner(System.in);
Integer.parseInt(scanner.nextLine());
What are the benefits as opposed to simply doing scanner.nextInt() ?
java.util.Scanner.java has the following in it:
public int nextInt() {
return nextInt(defaultRadix);
}
public int nextInt(int radix) {
// Check cached result
if ((typeCache != null) && (typeCache instanceof Integer)
&& this.radix == radix) {
int val = ((Integer)typeCache).intValue();
useTypeCache();
return val;
}
setRadix(radix);
clearCaches();
// Search for next int
try {
String s = next(integerPattern());
if (matcher.group(SIMPLE_GROUP_INDEX) == null)
s = processIntegerToken(s);
return Integer.parseInt(s, radix);
} catch (NumberFormatException nfe) {
position = matcher.start(); // don't skip bad token
throw new InputMismatchException(nfe.getMessage());
}
}
As I see it, Scanner calls Integer.parseInt() itself as well, on top of additional hocus pocus. Are there significant performance gains in doing simply Integer.parseInt(scanner.nextLine()) ? Are there on the other hand any drawbacks?
How about when scanning through a file with significant amount of data, and not a user input?
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 a new line character.
code :
int age=myScannerInstance.nextInt();
String name = myScannerInstance.nextLine();// here the actual name will not be read. The new line character will be read.
nextInt() will again go back to the underlying stream and read. IO calls take time (expensive). It will do lot of checks to get the next integer. nextLine() will do those checks only once. So, if you call nextLine() once and read 5 integers (as a single line String), split them and parse them as integers (using Integer.parseInt()), it will be faster and more efficient than reading each int individually.
Using nextLine() + parseInt() will give you enormous performance benefit when you are running a very large loop.
Usage :
Using nextInt() gives you an additional advantage wherein you will get an exception if the input text is not an integer. example 123 is accepted.. 123sdsa will throw an InputMismatchException. So, you can catch it and handle it appropriately.
Using nextLine() will read the entire line, so, it will read the entire String sada1231 and then fail with NumberFormatException if it cannot parse the String as a number. You will have to handle that exception.
Generally, one nextLine() / nextInt() call won't make much of a difference. If you have a loop or if you are reading lot of data, then using readLine() with parseInt() will be very efficient.
nextInt() reads a number, but doesn’t consume line separator. While nextLine() reads the String and consumes the new-line character. According to Java Docs:
… 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.
In other words when you enter a number then press Enter, input.nextInt() consumes only the number, not the "end of line", primitive data types like int, double etc does not consume "end of line", due which this "end of line" remain in buffer ane When input.next() executes, it consumes the "end of line" from buffer from the first input. So you professor is trying to get to the next line after he reads the user input. You have to look at the logic of his codes only then you can understand it.
I also used to face this problem often. So i use to code like this..
public static void main(String[] args) {
Scanner key= new Scanner(System.in);
String name;
int age;
age = key.nextInt();
key.nextLine();
name = key.nextLine(); //to carry the new line character left behind nextInt()
System.out.println("Age : "+age);
System.out.println("Name: "+name);
}
here as the key.nextInt() leaves a new line character we are using key.nextLine() to carry the new Line character and then move to the nextline where the actual data is present. As we discussed above using Integer.parseInt() will be more efficient than using nextInt(). But this is also one of the way to code to overcome the problem.
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.
public class Solution {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
double d = scan.nextDouble();
scan.nextLine();
String s = scan.nextLine();
scan.close();
System.out.println("String: " + s);
System.out.println("Double: " + d);
System.out.println("Int: " + i);
}
}
I have following code and am facing a problem if I use System.in.read() before Scanner.
Then the cursor moves at the end by skipping nextLine() function.
import java.util.Scanner;
public class InvoiceTest{
public static void main(String [] args) throws java.io.IOException {
System.out.println("Enter a Charater: ");
char c = (char) System.in.read();
Scanner input = new Scanner(System.in);
System.out.println("Enter id No...");
String id_no = input.nextLine();
System.out.println("Charater You entered "+ c +" Id No Entered "+ id_no);
}
}
You are not consuming the newline character upon entering your character(System.in.read()) thus the input.nextLine() will consume it and skip it.
solution:
consume the new line character first before reading the input of for the id.
System.out.println("Enter a Charater: ");
char c = (char) System.in.read();
Scanner input = new Scanner(System.in);
System.out.println("Enter id No...");
input.nextLine(); //will consume the new line character spit by System.in.read()
String id_no = input.nextLine();
System.out.println("Charater You entered "+c+" Id No Entered "+id_no);
}
EJP comments thus:
Don't mix System.in.read() with new Scanner(System.in). Use one or the other.
Good advice!
So why is it a bad idea to mix reading from the stream and using Scanner?
Well, because Scanner operations will typically read ahead on the input stream, keeping unconsumed characters in an internal buffer. So if you do a Scanner operation followed by a call to read() on the stream, there is a good chance that the read() will (in effect) skip over characters. The behaviour is likely to be confusing and unpredictable ... and dependent on where the input characters are actually coming from.
The problem is I cant read the variable input with next() cause when I try to split (.split" ") every whitespace then the array just get the first two words I type so I had to use keyboard.nextLine() and the splitting process works the way it should work and I get all the words in the array but the problem is that If I use nextLine() then I have to create another keyboard object to read the first variable (answer) and that is the only way I can make it work here is the code
Scanner keyboard=new Scanner(System.in);
Scanner keyboard2=new Scanner(System.in);//just to make answer word
int answer=keyboard.nextInt();//if I don't use the keyboard2 here then the program will not work as it should work, but if I use next() instead of nextLine down there this will not be a problem but then the splitting part is a problem(this variable counts number of lines the program will have).
int current=1;
int left=0,right=0,forward=0,back=0;
for(int count=0;count<answer;count++,current++)
{
String input=keyboard.nextLine();
String array[]=input.split(" ");
for (int counter=0;counter<array.length;counter++)
{
if (array[counter].equalsIgnoreCase("left"))
{
left++;
}
else if (array[counter].equalsIgnoreCase("right"))
{
right++;
}
else if (array[counter].equalsIgnoreCase("forward"))
{
forward++;
}
else if (array[counter].equalsIgnoreCase("back"))
{
back++;
}
}
}
}
Thanks :)
Put keyboard.nextLine() after this line:
int answer=keyboard.nextInt();
This is a common problem that usually happens when you use nextLine() method after nextInt() method of Scanner class.
What actually happens is that when the user enters an integer at int answer = keyboard.nextInt();, the scanner will take the digits only and leave the new-line character \n. So you need to do a trick by calling keyboard.nextLine(); just to discard that new-line character and then you can call String input = keyboard.nextLine(); without any problem.