Is there a website where I can run Java or C++ code online that lets me use cin in c++? Codepad and many others just let you see your program but don't let the user provide input, and I wanted to know if I could find a website that allows that.
Ideone doesn't seem to do the input right : This is supposed to be a game where you enter a word then the letter changes places, and then you need to figure out the word. The code is correct but the website doesnt read the cin right ! http://ideone.com/0PmDX
Ideone does just that this is also in the Info section of the c++ tag.
In your provided code: http://ideone.com/0PmDX you call return main(); which is undefined behaviour and is causing your infinite loops in your code. Try return 0 instead.
Code: http://ideone.com/ymXeH
Related
The standard JVM method for reading a password from the command line without showing it is java.io.Console.readPassword(). This, however, shows nothing while the user is typing; users accustomed to graphical programs will expect symbols such as "•" or "*" to appear in place of the characters they type. Naturally, they will also want backspacing, inserting, and so on to work as normal, just with all the characters being operated on replaced with the same symbol.
In 2019, is there a generally accepted JVM procedure for showing "*******" when the user types "hunter2" in a console application? Can this even be done properly without a GUI? A 2011 SO question on the topic got an answer linking to this article on the topic; can we do better nowadays than the rather elaborate solution shown therein?
(I happen to be using Kotlin as my language of choice, so a Kotlin-specific solution will satisfy if there is one.)
hunter2? Wow. Reference acknowledged.
There is no easy way. The primary problem is that the standard System.in doesn't give you any characters at all until the user has pressed enter, so there's no way to emulate it (if you try to read char-for-char from System.in and emit a * every time a key is pressed, that won't work).
The lanterna library at https://github.com/mabe02/lanterna can do it, though. If you want to emulate it, it's.. very complicated. It has branching code paths for unix and windows. For example, on unix, it uses some hackery to figure out what tty you're on, and then opens the right /dev/tty device. With lanterna, writing this yourself would be trivial.
It's that or accept Console.readPassword()'s blank nothingness, really. Or, write a web interface or a swing/awt/javafx GUI.
I think answer to your question can be found here in stackoverflow itself.
please see this:
masking-password-input-from-the-console-java
sample code from there:
import java.io.Console;
public class Main {
public void passwordExample() {
Console console = System.console();
if (console == null) {
System.out.println("Couldn't get Console instance");
System.exit(0);
}
console.printf("Testing password%n");
char passwordArray[] = console.readPassword("Enter your secret password: ");
console.printf("Password entered was: %s%n", new String(passwordArray));
}
public static void main(String[] args) {
new Main().passwordExample();
}
}
hope this is helpful. :)
I have the following piece of code called Code1.
http://pastebin.com/tc0Vd8xh
When I run this, the sketch does not work.
However when I replace "i=+50" for "i = i + 50" the code works.
My question is why the "i=+50" bit does not work?
As far as I know "i=+50" is proper Java and Processing is based on Java.
I tried to Google about "i=+50" but Google does not process non-alphanumeric characters.
So I came here and I searched in previous questions before asking here. Anyone, any idea why "i=+50" does not work?
The statement i=+50 is the assignment of positive 50 to i. That is why it compiles, but doesn't add 50 to i on each loop. As #RoelHarbers and #ByoTic mentioned, you actually want i += 50
You're using =+, which is not a java operator (or an operator in any other language I know of)
The proper syntax is:
i+=50
Because it's i+=50 and not i=+50.
i =+ 50 is not going to do what you want, it is going to initialize i with 50. Instead, use i+=50, this going to add the 50 to whatever value that i holds.
My class got a Java programming project today in class and the program we are using is jGrasp, I know that I can handle the pieces of this project except for one aspect.
The project requires the user to enter data into the program (i.e. answer a question), but for this Semester we are not using GUIs so I can't create a GUI for the user to input the data or answer.
I'd like to know how could the user enter data into jGrasp without using a GUI?
Thanks for taking the time to read this post, and I'd greatly appreciate any help you could give me.
I hope I'm helpful :) it's funny that I want to know about the same thing in GUI & you want to know without it :D ....well i can solve your matter I believe :)
From the starting,
Example:
import java.util.*;
public class Assignment //class name
{
static Scanner stdIn = new Scanner(System.in);
public static void main(String[]args)
{
int A, B; //Intialize variables as normal
System.out.print("Enter a question you want to ask "); //Between the double quotes " " you type anything you want to type :)
A = stdIn.nextInt();
: //
: // complete your program
System.out.println("Add the statement for giving the result "+B); // "+B" for inserting the answer along with your result statement
}
}
Note: In the statements,
System.out.print();
System.out.println();
These statement work similarly.... just the difference is statement with "print" is displayed on the same line & Statement with "println" is displayed in next line....when you will try you will understand ...just for avoiding confusion why I have written so I'm mentioning it :)
I hope this helps...tczz :)
I'm in the final stages of testing before I release the alpha version of a program I am writing. This may be a silly question but I just can't seem to figure it out. I understand what drjava is telling me, that I'm missing a variable, but I also don't understand because I never made a variable under the name "()". I'm not even sure you can set any type of variable to a open-close parenthesis. Anyways I was testing and while it works, it doesn't the way I want it to. I entered into the scanner "Mr. B." without the quotes of course. The program did not print the B. I'm thinking it might be the space in between Mr. and B, because other inputs with a space did the same. I can not release a version of my program knowing there is a GIANT glitch in the code. I'm wondering why, and I tried to fix it by changing ownersname.next(); to ownersname.nextLine and ownersname.next and ownersname.nextScanner and ownername.nextScanner. This is where the error comes in, when it says it can't find the variable until I change it back to it's original code, which is below.
Scanner ownersname = new Scanner(System.in);
String sownersname = ownersname.next();
System.out.println(sownersname + "? That is a nice name.");
I'm in the final stages of testing before I release the alpha version of a program I am writing.
You're creating a professional application? Please do tell us more about this.
I understand what drjava is telling me, that I'm missing a variable, but I also don't understand because I never made a variable under the name "()". I'm not even sure you can set any type of variable to a open-close parenthesis.
When posting questions here, if you have an error message from the compiler, please post the entire error message with your question. Don't paraphrase it. And indicate by obvious comment in your code, i.e., // ****** error here ***** where the error is occurring.
Anyways I was testing and while it works, it doesn't the way I want it to. I entered into the scanner "Mr. B." without the quotes of course. The program did not print the B. I'm thinking it might be the space in between Mr. and B, because other inputs with a space did the same.
Don't use Scanner#next() which gets only the next token -- the next word before reaching whitespace (here, Mr.), and will not get the rest of the text on the line. Instead use Scanner#nextLine() which gets you the whole line.
For example:
Scanner ownersname = new Scanner(System.in);
// String sownersname = ownersname.next(); // *** not this ***
String sownersname = ownersname.nextLine(); // *** but rather this ***
System.out.println(sownersname + "? That is a nice name.");
I can not release a version of my program knowing there is a GIANT glitch in the code.
Seriously, you're creating a professional application? I'm not yet at that stage, which is why I ask.
I'm wondering why, and I tried to fix it by changing ownersname.next(); to ownersname.nextLine and ownersname.next and ownersname.nextScanner and ownername.nextScanner. This is where the error comes in, when it says it can't find the variable until I change it back to it's original code, which is below.
I'd be curious to see your nextLine() method attempt, because that is the solution. Perhaps you were trying to call the method without using the method parenthesis.
I also assume that you're familiar with the Java API and have looked up the Scanner entry for it. If you did, you would see right away that there is no nextScanner() method for this class. This is one reason I have to wonder about your making a professional application at your stage. Again, I don't feel that I'm at the stage yet to create one yet, so please don't take this as an insult, just a curiosity.
As you can probably understand from the question itself, I'm new to Java.
I was given an exercise to write a Java program which receives a character, prints it and the next character in the Unicode table.
Now, I have the solution to this exercise:
public static void main(String[] args){
char c = args[0].charAt(0);
char c1 = (char)(c + 1);
System.out.println(c + "\t" + c1);
}
I understand basic idea of this code, but I'm trying to run this code in Eclipse I get an annoying error:
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 0
at MainClass.main(MainClass.java:9)
Note: I have yet to run a Java program that actually receives something as a parameter so I guess it's a stupid beginners' mistake... Here is the full code that I tried to compile in Eclipse:
public class MainClass {
/**
* #param args
*/
public static void main(String[] args){
char c = args[0].charAt(0);
char c1 = (char)(c + 1);
System.out.println(c + "\t" + c1);
}
}
Thanks in advance
Select "Run -> Run Configurations" from the menu.
Search for you project in the list on the left and select it.
Select the "Arguments" tab on the right.
Write the argument you want to pass to the programm in "Programm arguments".
Click "Run"
Right click on your java file in project explorer of your eclipse. Then Run As> Run Configuration
Then you will get a window. Like-
Click on Arguments Tabs, and then write some text there, may be a character.
And then Click on Apply button and Run Button.
The default run configuration in Eclipse runs a Java program without any arguments, hence the ArrayIndexOutOfBoundsException. Your code is trying to get first element of the args array when there aren't any!
You can edit the run configuration to provide the arguments to run your program with. Then it should not throw this exception.
However, a good practice is to check the size of array before accessing it's elements, more so when the array is coming as an argument from outside of your code.
This is a great question with some very good answers. I would like to add some pointers about how to debug your own program. Debugging is as important (if not more important) than writing code.
For one thing, Eclipse has some great debugging features. You can use this debugger to find problems in your code. I suggest that you learn how to use it. In particular, you can set watches for variables to see what value they have as you step through the execution of your code.
Alternatively, you can add calls to System.out.println() to print out the values of any variables. For example, adding the following line at the beginning of your code might help you narrow down the problem:
System.out.println(args[0]);
This would also give an ArrayIndexOutOfBoundsException if no command-line arguments are given. Then you could do something like
System.out.println(args.length);
which would print out 0. This then gives you a clue as to where the problem is.
Of course, even when you get to this point, you still might not know how to solve the problem. This is where sites like StackOverflow come in handy.
Good luck with your Java experience. Please come back when you need more help.
If your Run Configurations are in place (as already shown in above answers):
Shortcut to Run a class is:
Ctrl + F11