Problem with very simple program

[iratus]

Hello everybody

I am very new to java and my professor have given me an exercise for the weekend.

The exercise says the following things

Write a program which reads an arbitrary amount of characters. As the termination criterion use the input of a dot. (Realization once with the while-loop and once with the repeat-loop.)

I tried to make it (with simple commands that i know), but it doesn't complies it because I have some mistakes and I don't know how to correct them.

Please help me if you can

code

class kastora {

public static void main(String[] args) {

char word='a';

while ( word!=".") {

System.out.println("Enter a Character:");

word = System.in.read();

System.out.print("You entered ");

System.out.println(word);
}
}
}

[Zack]

but it doesn't complies it because I have some mistakes and I don't know how to correct them.

Please paste the full errors, including line numbers, here.

I suspect it has to do with your code word!="a". word is a char type and "a" is a String type.

[j2me64]

Hello everybody

I am very new to java and my professor have given me an exercise for the weekend.

The exercise says the following things

Write a program which reads an arbitrary amount of characters. As the termination criterion use the input of a dot. (Realization once with the while-loop and once with the repeat-loop.)

I tried to make it (with simple commands that i know), but it doesn't complies it because I have some mistakes and I don't know how to correct them.

Please help me if you can

code

class kastora {

public static void main(String[] args) {

char word='a';

while ( word!=".") {

System.out.println("Enter a Character:");

word = System.in.read();

System.out.print("You entered ");

System.out.println(word);
}
}
}

1. first of all the method read() returns a int. so define your word as type int.
2. the method read() throws an IOException, so you must catch this exception.
3. adapt all your assignment and operations for word to be a type int.

[iratus]

Hello again

I can not copy the errors cause i am working with CMD (not eclipse)---sorry

Questions

1) so if i define the 'word' as int will be able somebody to type characters ???
2) I don't really know what is an exception and IOException (too newbie) !!!

[JosAH]

Hello again

I can not copy the errors cause i am working with CMD (not eclipse)---sorry

Yes you can: right click the title bar and see what shows up.

kind regards,

Jos

[iratus]

Microsoft Windows [Version 6.1.7600]
Copyright (c) 2009 Microsoft Corporation. All rights reserved.

C:\Users\FUCK-OFF>cd C:\Master_Geomatics\Semester_1\Java\excersises\Arr ay

C:\Master_Geomatics\Semester_1\Java\excersises\Arr ay>C:\Master_Geomatics\Semeste
r_1\Java\JAVA_COMPILER\jdk1.6.0_22\bin\javac kastora.java
kastora.java:7: incomparable types: char and java.lang.String
while ( word!=".") {
^
kastora.java:11: possible loss of precision
found : int
required: char
word = System.in.read();
^
2 errors

C:\Master_Geomatics\Semester_1\Java\excersises\Arr ay>




Thanks it worked !!!!

[Fubarable]

As usual the error message is telling you what is wrong:

kastora.java:7: incomparable types: char and java.lang.String
while ( word!=".") {

word is a char and "." is a String and the compiler is telling you that you can't compare the two with !=. Better to compare word with another char such as '.'.

Nice subdirectory name by the way. ;)

[iratus]

You confused me a little bit !!!

Can you explain this with one example please ???

Thanks

[Fubarable]

You confused me a little bit !!!

Can you explain this with one example please ???

Thanks

One example coming up:

public class Fu4 {
   public static void main(String[] args) {
      char word = 'z';
      
      if (word != 'a') {
         System.out.println("word is not 'a'");
      }
      
      if (word != "a") {
         System.out.println("this won't compile since word is not a String but rather a char");
      }
   }
}

Again the problem above and with your code is that "a" is not a char but a String, and you can't compare a char with a String but instead need to compare a char with a char.

¿Comprendes?

[ST34LTH]

instead of

char word = "a"

try

String word = "a"

I think, im not a pro so sorry if its wrong ;)

[iratus]

So the dot '.' is a string or a char ???

[Fubarable]

One other problem is that you need to use a Scanner or a BufferedReader to get user input and shouldn't directly get user input from System.in.

[ST34LTH]

ive tried it out and i came up with this

import java.util.Scanner;
public class Kastora 
{
	

		public static void main(String[] args) 
		{
		Scanner UserInput = new Scanner(System.in);
		
		String word = "a";
		String dot = ".";
		
		
		
		

		
		while (word != ".")
		{
			if (word.equals(dot))
			{
				System.out.println("Invalid Entry");
				break;
			}
			System.out.println("Please enter next character");
			
			word = UserInput.next();
			
			System.out.println("You entered " + word);
			
		}

		}
		 
}

You cant use

word != "."

to comapre letters I think you can only use it for numbers, since it will carry on even if you enter "."

I just dont know how to put it in the while condition

[Fubarable]

So the dot '.' is a string or a char ???

The quotes are what matters for Java:
'.' is a char
"." is a String.