Count number of digits in string using scanner

[wendysbiggy]

can someone please let me know what I need to change and why. I have been looking at it for a little while and cannot figure it out on my own. Thank You.

import java.util.Scanner;

class NumberOfDigits2
{
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
System.out.print("Please enter a Number!");
in.nextLine();

System.out.print("The Number Of Digits Is " + NumberOfDigits2);
}
public static int NumberOfDigits2()
{
int n;
if(n<0) n = -n;
int c = 1;
while(n > 9)
{
c++;
n = n/10;
}
return c;
}
}

[justin1980]

I think you need numberOfDigits2() with the () after it.

[wendysbiggy]

My apologies i should have listed what the error was on compile... here it is...

NumberOfDigits2.java:12: cannot find symbol
symbol : variable NumberOfDigits2
location: class NumberOfDigits2
System.out.print("The Number Of Digits Is " + NumberOfDigits2);
^
1 error

[wendysbiggy]

Ok, i changed the method NumberOfDigits2() to NumberOfDigits2;
and now i recieve this error on compile...

NumberOfDigits2.java:24: return outside method
return c;
^
1 error

----jGRASP wedge2: exit code

[Eranga]

class NumberOfDigits2
{
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
System.out.print("Please enter a Number!");
in.nextLine();

System.out.print("The Number Of Digits Is " + NumberOfDigits2);
}

Your class name, use it in the print() method. It's not legal.

[wendysbiggy]

class NumberOfDigits2
{
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
System.out.print("Please enter a Number!");
in.nextLine();

System.out.print("The Number Of Digits Is " + NumberOfDigits2);
}

Your class name, use it in the print() method. It's not legal.

Sorry, I dont understand what you mean. can you explain please. Thanks

[Eranga]

class NumberOfDigits2

This is the class definition, NumberOfDigits2 is the name of your class. Hope you know that well.

System.out.print("The Number Of Digits Is " + NumberOfDigits2);

This is the system class' print() method. You pass the class name to there. It's not a valid argument to print() method. For a min, on that print() method line change as follows and see.

System.out.print("The Number Of Digits Is " + 100);

[justin1980]

n doesn't have a value?That might be a problem to.

[Eranga]

Yes, that's the second mistake our OP did in his code. Lets solve it one by one.

[justin1980]

Ok, i changed the method NumberOfDigits2() to NumberOfDigits2;
and now i recieve this error on compile...

No, don't change it to numberOfDigits; change the method to a new name like digits() and have the () after both

[wendysbiggy]

Ok so i wrote...
System.out.print("The Number Of Digits Is " + 100);
}
/*public static int NumberOfDigits2;
{
int n;
if(n<0) n = -n;
int c = 1;
while(n > 9)
{
c++;
n = n/10;
}
return c;
}*/
}

and it will print "The Number of Digits is 100"...
if i uncomment the last method i recieve "return outside method" error

[Eranga]

No need to change the method name all the time, except if it's constructor. Actual error we have here is the user the class name as a parameter and variable n is not initialize.

[Eranga]

if i uncomment the last method i recieve "return outside method" error

That's why it's good practice to use intend in coding.

import java.util.Scanner;

class NumberOfDigits2 {
	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		System.out.print("Please enter a Number!");
		in.nextLine();

		System.out.print("The Number Of Digits Is " + 100);
	}

	public static int NumberOfDigits2() {
		int n = 0;
		if(n<0) {
 			n = -n;
		}

		int c = 1;
		while(n > 9) {
			c++;
			n = n/10;
		}
		return c;
	}
}

[justin1980]

Methods have to have these () after them. and n needs a value so try something like this.

       

Scanner in = new Scanner(System.in);
System.out.print("Please enter a Number!");
in.nextLine();

System.out.print("The Number Of Digits Is " + Digits2());

 
}
  public static int Digits2()
{
int n = 77;
if(n<0) n = -n;
int c = 1;
while(n > 9)
{
c++;
n = n/10;
}
return c;

}
}

That still will not count the letters in a string or whatever you are trying to do, but I think there wont be any errors. What is it exactly you are trying to do?

[justin1980]

That's why it's good practice to use intend in coding.

What does intend mean?

[wendysbiggy]

I see, so now no matter what string is entered it will always print 100

[Eranga]

That still will not count the letters in a string or whatever you are trying to do, but I think there wont be any errors. What is it exactly you are trying to do?

Actually our OPs code not doing anything yet. He tries to count number of digits in the user specifies number, but not doing it in proper way.

[wendysbiggy]

What does intend mean?

I believe he meant Indent! haha

[Eranga]

I see, so now no matter what string is entered it will always print 100

That's the different thing pal. I just ask you to do it point out the error is.

First you get the user input and seem that you tried to count the number of digit in a different method. So you have to call the relevant method with the correct parameter. That method return value print to the console. That what you've to do basically.

[justin1980]

Originally Posted by justin1980 View Post
What does intend mean?
I believe he meant Indent! haha

Oh duh, I'm sorry!