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 01202010, 04:55 AM #1Member
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Count number of digits in string using scanner
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;
}
}
 01202010, 05:05 AM #2Member
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I think you need numberOfDigits2() with the () after it.
 01202010, 05:08 AM #3Member
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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
 01202010, 05:11 AM #4Member
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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
 01202010, 05:13 AM #5
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Java Code: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); }
 01202010, 05:17 AM #6Member
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 01202010, 05:20 AM #7
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Java Code:class NumberOfDigits2
Java Code:System.out.print("The Number Of Digits Is " + NumberOfDigits2);
Java Code:System.out.print("The Number Of Digits Is " + 100);
 01202010, 05:21 AM #8Member
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n doesn't have a value?That might be a problem to.
 01202010, 05:24 AM #9
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Yes, that's the second mistake our OP did in his code. Lets solve it one by one.
 01202010, 05:28 AM #10Member
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Ok, i changed the method NumberOfDigits2() to NumberOfDigits2;
and now i recieve this error on compile...
 01202010, 05:29 AM #11Member
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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
 01202010, 05:30 AM #12
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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.
 01202010, 05:31 AM #13
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That's why it's good practice to use intend in coding.
Java Code: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; } }
 01202010, 05:32 AM #14Member
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Methods have to have these () after them. and n needs a value so try something like this.
Java Code: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; } }
 01202010, 05:33 AM #15Member
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That's why it's good practice to use intend in coding.
 01202010, 05:33 AM #16Member
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I see, so now no matter what string is entered it will always print 100
 01202010, 05:34 AM #17
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 01202010, 05:34 AM #18Member
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 01202010, 05:36 AM #19
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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.
 01202010, 05:36 AM #20Member
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