CODEWITHSUNDEEP
javaarraysjava.util.scanner
Swarathesh Addanki
Swarathesh Addanki

Reputation: 451

reading a string from scanner and converting it to character array

here is a code i wrote in java, I'm still a beginner in a java so if this is a noob mistake please rectify it and also provide alternative solutions for reading the string and converting it to char array

import java.util.Scanner;
import java.io.*;

class TestClass {
public static void main(String args[] ) throws Exception {
   Scanner s = new Scanner(System.in);
   String str = s.nextLine();
   char[] c = str.toCharArray(str);
   int x=0,y=0;
   for(int i=0;i<=str.length;i++)
   {
      if(c[i]=='L')
        {
            x=x+1;
        }
    else if(c[i]=='R')
        {
            x=x-1;
        }
        else if(c[i]=='U')
        {
            y=y-1;
        }
        else if(c[i]=='D')
        {
            y=y-1;
        }
   }
    System.out.println(x+""+y);
}
}

I'm getting the following error

10: error: method toCharArray in class String cannot be applied to given types;

no arguments String actual and formal argument lists differ in length 12: error: cannot find symbol

variable length variable str of type String

Upvotes: 0

Views: 2269

Answers (3)

z7r1k3
z7r1k3

Reputation: 737

Combining answers (I don't take credit for these)

You put an argument in a function that doesn't take any:

char[] c = str.toCharArray(str);

Just use

char[] c = str.toCharArray();

-by Arc676

And also note that

it's str.length() not str.length

-by TomN

Both are correct answers

Also, I noticed that you have (I do take credit for these)

for(int i=0;i<=str.length;i++)

It should be '<', not '<=' to avoid an out of bounds exception

for(int i=0;i<str.length();i++)

Also, just a tip:

System.out.println(x+""+y);

I think you missed putting a space in those quotation marks.

Upvotes: 0

Ulises
Ulises

Reputation: 9625

Method str.toCharArray(str); doesn't exist. toCharArray takes no arguments. Replace with str.toCharArray()

Upvotes: 0

Arc676
Arc676

Reputation: 4465

You put an argument in a function that doesn't take any:

char[] c = str.toCharArray(str);

Just use

char[] c = str.toCharArray();

For more info, see the String documentation

Upvotes: 2

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