Tis is the program to check for anagram.What is the error at line 12 .

Question

public class anagram {
    public static void main(String [] args){
        String a="tanu";
        String b="anut";

      //  boolean isanagram=true;
        if(a.length() == b.length()){
            String[] s1=a.split("");
            for(int i=0;i<a.length();i++)

                for(int j=0;j<b.length();j++)
          /*Line12*/    if(s1[i]=b.charAt(j)){
                        System.out.println("anagram");
                    }
                    else
                        System.out.println("not an anagram");
                    }
           }}

Answer 1

When you are checking with strings to access the elements you need to follow this syntax 👇

name of the string.charAt(name of the iterator)

And for arrays 👇

Name of array[name of iterator]

Note :- This is valid when you are using loops for finding elements

Answer 2

if(s1[i]==b[j])

Use this please and then tell me if you don't get the expected output

Answer 3

Atul [Inactive]
this too is producing error

Answer 4

Tanu Jain My bad use a.charAt(i)==b.charAt(j)

Use this

Answer 5

Thanks
This worked but don't know what was the problem in first code