Why is there a arr cannot be resolved error

Question

0.00/5 (No votes)

See more:

<pre>package dsa;
import java.util.ArrayList;
public class PairSumR {
	public static boolean isValidPair(int[] arr, int n, int k, int m) 
	{
	ArrayList<Integer> list = new ArrayList<Integer>();
		int rem= Integer.MAX_VALUE;
	for(int i= 0; i<arr.length;i++){
		
		 rem=arr[i]%k;
		 if(list.contains(m-rem)) return true;
		if(!list.contains(rem)) list.add(rem);
	} 
	
	return false;
	}
	
	public static void res() {
		System.setOut(isValidPair(arr, 5, 3, 2));
	}

	public static void main(String [] args) {
		
	
		
		int[] arr= {3,5,2,6,4};
		PairSumR.res();
		
	}
}

What I have tried:

i have passed the array in the method call and provided it as a parameter

error shown :
arr cannot be resolved to a variable

Posted 20-Mar-23 10:01am

Shivam Siddharth

Updated 20-Mar-23 22:48pm

v3

Add a Solution

2 solutions

Solution 2

Look at the call to the res method in main:

Java

public static void main(String [] args) {
    int[] arr= {3,5,2,6,4};
    PairSumR.res(); // calling res without any parameters

Now look at the call from res to isValidPair

Java

public static void res() {
    System.setOut(isValidPair(arr, 5, 3, 2)); // trying to pass the variable arr
}

But arr does not exist anywhere in res so the compiler rejects the code. You need to rework the definition of res and the call to it in main:

Java

public static void res(int[] arr) {  // the variable arr is received from the caller
    System.setOut(isValidPair(arr, 5, 3, 2));
}
public static void main(String [] args) {
    int[] arr= {3,5,2,6,4};
    PairSumR.res(arr); // calling res with the correct parameter

Also, I am not sure why you are using System.setOut here, as it expects a valid stream reference not a boolean.

Posted 20-Mar-23 22:48pm

Richard MacCutchan

Updated 20-Mar-23 23:53pm

v3

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

OriginalGriff · Accepted Answer · 2023-03-20T10:08:00

Solution 1

arr is both a parameter to isValidPair and a local variable in main.
As such, it is only accessible inside those two methods - and because one is a parameter to a method, what's in it does not have to be the same array as declared in main as any array of integers can be passed in.

So it doesn't exist at all inside the res method - and that is what the error is telling you. Probably, you need it to be a parameter to the res method as well as the isValidPair method.

Posted 20-Mar-23 10:08am

OriginalGriff

Comments

Shivam Siddharth 20-Mar-23 16:55pm

why doesn't it consider it as an input since value is being given and its type is mentioned in the declaration

OriginalGriff 20-Mar-23 18:12pm

Because it is out of scope.
Variables only exist while they are in scope - the curly brackets within which they are declared. Once they go out of scope they no longer can be accessed.

Shivam Siddharth 21-Mar-23 4:16am

your thing worked ,but I am not using it outside its scope just passing the values like when called normally & how other variables are being accepted n and k

OriginalGriff 21-Mar-23 5:18am

Because they are parameters as well - so they exist inside the isValidPair method, and not outside it.

Think of each method as a separate room in your house: if you want to get some milk, you need to go to the kitchen and open the fridge - it's not accessible from your bedroom! Similarly, if you want a shower, you have top go to the bathroom - there is no source of hot water in your lounge.

Scope matters in software as in real life: you need to be "in the right place" to use something.