override equals

[abalfazl hossein]

I wrote a class myself , I want to override equals method in it.

public class equalsTest {
    int number;
    public equalsTest (int i){
    number= i;
    }
    public boolean equals(Object obj){
         if (this == obj){
              equalsTest a=(equalsTest)obj;
              if (a.number == this.number)
                  return true;
              else
                  return false;
         }if (!(obj instanceof equalsTest))
              return false;

    
    }
 }

public class Main {
	public static void main(String[] args) {
        equalsTest e = new equalsTest(10);
		equalsTest f = new equalsTest(10);
        System.out.print( e.equals(f));
       }
}

error is missing return statement.

Did I override equals correct?

[fred rosenberger]

Clearly if you get an error while compiling, you did not do something correct. Your primary error here is that there are ways that you can go through your code and never hit a return statement.

What happens if 'this' does NOT equal 'obj', and your obj IS an instance of equalsTest?

you'd skip your first if statement, skip the second one, and have nothing to return.

[Rahul Sudip Bose]

you are getting the missing return statement. Why ? If for some reason the execution reaches the last if(-----) AND the condition inside that if evaluates to false then no value is returned. Suppose there was an else with the last if() and it returned a boolean, that error will go.

PS : Please give a short description of your code. It is easier to go through the code after reading some info rather than figuring out on my own.

Here is a remedy for such a situation :

public boolean equals(Object obj){  

boolean result = true; // or false as per your need

         if (this == obj){  
              equalsTest a=(equalsTest)obj;  
              if (a.number == this.number)  
                  result= true;  
              else  
                  result = false;  
         }if (!(obj instanceof equalsTest))  
              result = false;
return result;  

That should make the compiler happy.

[Rob Spoor]

abalfazl hossein wrote:

         if (this == obj){
              equalsTest a=(equalsTest)obj;
              if (a.number == this.number)
                  return true;
              else
                  return false;

If obj == this then a == this and a.number == this.number per definition. You can replace the entire block by this:

if (this == obj) {
    return true;
}

The rest of the functionality, testing for number equality, should be done if the instanceof operator returns true - the part you're missing right now.

[David Byron]

or simply

return this == obj;

[abalfazl hossein]

public class equalsTest {
    int number;
    public equalsTest (int i){
    number= i;
    }
    public boolean equals(Object obj){
         if (this == obj ){
            
              equalsTest a=(equalsTest)obj;
              if (a.number == this.number)
                  return true;
         } else
                  return false;
         }if (!(obj instanceof equalsTest))
              return false;

    
    }
 }

At first I must be sure these objects has same type, Then check their fields

if this == obj , then equalsTest a=(equalsTest)obj; if a.number == this.number return true. Right?

But it has error, equal method does not check type of its parameter.for this line:

public boolean equals(Object obj)

and has error for this:

if (!(obj instanceof equalsTest)

[Rob Spoor]

David Byron wrote:or simply

return this == obj;

Which completely ignores checking for actual equality. You might as well not override equals and inherit the method from Object, which does the same.

Abalfazl, I think you want something like this:

public boolean equals(Object obj) {
    if (this == obj) {
        return true;
    }
    if (!(obj instanceof equalsTest)) {
        return false;
    }
    // you now know 100% guaranteed that obj instanceof equalsTest so you can safely cast
    equalsTest a=(equalsTest)obj;
    // do whatever you need
}