Fixing the bad operand types error in Java can be frustrating if you aren’t experienced in debugging. Follow this guide, and you will learn how to make your code compile successfully again.

Common Bad Operand Types Error In Java

“&” Operator

Here is a snippet demonstrating the first common cause of this error. Its purpose is to check the parity of an integer (whether it is odd or even).

One way to do so is to carry out a bitwise AND operator between “1” and the target number. The operation should yield “0” as the result when the number is even and “1” when odd. Here is how many beginners would implement that algorithm in Java:

public class MyClass {
    public static void main(String args[]) {
      int x = 9;
      if( x & 1 == 1 ){
          System.out.println(x + ” is an odd number.”);
      }
      else{
          System.out.println(x + ” is an even number.”);
      }
    }
}

The compiler would give this warning:

/MyClass.java:5: error: bad operand types for binary operator ‘&’
      if( x & 1 == 1 ){
            ^
  first type:  int
  second type: boolean

Why Did The Error Happen?

The “&” operator is commonly used when processing bits of values or binary digits. 

However, the “&” operator has lower precedence than “==” (the equality operator) in Java. The compiler followed this rule and processed the “==” operator in the if condition first. The result was a true boolean value.

Now you can easily see why the bad operand types error occurred.

The operands of the “&” operator must be one of the integer types in Java: byte, char, short, int, and long. Here we had one int (9) and one boolean (true) value instead, prompting the Java compiler to stop.

How Can We Fix That Error?

From the clue above, the obvious solution is to tell the compiler to run the “&” operator first. You can override the precedence rules by adding parentheses:

The program should produce the correct answer now:

“&&” Operator

Putting the wrong type of operands into the sides of Java operators happens more than you might think. Likewise, the following fragment won’t run as expected:

public class MyClass {
    public static void main(String args[]) {
      int x = 8 + 9;
     
      if((x * 7) && (x > 25)){
          System.out.println(“True”);
      }
      else{
          System.out.println(“False”);
      }
    }
}

Output:

/MyClass.java:5: error: bad operand types for binary operator ‘&&’
      if((x*7) && (x > 25)){
              ^
  first type:  int
  second type: boolean

Remember that the conditional AND operator “&&” takes only boolean values as operands. But the “x * 7” expression on the left side creates an int value.

You can solve this by modifying the code and making it return a boolean value instead. For example, the if statement like this would be compiled without error:

Output:

“==”, “<=”, And “>=” Operators

These equality and relational operators demand operands of the same type. You will encounter the bad operand types error in Java when not adhering to this rule.

Example:

public class MyClass {
    public static void main(String args[]) {
      int x = 7;
      String y = “7”;
      int z = Integer.parseInt(y)
     
      if(x == z){
          System.out.println(“x and y is equal.”);
      }
      else{
          System.out.println(“x and y is not equal.”);
      }
    }
}

The program above can be compiled due to this error:

/MyClass.java:6: error: bad operand types for binary operator ‘==’
      if(x == y){
          ^
  first type:  int
  second type: String

The “==” operator can’t compare an int with a String. You can try converting the y variable to an int to make the comparison valid:

public class MyClass {
    public static void main(String args[]) {
      int x = 7;
      String y = “7”;
      int z = Integer.parseInt(y);
     
      if(x == z){
          System.out.println(“x and y are equal.”);
      }
      else{
          System.out.println(“x and y are not equal.”);
      }
    }
}

The code should run without issue:

Conclusion

The bad operand types error in Java happens when you somehow provide operators with the wrong type. This issue is easy to fix, however, because the compiler should give a detailed warning telling you where your code has gone wrong.