Java error bad operand types for binary operator

This article introduces bad operand types error in Java.
  1. Bad Operand Types for the & Operator in Java
  2. Bad Operand Types for the && Operator in Java
  3. Bad Operand Types for the == Operator in Java
  4. Bad Operand Types for the <= Operator in Java

Fix the Bad Operand Types Error in Java

This tutorial introduces the bad operand types error for binary operators in Java.

A binary operator is an operator which requires two operands. Operators such as arithmetic operators and relational operators are called binary operators.

Operators play a vital role in programming, and sometimes, a binary operator can give a bad operand types error due to improper use. The bad operand types error is a compile-time error when both types of operands are incompatible.

For example, we will get this error when comparing a string with an integer. This article will look at different examples of how this error occurs and how to resolve it.

Sometimes, the order of precedence of operators can also lead to incompatible operator types and result in an error to the console.

Bad Operand Types for the & Operator in Java

Let’s first understand the most common case where bad operator error can occur in Java code. Look at an example code below:

public class MyClass {
    public static void main(String args[]) {
      int x = 43;
      if( x & 21 == 1){
          System.out.println("if block executing");
      }
      else{
          System.out.println("else block executing");
      }
    }
}

Output:

MyClass.java:4: error: bad operand types for binary operator '&'
      if( x & 21 == 1){
            ^
  first type:  int
  second type: boolean
1 error

This error occurs because the precedence of the == (equals) operator is more than that of the & operator. This leads to the 21 == 1 evaluation, which gives us a boolean value.

Now, notice the & has one integer operand and one boolean. Since both the operators are of different types, the & operator cannot work, so we get an error.

We will use parenthesis to indicate that x & 21 needs to be evaluated first to resolve this error. Look at the modified code below:

public class MyClass {
    public static void main(String args[]) {
      int x = 43;
      if( (x & 21) == 1){
          System.out.println("if block executing");
      }
      else{
          System.out.println("else block executing");
      }
    }
}

Output:

Bad Operand Types for the && Operator in Java

Similarly, if you are working with logical && (and) operator, then you may face bad operand types error in some cases like the below example code:

public class MyClass {
    public static void main(String args[]) {
      int x = 43;
      if((x > 10) && (x*5)){
          System.out.println("if block executing");
      }
      else{
          System.out.println("else block executing");
      }
    }
}

Output:

MyClass.java:4: error: bad operand types for binary operator '&&'
      if((x > 10) && (x*5)){
                  ^
  first type:  boolean
  second type: int
1 error

This error occurs because the && operand expects two boolean operands.

Here, the expression x * 5 gives an integer value. Hence, the && operator here has an integer operand and gives us the bad operand types error.

To resolve this error, we will modify this code such that x * 5==21 which returns a boolean value. Look at the modified code below:

public class MyClass {
    public static void main(String args[]) {
      int x = 43;
      if((x > 10) && (x*5 == 21)){
          System.out.println("if block executing");
      }
      else{
          System.out.println("else block executing");
      }
    }
}

Output:

Bad Operand Types for the == Operator in Java

You may get the same error when working with the == equals operator. It can give a bad operator error if both the operands passed are of different types.

Look at the example below:

public class MyClass {
    public static void main(String args[]) {
      int x = 43;
      String y = "43";
      if(x == y){
          System.out.println("if block executing");
      }
      else{
          System.out.println("else block executing");
      }
    }
}

Output:

MyClass.java:5: error: bad operand types for binary operator '=='
      if(x == y){
           ^
  first type:  int
  second type: String
1 error

This error occurs because the operands of the == equals operator are of different types. One is a string, and the other one is an integer.

To resolve this error, we will have to convert one of them to get the same data types. If we convert integers into strings, the comparison will occur in lexicological order.

So, we will convert string to int. Look at the modified code below:

public class MyClass {
    public static void main(String args[]) {
      int x = 43;
      String y = "43";
      if(x == Integer.parseInt(y)){
          System.out.println("if block executing");
      }
      else{
          System.out.println("else block executing");
      }
    }
}

Output:

Bad Operand Types for the <= Operator in Java

Like the previous case example, the <= (less than equals to) operator can also give a bad operator types error when both the operands are of different types. Look at the example below:

public class MyClass {
    public static void main(String args[]) {
      int x = 43;
      String y = "43";
      if(x <= y){
          System.out.println("if block executing");
      }
      else{
          System.out.println("else block executing");
      }
    }
}

Output:

MyClass.java:5: error: bad operand types for binary operator '<='
      if(x <= y){
           ^
  first type:  int
  second type: String
1 error

To resolve this error, we will have to convert one of them to get the same data types. Look at the modified code below:

public class MyClass {
    public static void main(String args[]) {
      int x = 43;
      String y = "43";
      if(x <= Integer.parseInt(y)){
          System.out.println("if block executing");
      }
      else{
          System.out.println("else block executing");
      }
    }
}

Output:

Источник

In this post, we will learn how to resolve error bad operand types for binary operator &, bad operand types for binary operator &&, bad operand types for binary operator ==, and bad operand types for binary operator <= in java.

Many developers get confused with & bitwise AND operator and && logical AND operator. Both operators return true if all the conditions are true, if any of the given condition is false then they will return false.

When using boolean operands, the main difference between them is that && operator does not evaluate the next condition if the condition before it is false whereas & operator evaluates all conditions even if they are false.

First, we will produce the error before moving on to the solution.

Read Also: char cannot be dereferenced error in java

1. Bad operand types for binary operator &

Example 1: Producing the error by using if condition

We can easily produce this error by using & in the if condition as shown below:

public class JavaHungry {
    public static void main(String args[]) {
      int x=100;
      
      // Using & operator in if condition

if(x & 100 == 1)
      {
          System.out.println("inside if condition");
      }
      else 
      {
          System.out.println("inside else condition");
      }
    }
}

Output:
/JavaHungry.java:5: error: bad operand types for binary operator ‘&’
if(x & 100 == 1)
^
first type: int
second type: boolean
1 error

Explanation:

The cause of this error is due to the precedence of operators. == operator has higher precedence over & operator. As a result, 100==1 will be calculated first and return the boolean value. If you notice & operator has two operands now, one is int, and the other is boolean. Since both operands are different it will give the compilation error as shown above.

Solution:

The above compilation error can be resolved by using parenthesis properly.

public class JavaHungry {
    public static void main(String args[]) {
      int x=100;
      
      // Using & operator in if condition
      
    if((x & 100) == 1)
      {
          System.out.println("inside if condition");
      }
    else 
      {
          System.out.println("inside else condition");
      }
    }
}

Output:
inside else condition

2. Bad operand types for binary operator &&

Example: Producing the error by using if condition

Just like above, we will produce the error first before moving on to the solution.

public class JavaHungry {
    public static void main(String args[]) {
      int x=1000;
      

if( (x > 100) && (x/2))
      {
          System.out.println(x);
      }
    }
}

Output:

/JavaHungry.java:5: error: bad operand types for binary operator ‘&&’
if( (x > 100) && (x/2))
^
first type: boolean
second type: int
1 error

Explanation:

The cause of this error is that (x/2) is a numeric expression that will return an integer value. We all know && is the logical AND operator. So, it expects boolean values on both sides. If you look at the if condition now, && operator has two operands: one is boolean and the other is int. The same is also mentioned in the compilation error.

Solution:

The above compilation error can be resolved by converting the second operand into a boolean type

public class JavaHungry {
    public static void main(String args[]) {
      int x=1000;     
      
    if( (x > 100) && (x/2 > 0))
      {
          System.out.println(x);
      }
    }
}

Output:
1000

3. Bad operand types for binary operator ==

Example: Producing the error by using if condition

We will produce the error bad operand types for binary operator == first before moving on to the solution.

public class BadOperandTypes {
    public static void main(String args[]) {
      int x = 10;
      String y = "100";
      // Using == operator in if condition      
      if (x == y) 
      {
          System.out.println("Executing if block");
      }
      else {
          System.out.println("Executing else block");
      }
    }
}

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

Explanation:

The cause of this error is due to the operands passed are of different types. If you notice == operator has two operands now, one is int, and the other is String. Since both operands are different it will give the compilation error as shown above.

Solution:

The above compilation error can be resolved by converting one of the operands to the same data types. If we convert int to String then the comparison will occur in lexicological order. Below, we are converting String to int.

public class BadOperandTypes {
    public static void main(String args[]) {
      int x = 10;
      String y = "100";
      // Using == operator in if condition      
      if (x == Integer.parseInt(y))
      {
          System.out.println("Executing if block");
      }
      else {
          System.out.println("Executing else block");
      }
    }
}

Output:
Executing else block

4. Bad operand types for binary operator <=

Example: Producing the error by using if condition

We will produce the error bad operand types for binary operator <= first before moving on to the solution.

public class BadOperandTypes2 {
    public static void main(String args[]) {
      int x = 5;
      String y = "500";
      // Using <= operator in if condition      
      if (x <= y) 
      {
          System.out.println("Executing if block");
      }
      else {
          System.out.println("Executing else block");
      }
    }
}

Output:
/BadOperandTypes2.java:6: error: bad operand types for binary operator ‘<=’
              if (x <= y) {
                      ^
    first type: int
    second type: String
1 error

Explanation:

Just like above, the cause of this error is due to the operands passed are of different types. If you notice <= operator has two operands now, one is int, and the other is String. Since both operands are different it will give the compilation error as shown above.

Solution:

The above compilation error can be resolved by converting one of the operands to the same data types. Please find below the modified code:

public class BadOperandTypes2 {
    public static void main(String args[]) {
      int x = 5;
      String y = "500";
      // Using <= operator in if condition      
      if (x <= Integer.parseInt(y))
      {
          System.out.println("Executing if block");
      }
      else {
          System.out.println("Executing else block");
      }
    }
}

Output:
Executing if block

That’s all for today. Please mention in comments in case you are still facing the error bad operand types for binary operator in java.

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Содержание

  1. Fix the Bad Operand Types Error in Java
  2. Bad Operand Types for the & Operator in Java
  3. Bad Operand Types for the && Operator in Java
  4. Bad Operand Types for the == Operator in Java
  5. Bad Operand Types for the Operator in Java
  6. Java Hungry
  7. [Solved] bad operand types for binary operator in java
  8. Example 1: Producing the error by using if condition
  9. Explanation:
  10. Solution:
  11. 2. Bad operand types for binary operator &&
  12. Example: Producing the error by using if condition
  13. Explanation:
  14. Solution:
  15. 3. Bad operand types for binary operator ==
  16. Example: Producing the error by using if condition
  17. Explanation:
  18. Solution:
  19. 4. Bad operand types for binary operator
  20. Example: Producing the error by using if condition

Fix the Bad Operand Types Error in Java

This tutorial introduces the bad operand types error for binary operators in Java.

A binary operator is an operator which requires two operands. Operators such as arithmetic operators and relational operators are called binary operators.

Operators play a vital role in programming, and sometimes, a binary operator can give a bad operand types error due to improper use. The bad operand types error is a compile-time error when both types of operands are incompatible.

For example, we will get this error when comparing a string with an integer. This article will look at different examples of how this error occurs and how to resolve it.

Sometimes, the order of precedence of operators can also lead to incompatible operator types and result in an error to the console.

Bad Operand Types for the & Operator in Java

Let’s first understand the most common case where bad operator error can occur in Java code. Look at an example code below:

This error occurs because the precedence of the == (equals) operator is more than that of the & operator. This leads to the 21 == 1 evaluation, which gives us a boolean value.

Now, notice the & has one integer operand and one boolean. Since both the operators are of different types, the & operator cannot work, so we get an error.

We will use parenthesis to indicate that x & 21 needs to be evaluated first to resolve this error. Look at the modified code below:

Bad Operand Types for the && Operator in Java

Similarly, if you are working with logical && (and) operator, then you may face bad operand types error in some cases like the below example code:

This error occurs because the && operand expects two boolean operands.

Here, the expression x * 5 gives an integer value. Hence, the && operator here has an integer operand and gives us the bad operand types error.

To resolve this error, we will modify this code such that x * 5==21 which returns a boolean value. Look at the modified code below:

Bad Operand Types for the == Operator in Java

You may get the same error when working with the == equals operator. It can give a bad operator error if both the operands passed are of different types.

Look at the example below:

This error occurs because the operands of the == equals operator are of different types. One is a string, and the other one is an integer.

To resolve this error, we will have to convert one of them to get the same data types. If we convert integers into strings, the comparison will occur in lexicological order.

So, we will convert string to int. Look at the modified code below:

Bad Operand Types for the Operator in Java

Like the previous case example, the (less than equals to) operator can also give a bad operator types error when both the operands are of different types. Look at the example below:

To resolve this error, we will have to convert one of them to get the same data types. Look at the modified code below:

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Java Hungry

Java developers tutorials and coding.

[Solved] bad operand types for binary operator in java

Example 1: Producing the error by using if condition

Output:
/JavaHungry.java:5: error: bad operand types for binary operator ‘&’
if(x & 100 == 1)
^
first type: int
second type: boolean
1 error

Explanation:

The cause of this error is due to the precedence of operators. == operator has higher precedence over & operator. As a result, 100==1 will be calculated first and return the boolean value. If you notice & operator has two operands now, one is int, and the other is boolean. Since both operands are different it will give the compilation error as shown above.

Solution:

The above compilation error can be resolved by using parenthesis properly.

Output:
inside else condition

2. Bad operand types for binary operator &&

Example: Producing the error by using if condition

Just like above, we will produce the error first before moving on to the solution.

Output:
/JavaHungry.java:5: error: bad operand types for binary operator ‘&&’
if( (x > 100) && (x/2))
^
first type: boolean
second type: int
1 error

Explanation:

The cause of this error is that (x/2) is a numeric expression that will return an integer value. We all know && is the logical AND operator. So, it expects boolean values on both sides. If you look at the if condition now, && operator has two operands: one is boolean and the other is int. The same is also mentioned in the compilation error.

Solution:

The above compilation error can be resolved by converting the second operand into a boolean type

3. Bad operand types for binary operator ==

Example: Producing the error by using if condition

We will produce the error bad operand types for binary operator == first before moving on to the solution.

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

Explanation:

The cause of this error is due to the operands passed are of different types. If you notice == operator has two operands now, one is int, and the other is String. Since both operands are different it will give the compilation error as shown above.

Solution:

The above compilation error can be resolved by converting one of the operands to the same data types. If we convert int to String then the comparison will occur in lexicological order. Below, we are converting String to int.

Output:
Executing else block

4. Bad operand types for binary operator

Example: Producing the error by using if condition

The above compilation error can be resolved by converting one of the operands to the same data types. Please find below the modified code:

Output:
Executing if block

That’s all for today. Please mention in comments in case you are still facing the error bad operand types for binary operator in java.

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I am writing a BST Program. I get the error:

«Bad Operand Types for Binary Operator «>»

first type: java.lang.Object

second type: java.lang.Object»

This is the method where it gives me the error:

public void placeNodeInTree(TreeNode current, TreeNode t)                                                                    
{   
    if(current == null)
        current = t;
    else{
       if(current.getValue() > t.getValue()) 
            current.setRight(t);
       if(current.getValue() < t.getValue()) 
            current.setLeft(t);  
        }
}

getValue() has a return type of Object, thus the java.lang.Object types. This is the first time I have ever seen this error. Can anyone give me some background on this error? Thanks

asked Feb 22, 2012 at 21:02

svsav's user avatar

1

Sure — you simply can’t apply the > operator between objects. What would you expect it to do? You can’t apply any of the other binary operators either — +, -, / etc (with the exception of string concatenation).

Ideally, you should make your TreeNode generic, and either have a Comparator<T> which is able to compare any two instances, or make T extend Comparable<T>. Either way, you can then compare them with:

int comparisonResult = comparator.compare(current.getValue(), t.getValue());
if (comparisonResult > 0) {
  // current "greater than" t
} else if (comparisonResult < 0) {
  // current "less than" t
} else {
  // Equal
}

or

int comparisonResult = current.getValue().compareTo(t.getValue());
// Code as before

Without generics you could cast the values to Comparable or still use a general Comparator… but generics would be a better bet.

answered Feb 22, 2012 at 21:04

Jon Skeet's user avatar

Jon SkeetJon Skeet

1.4m851 gold badges9045 silver badges9133 bronze badges

1

Java doesn’t support operator overloading, so the < operator is not defined for non-primitive types. You probably want to use the Comparable<T> interface instead.

answered Feb 22, 2012 at 21:06

Daniel Pryden's user avatar

Daniel PrydenDaniel Pryden

58.6k15 gold badges98 silver badges134 bronze badges

You cannot compare objects using > or <. You need to compare them using some method, like compareTo (which you need to implement).

answered Feb 22, 2012 at 21:05

MByD's user avatar

MByDMByD

135k26 gold badges266 silver badges275 bronze badges

You cannot compare two arbitraty objects with the > operator. The > operator can only be used (directly) on primitive integer types.

You could make the objects you intend to compare implement interface java.lang.Comparable, and then call the compareTo method on them to compare them.

Comparable left = (Comparable)current.getValue();
Comparable right = (Comparable)t.getValue();

if (left.compareTo(right) > 0)
    current.setRight(t);

// etc.

answered Feb 22, 2012 at 21:07

Jesper's user avatar

JesperJesper

200k45 gold badges318 silver badges347 bronze badges

The default implementation of equals only cares about reference equality. An object does not know if Cat is greater than Apple nor would it care. You should provide a concrete implementation that overrides both equals and hashcode as well as implements the Comparable interface. This will allow you to determine if in fact Cat is greater than Apple.

answered Feb 22, 2012 at 21:08

Woot4Moo's user avatar

Woot4MooWoot4Moo

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I think u can just use another primitive variables like

int temp=(int)current.getValue();

Syscall's user avatar

Syscall

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answered Mar 5, 2021 at 16:44

user15338093's user avatar

2

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Hi,

I am trying to subtract two integers in Java. I am using NetBeans 8.1. I am first reading the integers from text field. Then converting them into integer. Then I am subtracting them But I am getting error:

     

bad operand types for binary operator ‘-‘

first type: String

second type: int

I have written the following code:

Somebody please guide me how to solve this problem.

Zulfi.


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easy — try to build a string to set a numeric value in a method that require String

what happens:

«» + int + int

Java evaluates from left to right — so the first «» is a String — String + int becomes a String — then you try the arithmetic String + int — wich doesn’t work.

This is an example, how you should do this:

As it looks like you’re a beginner — don’t even think about this crap «»+something to get a String — look in the API doc for correct methods how to create a String the right way.

Saloon Keeper

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Matt Wong wrote:Java evaluates from left to right — so the first «» is a String — String + int becomes a String

The reason for this is operator associativity, not direction of evaluation.

+

and

—

have the same precedence, so then associativity kicks in, which for

+

and

—

is left-to-right.

Marshal

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. . . but you are not allowed to apply the − operator to Strings. That is why you are getting the compile‑time error.

Zulfi Khan

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Hi,

Thanks all for your responses.

Actually the same thing works for Multiply and divide. If its the question of «String conversion» then why its not doing on Multiply & divide.

Is this a specialized behavior towards «subtract» operator?

Zulfi.


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No. Remember that Stephan said:

Stephan van Hulst wrote:+ and — have the same precedence, so then associativity kicks in, which for + and — is left-to-right.

But

+

and

*

do not have the same precedence. In fact

*

has higher precedence so it’s evaluated before

+

, with the result as you observe in your example.


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Sorry, side question, Zulfi: why are you coding a different method for each math operation?

Paul Clapham

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And all this business of operator precedence aside, interesting as it may be, it’s better to use code like what Matt Wong posted earlier in the thread rather than using

«» + number

to make a String representation of a number.

Stephan van Hulst

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Paul Clapham wrote:In fact * has higher precedence so it’s evaluated before +

Precedence doesn’t affect order of evaluation. Evaluation is always performed as a left-to-right depth-first iteration over the expression tree. Operator precedence determines how the expression tree is built up.

Paul Clapham

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Stephan van Hulst wrote:Precedence doesn’t affect order of evaluation. Evaluation is always performed as a left-to-right depth-first iteration over the expression tree. Operator precedence determines how the expression tree is built up.

I guess that’s one way of looking at it. But given the expressions

a + b + c

and

a + b * c

it’s a fact that

a + b

is evaluated first in the first expression and

b * c

is evaluated first in the second expression. There’s a lot of people who would identify that difference as being a difference in the order of evaluation.

Stephan van Hulst

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(b * c)

finishes evaluating first, but not before

(a + (b * c))

starts evaluating.

I think it’s important to be pedantic on this subject, in light of the many questions we get about why

(x + ++x)

yields

3

and not

4

when

x

starts out as

1

.

Zulfi Khan

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Hi,

Thanks for providing solution to this problem.

Operator precedence seems to be right a answer. However the truth is that I did not grasp the other answer.

Believe me I realized this answer sometime back but I could not post my clarification on this.

Sorry, side question, Zulfi: why are you coding a different method for each math operation?

Okay i would try it using switch-case but in case of event driven programming, handler function is a easy option.

Good team work.

God bless you guys.

Zulfi.

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Zulfi Khan wrote:Sorry, side question, Zulfi: why are you coding a different method for each math operation?

Okay i would try it using switch-case but in case of event driven programming, handler function is a easy option.

Yes, I agree with that. If you see old Swing code you will quite often see a single handler function with switch-case code in it, but after a few years people realized that having a different handler for each event was a much better design. Each method (including event handlers) should be designed to do only one thing, and that’s exactly what you have done there.

(Okay, it looks like Netbeans made that decision on your behalf but you have still explained the decision correctly.)

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