二元运算符'<='的坏操作数类型

Question

class First {  
   public static void main(String[] arguments) { 
      int x =60;  
      if (51 <= x <= 9) {   
         System.out.println("Let's do something using Java technology.");
      } else { 
         System.out.println("Let's");
      }
   } 
}

I am getting the error and I cant understand why as I am new to Java and programming.

bad operand types for binary operator <=

if (51 <= x <= 9) {
   first type:  boolean 
   second type: int
1 error

Answer 1

Use this code instead of yours to fix problem

51 <= x && x <= 9

Your trouble is because first comparison returns boolean value and after you compare it with int value. It's wrong.

Comparison is a binary operation, that processed from left to right one by one.

Answer 2

How Java works for: 51 <= x <= 9 51 <= x is calculated first which results to a false (boolean) in your code. Then the result of that is tried with <= 9 . And therefore the error, "<= not valid for boolean and int".

As suggested in other answers you will have to use && (and) operator. For Example:

if (x <= 51 && x >= 9) {
    //do something
}

As you can see in my answer I have used both less than and greater than, which helps in reading code. Read as if x is less than equal to 51 and x is greater than equal to 9 then.

Hope this helps in explanation.

Answer 3

You should use logical operator && meaning and . Others are || meaning or and negation ! . You can make various combinations using these operators and brackets () .

Your condition should be like this one:

if (51 <= x && x <= 9) {   

Answer 4

You are using syntax from another language (Python maybe) in Java you need to do:

if (51 <= x && x <= 9)

example

int x = 60;
        if (51 <= x && x <= 9) {
            System.out.println("Let's do something using Java technology.");
        } else {
            System.out.println("Let's");
        }