Explain operator precedence in java ?

It's probable to combine multiple arithmetic expressions in one statement. For example the subsequent line adds the numbers one by five:

int m = 1 + 2 + 3 + 4 + 5;

A slightly more interesting instance: the subsequent program calculates the energy equivalent of an electron by using Einstein's famous formula E = mc².

class mc2 {

  public static void main (String args[]) {

     double mass = 9.1096E-25;

    double c = 2.998E8;

    double E = mass * c * c;

    System.out.println(E);

  }

}

Here's the output:

$ javac mc2.java

$ java mc2

8.18771e-08

This is all very obvious. Therefore if you use various operators on the similar line it's not always clear what the result will be. For example consider the subsequent code fragment:

int n = 1 - 2 * 3 - 4 + 5;

Is n equal to -2? You might think so if you just compute from left to right. Therefore if you compile this in a program and print out the result you'll search out that Java thinks n is equal to -4. Java got in which number since it performs all multiplications before it performs any additions or subtractions. If you such as you can think of the calculation Java did as being:

int n = 1 - (2 * 3) - 4 + 5;

This is an subject of order of evaluation. Within the limited number of operators you've learned so far here is how Java calculates:

1. *, /, % Do all multiplications, divisions and remainders from left to right.
2. +, - Do additions and subtractions from left to right.
3. = Assign the right-hand side to the left-hand side