Java Operator Precedence Comparison

Operator precedence the way you defined it, while common, is not a universal truth that the Java language should recognize. Therefore no, Java language itself does not have such comparison. It is of course easy to write your own:

int precedenceLevel(char op) {
    switch (op) {
        case '+':
        case '-':
            return 0;
        case '*':
        case '/':
            return 1;
        case '^':
            return 2;
        default:
            throw new IllegalArgumentException("Operator unknown: " + op);
    }
}

Then given char op1, op2, just compare precedenceLevel(op1), precedenceLevel(op2).

You can also use if-else, or ternary operators instead of switch if you only have very few operators. Another option would be to use an enum Operator implements Comparable<Operator>, but depending on what you're doing, perhaps a parsing tool like ANTLR is the better.

Note that the above example puts ^ at the highest precedence, implying that perhaps it's used to denote exponentiation. In fact, ^ in Java language is the exclusive-or, and it has a lower precedence than +.

    System.out.println(1+2^3);   // prints 0
    System.out.println(1+(2^3)); // prints 2
    System.out.println((1+2)^3); // prints 0

This just goes to show that the precedence and even semantics of these symbols are NOT universal truths.

See also:

• What does the ^ operator do in Java?