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javaevaluationarithmetic-expressions
ncw
ncw

Reputation: 1733

Java precedence for multiple + and - operators

This is more of a theoretical question to understand Java's evaluation of arithmetic operations. Since + and - have the same precedence, I don't quite understand how Java evaluates the following expressions (where there are more than one + and - operators between the two operands).

public static void main(String[] args) {
    int a = 1;
    int b = 2;
    System.out.println(a+-b);    // results in -1
    System.out.println(a-+b);    // results in -1
    System.out.println(a+-+b);   // results in -1
    System.out.println(a-+-b);   // results in  3
    System.out.println(a-+-+b);  // results in  3
    System.out.println(a+-+-b);  // results in  3
    System.out.println(a-+-+-b); // results in -1
    System.out.println(a+-+-+b); // results in  3
}

From the Java 8 Language Specification (§15.8.2):

The binary + operator performs addition when applied to two operands of numeric type, producing the sum of the operands.
The binary - operator performs subtraction, producing the difference of two numeric operands.
[...]
Addition is a commutative operation if the operand expressions have no side effects.
Integer addition is associative when the operands are all of the same type.

What I also noticed, is that everytime the #operators is even, the result is the same and the order doesn't matter. But when the #operators is odd, this doesn't necessarily influence the result. E.g. in the following two expressions there is one more - than +, however the result is different.

System.out.println(a-+-b);   // results in 3
System.out.println(a-+-+-b); // results in -1

With all that information I still don't see the pattern or the way how this works.

Upvotes: 9

Views: 460

Answers (4)

Osama Salama
Osama Salama

Reputation: 647

Its evaluated as it is in how you do calculations in math with addition and subtraction. In your example all your calculations start with the smaller number a=1, then you have -, and + signs between a small number and a bigger number b=2. for example when you do 1 -- 2, you are doing 1+2 because 2 - signs cancel each other becomes 1 + 2, 1 - 2 will be -1 because 2 ( larger number has a - sign, which makes it a negative integer, adding a 1 to a negative integer increases its value by one

Upvotes: 2

janos
janos

Reputation: 124666

In math, how would you evaluate this?

a - + - b

Adding some brackets helps:

a - (+ (-b))

We can do this, because this doesn't violate the rules of precedence.

Then we can start reducing: + (-b) is really -b, and a - -b is really a + b, so the result is 1 + 2 = 3.

Let's see the second one:

a - + - + - b
a - (+ (- (+ (-b))))
a - (+ (- (-b)))
a - (+ b)
a - b
1 - 2 = -1

So simple rules of math work naturally.

Upvotes: 9

Lord Thanatos
Lord Thanatos

Reputation: 27

This appears to be a mathematical issue... (- & - = +), (- & + = -) and (+ & + = +)

int a = 1;
int b = 2;
System.out.println(a+-b);    // 1 +(-2) = 1-2 = -1
System.out.println(a-+b);    // 1-(+2) = 1-2 = -1
System.out.println(a+-+b);   // 1+-(+2) = 1-(+2) = 1-2 = 1
System.out.println(a-+-b);   // 1- + (-2) = 1- (-2) = 1+2 =3 
System.out.println(a-+-+b);  // 1 - + - (+2) = 1 - - (+2) = 1-2 = -1
System.out.println(a+-+-b);  // 1+ - + (-2) =1 - + (-2) = 1+2 = 3 
System.out.println(a-+-+-b); // 1-+-+(-2) = 1 - - (-2) = 1 + (-2) = -1
System.out.println(a+-+-+b); // 1 +- +- (+2) = 1 -- (+2) = 1+2 = 3

Upvotes: 2

ItamarG3
ItamarG3

Reputation: 4122

I believe that the - sign negate expressions:

int a = 1;
int b = 2;
System.out.println(a+(-b));
System.out.println(a+-b);

both print -1. This is from oracle docs:

Unary minus operator; negates an expression

the minus simply changes the sign of the number after it. If b were negative, then a+-b would return 3.

Upvotes: 2

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