What is the right precedence of the math expressio

问题:

What is the correct sequence of the math operations in this expression in Java:

    a + b  * c / ( d - e )
1.    4    1   3     2
2.    4    2   3     1

I understand that result is the same in both answers. But I would like to fully understand the java compiler logic. What is executed first in this example - multiplication or the expression in parentheses? A link to the documentation that covers that would be helpful.

UPDATE: Thank you guys for the answers. Most of you write that the expression in parentheses is evaluated first. After looking at the references provided by Grodriguez I created little tests:

int i = 2;
System.out.println(i * (i=3)); // prints '6'
int j = 2;
System.out.println((j=3) * j); // prints '9'

Could anybody explain why these tests produce different results? If the expression in parentheses is evaluated the first I would expect the same result - 9.

回答1:

Almost everybody so far has confused order of evaluation with operator precedence. In Java the precedence rules make the expression equivalent to the following:

a + (b  * c) / ( d - e )

because * and / have equal precedence and are left associative.

The order of evaluation is strictly defined as left hand operand first, then right, then operation (except for || and &&). So the order of evaluation is:

  a
      b
      c
    *
      d
      e
    -
  /
+

order of evaluation goes down the page. Indentation reflects the structure of the syntax tree

Edit

In response to Grodriguez's comments. The following program:

public class Precedence 
{
    private static int a()
    {
        System.out.println("a");
        return 1;
    }   
    private static int b()
    {
        System.out.println("b");
        return 2;
    }
    private static int c()
    {
        System.out.println("c");
        return 3;
    }
    private static int d()
    {
        System.out.println("d");
        return 4;
    }
    private static int e()
    {
        System.out.println("e");
        return 5;
    }

    public static void main(String[] args) 
    {
        int x = a() + b() * c() / (d() - e());
        System.out.println(x);
    } 
}

Produces the output

a
b
c
d
e
-5

which clearly shows the multiplication is performed before the subtraction.

回答2:

As JeremyP has nicely shown us, the first answer is correct.

In general, the following rules apply:

• Every operand of an operator is evaluated before the operation itself is performed (except for ||, &&, and ? :)
• Operands are evaluated left to right. The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.
• The order of evaluation respects parentheses and operator precedence:
  • Parentheses are evaluated first.
  • Operators are evaluated in order of precedence.
  • Operators with equal precedence are evaluated left-to-right, except for assignment operators which are evaluated right-to-left.

Note that the first two rules explain the result in your second question:

int i = 2;
System.out.println(i * (i=3)); // prints '6'
int j = 2;
System.out.println((j=3) * j); // prints '9'

Reference documentation:

http://java.sun.com/docs/books/jls/second_edition/html/expressions.doc.html#4779

Tutorial:

http://download.oracle.com/javase/tutorial/java/nutsandbolts/operators.html

回答3:

It evaluates the expressions in the following order. Variable names are expressions that need to be evaluated.

a + b * c / (d - e)
    2   3    5   6
      4        7
1         8
  9

So, the answer to your question is #1. The order of operations determines the shape of the expression tree (what is the left side of the tree, and what is the right), but the left side is always evaluated first (and the root is evaluated last).

回答4:

I would imagine that it might evaluate something like this evaluating from left to right.

a+b*c/(d-e)

Action           Left Value      Right Value
Start Add        a               b*c/(d-e)
Start Multiply   b               c
Calc Multiply (since it can)    
Start Divide     b*c             (d-e)
Start Subtract   d               e
Calc Subtract
Calc Divide
Calc Add

This can be thought of as creating a binary tree representing the calculations and then working from the leaf nodes, left to right, calculating things. Unfortunately my ascii art isn't great but here's an attempt at representing the tree in question:

   Add
    /\
   /  \
  a    \
     Divide
       / \
      /   \
     /     \
    /       \
Multiply  Subtract
  /\         /\
 /  \       /  \
b    c     d    e

And I did some tests in C# (I know its not the same but that's where my interests lie and the tests can be easily adapted) as follows:

        f = 1;
        Console.WriteLine((f=2) + (f) * (f) / ((f) - (f)-1));
        Console.WriteLine(2 + 2 * 2 / (2 - 2 - 1));
        f = 1;
        Console.WriteLine((f) + (f=2) * (f) / ((f) - (f)-1));
        Console.WriteLine(1 + 2 * 2 / (2 - 2 - 1));
        f = 1;
        Console.WriteLine((f) + (f) * (f = 2) / ((f) - (f)-1));
        Console.WriteLine(1 + 1 * 2 / (2 - 2 - 1));
        f = 1;
        Console.WriteLine((f) + (f) * (f) / ((f=2) - (f)-1));
        Console.WriteLine(1 + 1 * 1 / (2 - 2 - 1));
        f = 1;
        Console.WriteLine((f) + (f) * (f) / ((f) - (f=2)-1));
        Console.WriteLine(1d + 1d * 1d / (1d - 2d - 1d));

The pairs of console.writeline statements are the algebraic one (using the set a number trick) and a numerical representation showing what the calculation actually does. The pairs produce the same result as each other.

As can be seen the arguments are evaluated in order with any after the assignment being 2 and those before being one. So the order of evaluation of things is simple left to right I think but the order of calculations is as you would expect it to be.

I assume this can be run almost with copy and paste to test in JAVA...

There may be some unnoticed assumptions in here so if anybody does spot logic flaws in here please do call me on them and I'll work them through.

回答5:

i am assuming that your expression would be something like

x = a + b * c / ( d - e )

the equality operator has right to left order of evaluation. so the expression on the right of = will be evaluated first.

if your refer this precedence chart: http://www.java-tips.org/java-se-tips/java.lang/what-is-java-operator-precedence.html

1) the brackets will be evaluated (d-e), lets say (d - e) = f so the expression then becomes x = a + b * c / f.

2) Now * and / have same precedence, but the order of evaluation is left to right to * will be evaluated first so lets say b * c = g, so the expression becomes x = a + g /f

3) Now / has the next precedence so g / f will be evaluated to lets say h so the expression will be come x = a + h,

4) lastly evaluating a + h

回答6:

In your second question, it seems Java is evaluating the part in parenthesis as an assignment, not an mathematical expression. This means that is will not perform parenthetical assignments in the same order as operations in parenthesis.

回答7:

The results of the calculation are defined by the Operator Order of Precedence. So parentheses have highest precedence here, multiplication and division next highest, and addition and subtraction lowest. Operators with equal precedence are evaluated left to right. So the expression in the question is equivalent to:

    a + (b  * c) / ( d - e ))

However there is a slight difference between what is normally meant by "being evaluated first" and the operator precedence for getting the correct answer.

"d-e" is not necessarily actually calculated before "a" is calculated. This pretty much doesn't make any difference unless one of the 'variables' in the expression is actually a function. The Java standard does not specify the order of evaluation of components of an expression.

回答8:

a + b * c / ( d - e )
      1         1
          2
  3

The whole point of operator precedence is to convert the expression into a syntax tree. Here the * and - are at the same level of the tree. Which one is evaluated first is irrelevant for the result and not warrantied.

Edit: Sorry, I got confused by my C background. As others have pointed out, Java has an "Evaluate Left-Hand Operand First" rule. Applying this rule to / tells you that * is evaluated first (your first answer).