CodeGym/Help with Java Tasks/I'm confused about operator precedence

Level 14

Madrid

04.11.2023
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I'm confused about operator precedence

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I'm confused about a table showing operator precedence. The table is found in this CodeGym article

Why does expr++ have more precedence than ++expr? I thought that ++expr increments first and foremost, whereas expr++ updates its value AFTER being used in the operation. Does anyone understand this? For instance:

int a = 2;
int b = 3;
int c = ++a + b--;

System.out.println(c); // c prints 6
System.out.println(b); // b prints 2
System.out.println(a); // a prints 3

If postfix (expr++) has the highest precedence, as shown in the table, then why does the expression ++a (which has lower precedence) increment the value first? If the postfix expression b-- has the maximum precedence, then why does the value of b increment after the operation has taken place?

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Thomas

Level 41 , Bayreuth, Germany

4 November 2023, 12:45

solution

That behaviour has nothing to do with operator precedence for the whole expression. That's an implementation detail of the postfix increment/ decrement operator only (compare JLS) The value of the postfix increment expression is the value of the variable before the new value is stored. In a case like here:

int a = 1;
int b = 10 + a++;

a is incremented to 2. But the expression a++ still is evaluated to the old value of a which is 1. In the end b is 11 and a is 2. If you want the expression a++ to evaluate to 2, then you need to use prefix. That uses as value for the expression ++a the value of a after incrementing it (or just use a + 1)

Guadalupe Gagnon

Level 37 , Tampa, United States

4 November 2023, 15:34

useful

To expand a bit on what Thomas said Java (and math in general) as a very specific way that expressions are evaluated. Take this larger example:

int a = 4, b = 2, c = 2, d = 1, e = 5;

int result = a++ + b-- + +c + ++d - --e;

The very first step is going to be to evaluate the urinary expressions that are post fixed, from left to right, which sets the value in the full equation AND THEN increments the variable by 1. After this step your equation would look like this:

int result = 4 + 2 + +c + ++d - --5;

if you could stop the program at this exact point and see the current value of a and b, their values would be 5 and 1 respectively. The next step would be to evaluate the prefix urinary variables which first increments the variable by 1 and then sets the value in the equation, also starting from left to right. So after this step the equation would look like this:

int result = 4 + 2 + 2 + 2 - 4;

And the result would be 6.

Guadalupe Gagnon

Level 37 , Tampa, United States

4 November 2023, 15:37

useful

Jesú, I thought the same thing as you when I was learning and saw the actual precedence because it doesn't make sense that the post fix would evaluate before the prefix. It made more sense to me when I realized that those expressions are for variables only. You can't use them with straight values. This is an illegal equation:

int result = 4++ + 2-- + +2 + ++1 - --5;

As variables they are individual equations that can be independently evaluated in steps, according to the operator precedence, in a larger equation.

Guadalupe Gagnon

Level 37 , Tampa, United States

4 November 2023, 15:53

useful

One last example:

int a = 4;

int result = a++ + a;

System.out.println(result);

If you think that postfixed ++ on the first a is evaluated after the expression then you would get the answer of 8; however if you run the code you will see that 9 is printed for result. Once again this is because the total expression a++ is evaluated completely before moving to the next step. This would mean the value 4 is plugged into the overall equation and then a is incremented by 1. After that first step the equation would look like this:

int result = 4 + a;

The value of a at this point is now 5 so that would be plugged in for the remaining a:

int result = 4 + 5;

and then 4 + 5 would evaluate to set result to 9.

Jesú

Level 14 , Madrid, Spain

4 November 2023, 17:58

So, if I understand this correctly, a postfix expression has maximum precedence when it comes to being... just the variable without increasing. It would have as much precedence as a variable with no postfix or prefix, just a plain variable. Then, the increment is applied. Thank you!

Jesú

Level 14 , Madrid, Spain

4 November 2023, 19:28

Thank you man, that was really helpful