How does prefix and postfix change the answer?

[Cj Hooper]

Hello!

Will the postfix x++ and the prefix --y change the answer for this question?



If x has the value 10 and so does y, then what is the value of (x ++) * (-- y)?

Answer: 11*9

Thank you so much!

[Junilu Lacar]

Not exactly sure what you're looking for here. Is that "answer" your answer? Is that what you think the expression evaluates to?

The best way to figure this out is to put it in a program and run it. Then you can see if what you think should happen actually does happen.

[Stefan Evans]

Prefix == apply the increment/decrement and then use the value
Postfix == use the value and then apply the increment/decrement

I don't think the answer to (x++) + (--y) is 11 * 9
Yes x ends with a value of 11, and Y ends with a value of 9
But those wouldn't be the values used in the expression.

[Cj Hooper]

Thank you both! When I run the program, I receive the output 10*9. The problem doesn't specify after the program is executed and that is what is throwing me off. The answer is 11*9 according the quiz. This is the coding I ran to test:

class test{
public static void main(String args[]){
int x = 10, y = 10;

System.out.println(x++); 


System.out.println(--y); 
}}

[Guillermo Ishi]

with prefix the ++ or -- happens before the operation, and with postfix after it.

int a = 1;

b = --a;
//a and b both zero at this line

a = 1;
b = a--;
//b is 1 and a is 0 at this line.

postfix is more traditional probablybecause it works well incrementing a zero based index.
Be careful in the last field of a for loop, because the are both the same! Either way the ++ or -- happens at the bottom of the loop.

[Junilu Lacar]

Cj Hooper wrote:Thank you both! When I run the program, I receive the output 10*9. The problem doesn't specify after the program is executed and that is what is throwing me off. The answer is 11*9 according the quiz. This is the coding I ran to test:

Well, the answer according to the quiz is wrong and you can take that to your instructor. Here's a variation of your test:

public static void main(String args[]){
    int x = 10, y = 10;
    int ans = (x++) * (--y);  // evaluates as 10 * 9
    System.out.printf("ans: %d, x: %d, y: %d", ans, x, y);
}

Edit: BTW, the output of the above is
ans: 90, x: 11, y: 9

[Ash Jon]

These are a lot easier to understand than you think they are, they just look a bit daunting...

I always found it best described using for-loops:

                // The postfix operator ++ indicates that you want to
		// increment the variable by one, but that it is
		// not needed for the current iteration
		
		// So on the first loop, i is still equal to 0
		// Once the body of the first loop has executed,
		// the i is then incremented, post execution
		
		for (int i = 0; i < 5; i++) {
			// body
		}
				
		// This time, you indicate that the increment
		// is important for the first loop. It is basically
		// a statement to say, "don't wait, do it now before
		// the variable is used"
		// On the first iteration, i will be 1, because it was
		// incremented by one before the loop body executed
		
		for (int i = 0; i < 5; ++i) {
			// body
		}
		
		// This 'wait - don't wait' applies wherever it is used
		// For example:
		
		int i = 1;
		int j = 2;
		
		j = j + i++; // = 3 because i is incremented AFTER the execution
		System.out.println(j);
		
		
		i = 1;
		j = j + ++i; // = 4 because i was incremented before it was used
		System.out.println(j);
		
		// This also applies to the negative prefix/postfix operator --

Hope it helps

[Campbell Ritchie]

I don't think this is actually easy to understand. It causes no end of confusion to beginners. The problem is that there are two values to consider in i++.
There is the value of i and there is the value of the whole expression. The value of the whole expression i++ is the same as the old value of i. We have an FAQ about that.
The same applies to ++i but when the value of ++i is equal to the new value of i, most people find that easy to understand.

[Junilu Lacar]

Sorry, Ash, but your description regarding the for-statement and prefix/postfix is incorrect.

Ash Jon wrote:
I always found it best described using for-loops:

                // The postfix operator ++ indicates that you want to
		// increment the variable by one, but that it is
		// not needed for the current iteration
		
		// So on the first loop, i is still equal to 0
		// Once the body of the first loop has executed,
		// the i is then incremented, post execution
		
		for (int i = 0; i < 5; i++) {
			// body
		}
				
		// This time, you indicate that the increment
		// is important for the first loop. It is basically
		// a statement to say, "don't wait, do it now before
		// the variable is used"
		// On the first iteration, i will be 1, because it was
		// incremented by one before the loop body executed
		
		for (int i = 0; i < 5; ++i) {
			// body
		}

This

		System.out.println("for (int i; i < 5; i++)");
		for (int i = 0; i < 5; i++) {
			System.out.println(i);
		}

		System.out.println("for (int i; i < 5; ++i)");
		for (int i = 0; i < 5; ++i) {
			System.out.println(i);
		}

produces the output
for (int i; i < 5; i++) 0 1 2 3 4 for (int i; i < 5; ++i) 0 1 2 3 4
As you can see, the output from both loops are identical. The use of the postfix operator on the index of a for-statement is idiomatic. That is to say, that's just how most people normally write it. Using a prefix increment is similar to saying it how Yoda, the character from Star Wars, would: "incremented i is, hmmm" instead of the normal way, "i is incremented." Semantically, however, there is no difference between i++ and ++i as the increment expression of the for-statement. This is because, contrary to what you described, the increment is always executed after the body of the loop is executed.

From https://docs.oracle.com/javase/tutorial/java/nutsandbolts/for.html:

The increment expression is invoked after each iteration through the loop;

The rest of your example where postfix/prefix is used in an assignment statement is correct though.

[Junilu Lacar]

I wrote:The rest of your example where postfix/prefix is used in an assignment statement is correct though.

Except maybe for the second part, which might confuse people if they think that execution continues from the first part:

int i = 1;
int j = 2;
 
j = j + i++; // = 3 because i is incremented AFTER the execution
System.out.println(j);
 
i = 1;
j = j + ++i; // = 4 because i was incremented before it was used
System.out.println(j);

Your comment on line 8 can be misleading because it's based on the assumption that the variable j is still 2. However, j is already 3 after line 4, in which case line 9 would print 5, not 4 as your comment implies.

It might be clearer if we presented it this way instead:

int i = 1;
int j = 5;
 
j = j + i++; // evaluated as 5 + 1

System.out.println(j); // 6
System.out.println(i); // 2 - incremented value is "visible" now

j = j + ++i; // evaluated as 6 + 3

System.out.println(j); // 9
System.out.println(i); // 3 - same value used to evaluate line 9

[Ash Jon]

Thanks for the corrections Junilu. I seriously did not know this was the case with for-loops...
I assumed it was the case, as I've never had to increment prior to execution of the body before (How silly do I look!)

Regarding the second part. I had meant to change j back to 2, as I did with with i (i = 1).

Thanks! Interesting to know that It's appreciated.

Ash

[Junilu Lacar]

Not a problem, Ash. You should see how silly I look sometimes. But it's all good, we're all learners, even some of the old dogs around here.