Why does `variable++` increment the variable but `variable + 1` does not?

Question

Here's the problem in which I encountered this issue:

The function should compare the value at each index position and score a point if the value for that position is higher. No point if they are the same. Given a = [1, 1, 1] b = [1, 0, 0] output should be [2, 0]

fun compareArrays(a: Array<Int>, b: Array<Int>): Array<Int> {

    var aRetVal:Int = 0
    var bRetVal:Int = 0

    for(i in 0..2){
        when {
            a[i] > b[i] -> aRetVal + 1 // This does not add 1 to the variable
            b[i] > a[i] -> bRetVal++ // This does...
        }
    }
    return arrayOf(aRetVal, bRetVal)

}

The IDE even says that aRetVal is unmodified and should be declared as a val

Why do you expect aRetVal to change? The way you have it, the result of aRetVal + 1 is not assigned to a variable. It is popular in many languages (see C++, Java, PHP, JavaScript, etc) that varName++ implicitly assigns to itself. See the increment/decrement operators documentation and the arithmetic operators documentation. — ordonezalex
– ordonezalex, Commented Jun 28, 2019 at 8:16

user2340612 · Accepted Answer · 2021-06-11 10:22:33Z

7

What others said is true, but in Kotlin there's more. ++ is just syntactic sugar and under the hood it will call inc() on that variable. The same applies to --, which causes dec() to be invoked (see documentation). In other words a++ is equivalent to a.inc() (for Int or other primitive types that gets optimised by the compiler and increment happens without any method call) followed by a reassignment of a to the incremented value.

As a bonus, consider the following code:

fun main() {
    var i = 0
    val x = when {
        i < 5 -> i++
        else -> -1
    }

    println(x) // prints 0
    println(i) // prints 1

    val y = when {
        i < 5 -> ++i
        else -> -1
    }

    println(y) // prints 2
    println(i) // prints 2 
}

The explanation for that comes from the documentation I linked above:

The compiler performs the following steps for resolution of an operator in the postfix form, e.g. a++:

Store the initial value of a to a temporary storage a0;
Assign the result of a.inc() to a;
Return a0 as a result of the expression.

...

For the prefix forms ++a and --a resolution works the same way, and the effect is:

Assign the result of a.inc() to a;

Return the new value of a as a result of the expression.

edited Jun 11, 2021 at 10:22

answered Jun 27, 2019 at 14:19

user2340612

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2 Comments

Torbjörn Österdahl Over a year ago

For a++, I get the new value assigned to a. For a.inc I don't get the new value assigned to a. In fact I can do a.inc(), also when a is a 'val' (immutable). So I would suggest that a++ and a.inc() is not equivalent.

user2340612 Over a year ago

@TorbjörnÖsterdahl that's right. What I meant is that a++ calls a.inc() under the hood, but it's true that the former implies the resulting value is reassigned to a in the end. I'll edit my answer to make it more clear

Marvin · Accepted Answer · 2019-06-27 13:15:10Z

3

Because

variable++ is shortcut for variable = variable + 1 (i.e. with assignment)

and

variable + 1 is "shortcut" for variable + 1 (i.e. without assignment, and actually not a shortcut at all).

answered Jun 27, 2019 at 13:15

Marvin

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Comments

r2rek · Accepted Answer · 2019-06-27 13:15:10Z

2

That is because what notation a++ does is actually a=a+1, not just a+1. As you can see, a+1 will return a value that is bigger by one than a, but not overwrite a itself.

Hope this helps. Cheers!

answered Jun 27, 2019 at 13:15

r2rek

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Comments

s1m0nw1 · Accepted Answer · 2019-06-27 13:15:19Z

2

The equivalent to a++ is a = a + 1, you have to do a reassignment which the inc operator does as well.

This is not related to Kotlin but a thing you'll find in pretty much any other language

answered Jun 27, 2019 at 13:15

s1m0nw1

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