In Kotlin, How does the following code snippet work to swap variables values?

variable.also{ codeblock }

as

1. read the variable and store it in a temporary variable
2. execute the codeblock
3. return what you stored in step 1.

So in your case it's

1. read b (which has value 2) and store it in a temporary variable
2. execute the b = a. meaning that 1 is assigned to the variable b.
3. return what you stored in step 1, which is the value 2.
4. this value is then assigned to the variable a

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Here is the documentation of also:

inline fun <T> T.also(block: (T) -> Unit): T

Calls the specified function block with this value as its argument and returns this value.

From that, we can note the following things:

1. It is an extension function (thus has a receiver).

2. It is an inline function.

3. Its argument is a function type, which:

   • Accepts the receiver as an argument.

   • Will also be inlined.

4. It returns the receiver object.

That all means this:

fun swapDemo() {
    var a = 42
    var b = 117
    a = b.also { b = a }
}

Is approximately the same as writing this:

fun swapDemo() {
    var a = 42
    var b = 117

    // begin 'also'
    val receiver = b
    val it = receiver // the 'block' function's parameter (unused)
    b = a // the 'block' function
    // end 'also'

    a = receiver // 'also' "returns"
}

Ignoring the it variable (also see below), you can see this looks just like a "traditional" swap implementation, with receiver functioning as the temporary variable.

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Note that, from inspecting the byte code, writing the following:

a = b.also { _ -> b = a }

Instead of:

a = b.also { b = a }

Will cause the it local variable to be omitted. At least in Kotlin 1.9.22.