Interesting kotlin 0x09:extensions are resolved statically

Recently http://portal.kotlin-academy.com/#/ I see a lot about Kotlin An interesting topic . I think it's very suitable for Kotlin lovers , If you are interested, you can check it by yourself .
【 amusing Kotlin 】 Record your understanding of each question in the series .

0x09：Extensions are resolved statically

open class C

class D: C()

fun C.foo() = "c"

fun D.foo() = "d"

fun printFoo(c: C) {
    
    println(c.foo())
}

fun main(args: Array<String>) {
    
    printFoo(D())
}

Above code , What is the result of the operation ？ optional ：

1. Doesn’t Compile
2. Rutime error
3. "c"
4. "d"

Think about it , Record the answers in your mind .

analysis

Basic information

• class C yes class D Parent class of , And each has a name foo The extension function of . class C The extension function of returns a string “c” , class D The extension function of returns a string “d”;

• Defines a top-level function printFoo() , Parameter type is C;

• main Call inside function printFoo function , And the passed in parameter is D Example .

  

Habitual thinking

main() Function printFoo() The incoming parameter is D object , Nature calls D The extension function of D.foo() , So the answer to this question is —— Options 4：“d”.

Is this analysis right ？

If Kotlin The essence of extension function is member function , There is no problem with this analysis , But is the essence of extension function a member function ？

Kotlin What is the essence of extension function ？

If something is unknown, give an example

Define an extension function ,receiver by String.

fun String.suffix(suffix: String) = this + suffix

Decompile if something is unknown

• Bytecode

• Decompile Java Code

• Kotlin spread function stay Java Use

public class Test {
    

    public static void main(String[] args) {
    
        //  Two parameters and class member function 
        String result =
            Extensions_are_resolved_staticallyKt.suffix("Hello", " OpenCV or Android"); 
        System.out.println(result);
    }
}

Sum up , Kotlin The essence of extension function is static function , And receiver Is the first parameter of the corresponding static function in the bytecode . therefore Java Call in Kotlin When defining an extension function , Need to receiver As the first parameter .

Problem solving

Back to topic

fun printFoo(c: C) {
    
    println(c.foo())
}

Because the parameter type is C, So in the code c.foo() The corresponding is fun C.foo() = "c" Compiled into static functions . Although the runtime passes in D(), But it will still be forced to C And then execute fun C.foo() = "c" The corresponding static method . therefore , In question , The correct answer is ：

Options 3 ："c"

extend

We give classes C and class D Add member functions respectively foo()

package puzzlers

open class C {
    
    open fun foo() = "cc"       //  class  C  Member functions  foo
}

class D : C() {
    
    override fun foo() = "dd"   //  class  D  Member functions  foo
}

fun C.foo() = "c"

fun D.foo() = "d"

fun printFoo(c: C) {
    
    println(c.foo())
}

fun main(args: Array<String>) {
    
    printFoo(D())
}

What is the result of the operation ？ Friendship tips member always win.