How does JVM implement the varargs?

问题

I recently got interested in such a feature in Java, as functions with variable number of arguments. This is a very cool feature. But I'm interested:

void method(int x, String.. args) {
  // Do something
}

How is this actually implemented on the runtime level? What comes to my mind, is that when we have a call:

method(4, "Hello", "World!");

The last two arguments are internally transformed into an array, that is passed to the method. Am I right about this, or the JVM actually pushes in the stack refereneces to the strings, not just one reference to the array?

回答1:

It is implemented at compile time level. You method is compiled to bytecode as

varargs method(I[Ljava/lang/String;)V
...

which is equivalent to

void method(int x, String[] args) {
...

with varargs flag.

And

method(4, "Hello", "World!");

is compiled as

method(4, new String[] {"Hello", "World!"});

回答2:

such a method is converted into

void method(int x, String[] args) {
}

and its call

method(4, "Hello", "World!");

is converted into

method(4, new String[]{"Hello", "World!"});

Note that the last call can be written directly. This is useful when overriding vararg methods:

@Override
void method(int x, String... args) {
    String[] newArgs=new String[];
    ... // fill new args; then
    super.method(newArgs);
}

回答3:

The last two arguments are internally transformed into an array, that is passed to the method. Am I right about this,

Yes, Your understanding is correct. An array constructs and passes as a argument.

To make sure that If you see the byte code of that call you can see the array there. An array creates and and passes to the destination.

来源：https://stackoverflow.com/questions/21746663/how-does-jvm-implement-the-varargs

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java

internals