Question

I was wondering why it is not allowed in Java to overload Foo(Object[] args) with Foo(Object... args), though they are used in a different way?

Foo(Object[] args){}

所用名称如下:

Foo(new Object[]{new Object(), new Object()});

其他形式:

Foo(Object... args){}

所用名称如下:

Foo(new Object(), new Object());

是否有理由这样做?

Answer 1

http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#301183”rel=“noreferer”>15.12.2.5 选择最具体的方法就这样说,但很复杂。 e.g. Foo(Number... ints)和Foo(Integer... ints)之间的 Cho

为了落后的兼容,这些实际上是相同的。

public Foo(Object... args){} // syntactic sugar for Foo(Object[] args){}

// calls the varargs method.
Foo(new Object[]{new Object(), new Object()});

例如,你可以将主(主)定义为:

public static void main(String... args) {

使他们不同的方式是,在加拉尔之前就有一个论点。

public Foo(Object o, Object... os){} 

public Foo(Object[] os) {}

Foo(new Object(), new Object()); // calls the first.

Foo(new Object[]{new Object(), new Object()}); // calls the second.

它们并不完全相同。微妙的区别是,虽然你能够把一个阵列转至一个堡垒,但你可以把一个阵列参数作为空洞。

public Foo(Object... os){} 

public Bar(Object[] os) {}

Foo(new Object[]{new Object(), new Object()}); // compiles fine.

Bar(new Object(), new Object()); // Fails to compile.

此外,var必须是最后的参数。

public Foo(Object... os, int i){} // fails to compile.

public Bar(Object[] os, int i) {} // compiles ok.

Answer 2

问题最直接的答案是,如果两项声明都存在,方法研究逻辑就会模糊不清。如果你在同一级别上宣布,那么在这样说时,就没有办法说明你想要采用的方法:

Object[] a = new Object[10];
Foo(a); // array or vararg?

And Java requires that there always be a most specific method for every method invocation. For the why, see Peter s answer.

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