为什么f（Double x）比f（double ... x）更好匹配？ [英] Why is f(Double x) a better match than f(double... x)?

问题描述

今天我正在攻读Java考试，我遇到了这个问题：

Today I was studying for an incoming Java exam and I ran into this question:

让 A 是一个定义如下的类：

Let A be a class defined as follows:

class A {
    public void f(Double x) { System.out.println("A.f(Double)"); }
    public void f(double... x) { System.out.println("A.f(double...)"); }
}

指令产生的输出是什么 A a =新A（）; af（1.0）; ？

What is the output produced by the instruction A a = new A(); a.f(1.0);?

答案似乎是 Af（Double ）但我不明白为什么。有人可以给我一个正确的解释吗？

The answer seems to be A.f(Double) but I can't understand why. Could someone give me a proper explanation?

推荐答案

重载决议总是倾向于一个带有明确数量的参数的函数变量参数列表，即使这意味着 1.0 是自动装箱的。

Overload resolution always favours a function with an explicit number of arguments over a function with a variable argument list, even if that means that 1.0 is auto-boxed.

稍微详细一点，根据 JLS 15.12.2 ：

1. 类型加宽


2. 自动装箱


3. 可变参数

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