重载函数int ...和long ...同时 [英] Overload function int... and long... simultaneously

问题描述

我想创建两个函数，比如说

I want to create two functions, say

long min(long...);
int min(int...);

但是当我尝试调用第二个时，即 min（1,5） 我得到一个模糊的方法调用

But when I try to invoke the second i.e min(1, 5) one I get ambiguous method call

除重命名外是否有解决方法？

Is there workaround except renaming?

推荐答案

这是一个已知错误

您描述的行为是用Java 7修复的错误。请参阅发行说明中的详细信息，该部分称为大多数特定Varargs方法选择的变化 。

The behaviour you describe is a bug which has been fixed with Java 7. See details in the release notes, section called "Changes in Most Specific Varargs Method Selection".

它应该编译的原因

变量arity排在最后确定最具体的方法。在 JLS 15.12.2.4 - 这是一个摘录：

Variable arity comes last in determining the most specific method. The rules to determine which vararg method applies when there are several are defined in JLS 15.12.2.4 - here is an extract:

一个名为m的变量arity成员方法比另一个变量arity成员方法同名如下：

- [...]

- 一个成员方法有k个参数，另一个有n个参数，其中n≥k ，和：

One variable arity member method named m is more specific than another variable arity member method of the same name if either:
- [...]
- One member method has k parameters and the other has n parameters, where n ≥ k, and:

• 第一种方法的参数类型是U1，...，Uk-1，Uk []。


• 另一种方法的参数类型是T1，...，Tn-1，Tn []。


• 对于来自1的所有j到n，Uj<：Tj

在你的情况下，k = n和 U1 [] = int [] 和 T1 [] = long [] 所以如果<$ c $可以做出决定c> int< ;: long 或相反。

In your case, k = n, and U1[] = int[] and T1[] = long[] so the determination can be made if int <: long or the opposite.

换句话说，t考虑到的ype不是 int [] 与 long [] 但是 int vs long 。并且它发生 int< ;: long 所以应该选择 int ... 方法并且它应该编译。

In other words, the type taken into account is not int[] vs. long[] but int vs long. And it happens that int <: long so the int... method should be chosen and it should compile.

结论：

代码应该（和确实）用Java编译好但是不能用Java 5或6编译。下面的代码用Java 7打印 int ：

The code should (and does) compile fine with Java 7 but would not compile with Java 5 or 6. The code below prints int with Java 7:

public class Test1 {
    public static void main(String[] args) {
        new Test1().m(1, 2);
    }
    int m(int... i) {
        System.out.println("int");
        return 0;
    }
    long m(long... i) {
        System.out.println("long");
        return 0;
    }
}

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