如何定义一个可以在java中获取任意参数的方法？ [英] How to define a method that can take arbitrary arguments in java?

问题描述

如何定义一个可以在java中获取任意参数的方法？有演示吗？

How to define a method that can take arbitrary arguments in java? Is there a demo?

推荐答案

varargs 在Java 5中引入。

varargs were introduced in Java 5.

例如：

public String join(String... parts);

这实际上是一个快捷方式：

This is actually a shortcut for:

public String join(String[] parts);

parts 参数用作数组在方法中，但可以在不构造数组的情况下调用该方法（如 obj.join（new String [] {part1，part2，part3}））

the parts parameter is used as an array in the method, but the method can be called without constructing an array (like obj.join(new String[] {part1, part2, part3}))

但是要非常小心，因为可能会产生歧义。例如：

However be very careful with this, because ambiguities can arise. For example:

public void write(String author, String... words);
public void write(String... words);

如果 obj.write（Mike，跳）？编译器足够聪明，可以检测出歧义，但是我遇到过某些编译器没有发现这些问题的情况（无法准确回忆）

Which method will be called if obj.write("Mike", "jumps") ? The compiler is clever enough to detect the ambiguity, but I've had cases when some compilers did not spot such problems (can't recall exactly)

使用varargs是实用的当物体属于同一类型或至少具有相同的功能目标时。如果你想要不同的论点。例如：

Using varargs is practical when the objects are of the same type, or at least with the same functional aim. If you want different arguments. For example:

public String publishBook(String title, [String author], 
      [String isbn], [boolean hardcover]); // [..] would mean optional

然后你需要重载你的方法

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