方法的Java变量号或参数

It is possible to declare a method that will allow a variable number of parameters?

可以声明一个允许可变数量参数的方法吗？

What is the symbolism used in the definition that indicate that the method should allow a variable number of parameters?

定义中使用的符号是什么，表明该方法应该允许可变数量的参数？

Answer: varargs

答案：varargs

6 个解决方案

#1

210

That's correct. You can find more about it in the Oracle guide on varargs.

那是对的。您可以在varargs的Oracle指南中找到更多相关信息。

Here's an example:

这是一个例子：

void foo(String... args) {
    for (String arg : args) {
        System.out.println(arg);
    }
}

which can be called as

可称为

foo("foo"); // Single arg.
foo("foo", "bar"); // Multiple args.
foo("foo", "bar", "lol"); // Don't matter how many!
foo(new String[] { "foo", "bar" }); // Arrays are also accepted.
foo(); // And even no args.

#2

Yes, it's possible:

是的，这是可能的：

public void myMethod(int...numbers) { ... }

#3

Variable number of arguments

It is possible to pass a variable number of arguments to a method. However, there are some restrictions:

可以将可变数量的参数传递给方法。但是，有一些限制：

The variable number of parameters must all be the same type
可变数量的参数必须都是相同的类型
They are treated as an array within the method
它们被视为方法中的数组
They must be the last parameter of the method
它们必须是方法的最后一个参数

To understand these restrictions, consider the method, in the following code snippet, used to return the largest integer in a list of integers:

要了解这些限制，请考虑以下代码片段中的方法，该方法用于返回整数列表中的最大整数：

private static int largest(int... numbers) {
     int currentLargest = numbers[0];
     for (int number : numbers) {
        if (number > currentLargest) {
            currentLargest = number;
        }
     }
     return currentLargest;
}

source Oracle Certified Associate Java SE 7 Programmer Study Guide 2012

源Oracle认证助理Java SE 7程序员学习指南2012

#4

Yup...since Java 5: http://java.sun.com/j2se/1.5.0/docs/guide/language/varargs.html

是的...自Java 5起：http：//java.sun.com/j2se/1.5.0/docs/guide/language/varargs.html

#5

If you may use different types of arguments, then use :

如果您可以使用不同类型的参数，请使用：

public void foo(Object... x) {
    String first    =  x.length > 0 ? (String)x[0]  : "Hello";
    int duration    =  x.length > 1 ? Integer.parseInt((String) x[1])     : 888;
}   
foo("Hii", ); 
foo("Hii", 146);

for security, use like this:
if (!(x[0] instanceof String)) { throw new IllegalArgumentException("..."); }

为安全起见，请使用如下：if（！（x [0] instanceof String））{throw new IllegalArgumentException（“...”）; }

The main drawback of this approach is that if optional parameters are of different types you lose static type checking. Please, see more variations .

这种方法的主要缺点是，如果可选参数属于不同类型，则会丢失静态类型检查。请看更多变化。

#6

Yes Java allows vargs in method parameter .

是Java允许方法参数中的vargs。

public class  Varargs
{
   public int add(int... numbers)
   { 
      int result = 1; 
      for(int number: numbers)
      {
         result= result+number;  
      }  return result; 
   }
}

#1

210