在参数Java访问变量 [英] Java access variables in arguments

问题描述

我想能够访问传递给我的方法的变量。我知道的C可能有某种方式来东西的地址作为参数传递，则指的是作为一个指针，并能够使用变量。总之，这里是一个例子。

 无效myFunction的（地址变量）{
    *变量=榜样;
}
 

然后调用它为：

  myFunction的（安培; somevariable）;
 

解决方案

有没有办法通过引用或Java中的指针传递参数，从而达到您想要的传递效果的唯一途径的可变的对象，并改变它，你调用方法中值：

 公共类ChangeableInt {
    私人诠释数目;
    公共ChangeableInt（int值）{数=值;}
    公众诠释的get（）{返回号码;}
    公共无效集（int值）{数=值;}
}
 

使用这个类在手，你可以这样做：

 无效myFunction的（ChangeableInt ARG）{
   arg.set（42）;
}
...
ChangeableInt VAL =新ChangeableInt（123）;
myFunction的（VAL）;
的System.out.println（val.get（））; //输出42
 

I would like to be able to access a variable passed to my method. I know C might have has some kind of way to pass the address of something as an argument, then refer to that as a pointer and be able to use the variable. Anyway, here's an example.

void myFunction(address to variable) {
    *variable = "example";
}

Then call it as:

myFunction(&somevariable);

解决方案

There is no way to pass arguments by reference or by pointer in Java, so the only way to achieve the effect that you would like is passing a mutable object, and changing its value inside the method that you call:

public class ChangeableInt {
    private int number;
    public ChangeableInt(int value) {number = value;}
    public int get() {return number;}
    public void set(int value) {number = value;}
}

With this class in hand, you can do this:

void myFunction(ChangeableInt arg) {
   arg.set(42);
}
...
ChangeableInt val = new ChangeableInt(123);
myFunction(val);
System.out.println(val.get()); // prints 42

这篇关于在参数Java访问变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！