Can fields of the class and arguments of the metho

问题:

I have a class with a fields called "a". In the class I have a method and in the list of arguments of this method I also have "a". So, which "a" I will see inside of the method? Will it be the field or it will be the argument of the method?

public class myClass {
   private String a;
   // Method which sets the value of the field "a".
   public void setA(String a) {
     a = a;
   }
}

By the way, there is a similar situation. A method has some local (for method) variables whose names coincide with the names of the fields. What will the "see" the method if I refer to such a method-local variable inside the method (the field or the local variable)?

回答1:

The more local scope has the priority, so the parameter a will hide the field a. In effect, you set the value of parameter a to itself. The proper idiom to avoid name clashes (and improve readability) is to use this to explicitly mark the class member:

public void setA(String a) {
  this.a = a;
}

The same is true for local variables vs member variables: local variables hide member variables with the same name.

回答2:

To add to all the answers recommending:

public void setA(String a) {
   this.a = a;
}

it's important to realise that omitting the this will simply set the parameter to itself. By using final thus

public void setA(final String a) {
   this.a = a;
}

you can eliminate errors caused by omitting this. Using final is a good practise whenever specifying parameters and fields that aren't intentionally required to change.

回答3:

The closest one. That is,

 a = a;

inside the method has no effect since both refer to the argument a. To refer to the instance variable a you use the this keyword.

 this.a = a;

回答4:

The local version will "shadow" the instance variable by the same name. One pattern to get around this in accessors like your is this:

public void setA(String a) {
   this.a = a;
}

which uses the this keyword to be explicit about scope.

回答5:

You need to use this to access the class variable, otherwise it will always take the parameter variable.