Java中的“ this”关键字

Question

I have a question regarding this in the following code. In the following, this.name will set the name. We could also do this using name = name , so my question is should the this pointer be used. This is not an homework

import java.io.*;

public class Employee{
    String name;
    int age;
    String designation;
    double salary;

    //This is the constructor of the class Employee
    public Employee(final String name){ //EDIT: Made parameter final
        this.name = name;
        //name= name; this is also correct
    }

    //Assign the age of the Employee  to the variable age.
    public void empAge(int empAge){
        age =  empAge;
    }

    //Assign the designation to the variable designation.
    public void empDesignation(String empDesig){
        designation = empDesig;
    }

    //Assign the salary to the variable salary.
    public void empSalary(double empSalary){
        salary = empSalary;
    }

    //Print the Employee details
    public void printEmployee(){
        System.out.println("Name:"+ name );
        System.out.println("Age:" + age );
        System.out.println("Designation:" + designation );
        System.out.println("Salary:" + salary);
    }
}

Answer 1

  //      name= name; this is also correct

This is not correct. It'll assign your parameter to itself. By using the this keyword, you're declaring which name you're using (ie the field on your Employee object).

You may wish to name the field differently from the parameter. However this means that all works well until someone automatically refactors your code to (perhaps inadvertently) declare the field and parameter as the same name!

For this reason you'll often see method signatures defined thus:

public Employee(final String name)

The final keyword prevents reassignment, and stops you from mistakenly reassigning the input parameter, and consequently not assigning to your field. Note also that if you declare the field as final , then compilation will fail if you don't make an assignment to that field. Using final is a good way to trap such errors and also enforce the immutability of an object (often a good thing - it contributes to a more reliable solution, especially in a threaded environment).

Answer 2

Well to avoid the confusion in cases like u mentioned name=name we uses this pointer to make it clear the we here mean class variable name .

So to make understand the reader here in this case we use this though there can be many other cases where (this) is more useful.

In some compilers name=name gives error as well

Answer 3

Notice that there are two things called name in your code. Your class Employee has member variable called name , and the constructor takes a parameter that's also called name .

What you want to do in the constructor is set the member variable name to the same value as the parameter name . To access the member variable, you have to use this.name , because name refers to the parameter - because the variables have the same name, the parameter is hiding the member variable.

Note that name = name; does not do the same thing as this.name = name; .

When you do name = name; , you assign the value of the parameter name to the parameter itself - not to the member variable. The compiler does not magically know that the first name is supposed to mean the member variable, and the second name is supposed to mean the parameter.

So, you need this in this case to refer explicitly to the member variable, instead of the parameter that is hiding the member variable.

Answer 4

In this case this is just a scope resolution/disambiguation.

public Employee(String name){
  this.name = name;
  // the above line will assign a value of parameter to instance variable
  //      name= name; this is also correct
  // (**NO** the above line will assign a value of parameter to itself) 
}

Answer 5

当参数名称和类变量名称相同时，要区分它们，可以编写this.classVariable来标识类变量

Answer 6

When you call a variable, the pointing one is the one who is the closest to your current scope.

So, if program is currently containing in memory a local variable toto and the wrapping class containing a field variable toto , you have to precise this keyword in order to access the field's one.

Otherwise, the field variable is said to be shadowed by the local variable and so doing toto = toto assigns the local parameter to itself (never useful) and not what you are expecting => the field variable.

Answer 7

Why make it so hard on yourself?

Just rewrite the signature/method like this:

public Employee(String _name) {
   name = _name;
}

Always try to avoid variable hiding or other hard-to-read constructs. If you want your code to be maintainable, write it in such a way that everybody can understand it immediately. Even if you are the only one; you might forget what you meant in time.