blocks|key|850060|text|使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|850061|def+person+=+new+Person("Kumar",+12)+|code-block|syntax|javascript|850062|您正在定义一个函数/惰性变量，它总是返回一个新的Person实例，名称为"Kumar“，年龄为12岁。这是完全有效的，编译器没有理由抱怨。调用person.age将返回这个新创建的Person实例的年龄，该值始终为12。|850063|在写的时候|850064|person.age+=+45|850065|您为Person类中的age属性分配了一个新值，由于age声明为var，因此该值是有效的。如果您尝试使用新的Person对象重新分配person，编译器将会报错|offset|length|style|CODE|850066|person+=+new+Person("Steve",+13)++//+Error|850067|entityMap^0|0|0|0|0|0|W|3|1U|6|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|11|8|@$O|12|P|13|Q|R]|$O|14|P|15|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|V|$]]

With 

<pre><code>def person = new Person("Kumar", 12) 
</code></pre>

you are defining a function/lazy variable which always returns a new Person instance with name "Kumar" and age 12. This is totally valid and the compiler has no reason to complain. Calling person.age will return the age of this newly created Person instance, which is always 12.

When writing

<pre><code>person.age = 45
</code></pre>

you assign a new value to the age property in class Person, which is valid since age is declared as <code>var</code>. The compiler will complain if you try to reassign <code>person</code> with a new Person object like

<pre><code>person = new Person("Steve", 13) // Error
</code></pre>

blocks|key|1064608|text|正如Kintaro已经说过的，person是一个方法(因为def)，并且总是返回一个新的Person实例。正如您发现的那样，如果您将方法更改为var或val，它将会起作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1064609|val+person+=+new+Person("Kumar",12)|code-block|syntax|javascript|1064610|另一种可能性是：|1064611|def+person+=+new+Person("Kumar",12)
val+p+=+person
p.age=20
println(p.age)|1064612|但是，代码中的person.age=20是允许的，因为您从person方法中返回了一个Person实例，并且在这个实例上，您可以更改var的值。问题是，在那一行之后，您不再有对该实例的引用(因为每次调用person都会生成一个新实例)。|offset|length|style|CODE|1064613|这没什么特别的，您在Java中会有完全相同的行为：|1064614|class+Person{+
+++public+int+age;+
+++private+String+name;
+++public+Person(String+name;+int+age)+{
++++++this.name+=+name;++
++++++this.age+=+age;
+++}
+++public+String+name(){+return+name;+}
}

public+Person+person()+{+
++return+new+Person("Kumar",+12);+
}

person().age+=+20;
System.out.println(person().age);+//-->+12|1064615|entityMap^0|0|0|0|0|7|D|T|6|17|6|1U|3|2T|6|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@$M|11|N|12|O|P]|$M|13|N|14|O|P]|$M|15|N|16|O|P]|$M|17|N|18|O|P]|$M|19|N|1A|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$]]

As Kintaro already says, person is a method (because of def) and always returns a new Person instance. As you found out it would work if you change the method to a var or val:

<pre><code>val person = new Person("Kumar",12)
</code></pre>

Another possibility would be:

<pre><code>def person = new Person("Kumar",12)
val p = person
p.age=20
println(p.age)
</code></pre>

However, <code>person.age=20</code> in your code is allowed, as you get back a <code>Person</code> instance from the <code>person</code> method, and on this instance you are allowed to change the value of a <code>var</code>. The problem is, that after that line you have no more reference to that instance (as every call to <code>person</code> will produce a new instance).

This is nothing special, you would have exactly the same behavior in Java:

<pre><code>class Person{ 
 public int age; 
 private String name;
 public Person(String name; int age) {
 this.name = name; 
 this.age = age;
 }
 public String name(){ return name; }
}

public Person person() { 
 return new Person("Kumar", 12); 
}

person().age = 20;
System.out.println(person().age); //--&gt; 12
</code></pre>

blocks|key|850298|text|让我们看一下这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|850299|class+Person(val+name:String,var+age:Int+)
def+person+=new+Person("Kumar",12)
person.age=20
println(person.age)|code-block|syntax|javascript|850300|并用等价的代码重写它|850301|class+Person(val+name:String,var+age:Int+)
def+person+=new+Person("Kumar",12)
(new+Person("Kumar",+12)).age_=(20)
println((new+Person("Kumar",+12)).age)|850302|看，def是一种方法。它将在每次调用时执行，并且每次都将返回(a)+new+Person("Kumar",+12)。这些在“赋值”中没有错误，因为它不是真正的赋值，而只是对age_=方法(由var提供)的调用。|offset|length|style|CODE|850303|entityMap^0|0|0|0|0|2|3|Y|N|2E|5|2N|3|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]|$M|11|N|12|O|P]|$M|13|N|14|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|R|$]]

Let's take this:

<pre><code>class Person(val name:String,var age:Int )
def person =new Person("Kumar",12)
person.age=20
println(person.age)
</code></pre>

and rewrite it with equivalent code

<pre><code>class Person(val name:String,var age:Int )
def person =new Person("Kumar",12)
(new Person("Kumar", 12)).age_=(20)
println((new Person("Kumar", 12)).age)
</code></pre>

See, <code>def</code> is a method. It will execute each time it is called, and each time it will return (a) <code>new Person("Kumar", 12)</code>. And these is no error in the "assignment" because it isn't really an assignment, but just a call to the <code>age_=</code> method (provided by <code>var</code>).

<pre><code>class Person(val name:String,var age:Int )
def person = new Person("Kumar",12)
person.age = 20
println(person.age)
</code></pre>

These lines of code outputs <code>12</code>, even though <code>person.age=20</code> was successfully executed. I found that this happens because I used def in <code>def person = new Person("Kumar",12)</code>. If I use var or val the output is <code>20</code>. I understand the default is val in scala. This:

<pre><code>def age = 30
age = 45
</code></pre>

...gives a compilation error because it is a val by default. Why do the first set of lines above not work properly, and yet also don't error?

Use of def, val, and var in scala

编译器

Java

class Person(val name:String,var age:Int )def person = new Person("Kumar",12)person.age = 20println(person.age)即使成功执行了person.age=20，这些代码行也会输出12。我发现发生这种情况是因为我在de...

问在scala中使用def、val和var
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