blocks|key|760693|text|没有区别，这只是一个明示和含蓄的问题。在第二种情况下，如果从父类继承的话，它会隐式调用这个学生类的超级构造函数。读更多关于这个的文章。你可以找到更多的信息，如果你谷歌它。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|760694|public+class+A+{
++++//there+is+a+hidden+constructor.+Even+if+you+explicitly+write+it++
++++//public+A(){
++++//}
}

public+class+B+extends+A+{
+private+int+i;


public+B(int+x){
+++this.i+=+x;
}

}|code-block|syntax|javascript|760695|创建对象表单B时，首先隐式调用A的构造函数。但是您可以显式地指定它。|760696|作为|760697|public+B(int+x){
+++++++super();
+++++++this.i+=+x;
}|760698|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

No Difference, Its just a matter of explicit and implicit. In the second case, it implicitly calls the super constructor of this Students class if it is inherited from a parent class. Read more on this. You can find more info if you google it.

<pre><code>public class A {
 //there is a hidden constructor. Even if you explicitly write it 
 //public A(){
 //}
}

public class B extends A {
 private int i;


public B(int x){
 this.i = x;
}

}
</code></pre>

when you create an Object form B you first call A's constructor implicitly. You can explicitly specify it though.

as 

<pre><code>public B(int x){
 super();
 this.i = x;
}
</code></pre>

blocks|key|760706|text|如果没有对超类构造函数的显式调用，编译器将生成对super()的调用。正因为如此，你们的例子没有区别。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|760707|entityMap^0|O|7|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

If there is no explicit call to a super class constructor, the compiler will generate a call to <code>super()</code>. Because of this there is no difference between your examples.

blocks|key|1153660|text|super()只调用函数的父构造函数，并允许您传递参数(如super(someArgument)+)或用于调用父构造函数(如super.someMethod()+)的方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1153661|单独的super()只调用父构造函数，而不传递任何参数或引用父构造函数的任何方法。何时调用super()的实际决定取决于重写方法的用途以及在重写某个方法时基类所期望的行为。|1153662|查看Java™教程或JavaTPoint以了解更多信息。|1153663|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/tutorial/java/IandI/super.html|1|http://www.javatpoint.com/super-keyword^0|0|7|T|J|1Q|I|0|3|7|19|7|0|2|7|0|A|A|1|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|15|8|@]|D|@$9|16|A|17|1|18]|$9|19|A|1A|1|1B]]|E|$]]|$1|J|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|K|$L|$5|M|N|O|E|$P|Q]]|R|$5|M|N|O|E|$P|S]]]]

<code>super()</code> just calls the parent constructor of the function and lets you either pass arguments such as <code>super(someArgument)</code> or can be used to call methods of the parent constructor such as <code>super.someMethod()</code>. 

<code>super()</code> alone just calls the parent constructor without passing any arguments or referencing any methods of the parent. The actual decision on when to call <code>super()</code> depends on the purpose of the overridden method and also the behaviour expected by the base class when you are overriding a certain method.

Check out <a href="https://docs.oracle.com/javase/tutorial/java/IandI/super.html" rel="nofollow">The Java™ Tutorials</a> or <a href="http://www.javatpoint.com/super-keyword" rel="nofollow">JavaTPoint</a> to learn more.

blocks|key|757758|text|Java隐式调用超类的构造函数，如果您没有显式调用它(构造函数链接-搜索这个术语在google上获取更多信息)。即使您的类没有从任何类显式继承，java也隐式地使它继承对象类(并调用对象类的构造函数)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|757759|所以，他们之间没有区别。|757760|entityMap^0|R|6|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

Java implicitly calls the constructor of the super class, if you haven't called it explicitly (constructor chaining - search this term on google for more info). Even if your class is not explicitly inherited from any class, java implicitly makes it inherit the Object class (and makes a call to the constructor of the Object class).

So, there's no difference between them.

blocks|key|1153737|text|有超功能的构造和没有超功能的构造有什么区别？|type|blockquote|depth|inlineStyleRanges|entityRanges|data|1153738|它可以很容易的测试。在这个案件中，它们实际上是一样的。更重要的是，您可能想知道何时以及如何使用超级。|unstyled|offset|length|1153739|运行以下代码：|1153740|public+class+Test
{+++++++++++++
++++public+static+void+main(String[]+args)+
++++{+++++++
++++++++new+Students("a",+"b");
++++}+
}

class+Parent
{
++++public+Parent(){
++++++++System.out.println("Parent+class+constructor+invoked");
++++}
}+++++++

class+Students+extends+Parent
{
++++String+name;
++++String+address;

++++public+Students(String+name,+String+address)+{
++++++++super();
++++++++this.name+=+name;
++++++++this.address+=+address;
++++}
}|code-block|syntax|javascript|1153741|输出:调用父类构造函数|style|BOLD|1153742|现在我们删除super()；|CODE|1153743|class+Students+extends+Parent
{
++++String+name;
++++String+address;

++++public+Students(String+name,+String+address)+{
++++++++this.name+=+name;
++++++++this.address+=+address;
++++}
}|1153744|1153745|就像类是如何隐式扩展到Object一样，即使您没有将它们扩展到某些东西。“默认情况下”，如果您没有给出一个构造函数，那么每个构造函数中都会调用一个super();。|1153746|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/tutorial/java/IandI/super.html^0|0|1B|2|0|0|0|0|0|5|0|6|7|0|0|0|5|0|B|6|21|8|0^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|18|8|@]|9|@$E|19|F|1A|1|1B]]|A|$]]|$1|G|3|H|5|D|7|1C|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|1D|8|@]|9|@]|A|$L|M]]|$1|N|3|O|5|D|7|1E|8|@$E|1F|F|1G|P|Q]]|9|@]|A|$]]|$1|R|3|S|5|D|7|1H|8|@$E|1I|F|1J|P|T]]|9|@]|A|$]]|$1|U|3|V|5|K|7|1K|8|@]|9|@]|A|$L|M]]|$1|W|3|O|5|D|7|1L|8|@$E|1M|F|1N|P|Q]]|9|@]|A|$]]|$1|X|3|Y|5|D|7|1O|8|@$E|1P|F|1Q|P|T]|$E|1R|F|1S|P|T]]|9|@]|A|$]]|$1|Z|3|-4|5|D|7|1T|8|@]|9|@]|A|$]]]|10|$11|$5|12|13|14|A|$15|16]]]]

<blockquote>
 what is difference between a construct having super function and without it.
</blockquote>

Well, it can be easily tested. They are actually the same for this case.
More importantly, you probably want to know when and how to use <a href="https://docs.oracle.com/javase/tutorial/java/IandI/super.html" rel="nofollow">super</a>.

Running the following codes:

<pre><code>public class Test
{ 
 public static void main(String[] args) 
 { 
 new Students("a", "b");
 } 
}

class Parent
{
 public Parent(){
 System.out.println("Parent class constructor invoked");
 }
} 

class Students extends Parent
{
 String name;
 String address;

 public Students(String name, String address) {
 super();
 this.name = name;
 this.address = address;
 }
}
</code></pre>

OUTPUT:
Parent class constructor invoked

Now we remove the <code>super()</code>;

<pre><code>class Students extends Parent
{
 String name;
 String address;

 public Students(String name, String address) {
 this.name = name;
 this.address = address;
 }
}
</code></pre>

OUTPUT:
Parent class constructor invoked

Just like how classes are implicitly extending to <code>Object</code> even when you don't extend them to something. "By default" a <code>super();</code> will be called in every constructor if one is not given by you.

What is the difference between these two constructors.

<pre><code>public Students(String name, String address) {
 super();
 this.name = name;
 this.address = address;
}

/*Here is constructor without super function call.*/
public Students(String name, String address) {
 this.name = name;
 this.address = address;
}
</code></pre>

What is the difference between a constructor having super() and not having super() function call in a Java Class

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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这两个构造函数之间有什么区别。public Students(String name, String address) {    super();    this.name = name;    this.address = address;}/*Here is constructor without super fu...

问在Java类中，具有super()函数调用的构造函数与没有超级()函数调用的构造函数之间有什么区别？
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问在Java类中，具有super()函数调用的构造函数与没有超级()函数调用的构造函数之间有什么区别？EN