Why only 1 public class in Java file


Java Problem Overview

In any Java file, why can we have only one public class whose name is same as the Java file name?

Java Solutions

Solution 1 - Java

It forces all Java code to be organized in a certain way, which in the long run helps improve code readability.

The Java designers chose a strict approach that enforces their idea of good design practices, and this is part of that theme. Contrast that with the anything-goes attitude in Perl.

Solution 2 - Java

According to this source, it is for efficient compilation :

> In the sidebar it explains why: "This > restriction is not yet enforced by the > compiler, although it's necessary for > efficient package importation" > > It's pretty obvious - like most things > are once you know the design reasons - > the compiler would have to make an > additional pass through all the > compilation units (.java files) to > figure out what classes were where, > and that would make the compilation > even slower.

The same applies also for imports of source files in IDEs. Another reason would be reasonable source sizes.

Solution 3 - Java

These are the rules. Although it is not quite true. You can define internal classes inside you "main" class like this:

public class A {  
   public class B {  

Solution 4 - Java

Courtesy of Dr Heinz Kabutz and his excellent newsletter....

> Why is each public class in a separate > file? > > This is a question that I have > frequently been asked during my > courses. Up to now I have not had a > good answer to this question. In > section 1, we read: "Although each Oak > compilation unit can contain multiple > classes or interfaces, at most one > class or interface per compilation > unit can be public". > > In the sidebar it explains why: "This > restriction is not yet enforced by the > compiler, although it's necessary for > efficient package importation" > > It's pretty obvious - like most things > are once you know the design reasons - > the compiler would have to make an > additional pass through all the > compilation units (.java files) to > figure out what classes were where, > and that would make the compilation > even slower.

Solution 5 - Java

We can have only one top level public either class or interface in any java compilation unit ( .java source file ).

But there can be any number of default classes/interfaces per src file.


JLS leaves the option to the java compiler. And most of the compiler implementations force to have file name same as :

(1) the public class/interface name

(2) if there is a main method and no public class then any name

(3) If there is main method and public class then main method should be in that public class

(4) if there is no public class and no main method then any valid name which may or may not be matching with the class/interface names in the file.

From (2): If two public classes allowed, we should give the file two names which is terribly meaningless to file system. From (3): If two public classes allowed, we should have two main methods which is terribly meaningless to java

Hence a Java source file can have one only public class.

I think the above 4 points are forced by compiler to make the job of both compiler and jvm to find particular java source file or class file easy-so-quick for the compilation/loading/linking. Java has such built in restrictions which developers should follow to have better programming.

Source: My readings and understanding.

Solution 6 - Java

To understand the basic reason behind these restrictions, let's suppose compiler doesn't give compile error for not naming file name same as public class name.

Suppose there is a package A

      /   \
file1.java   file2.java


package A;

class file1
  public static void main(String args[])


public class file3
 public static void main(String args[])


Now as we know that a public class can also be accessed outside the package, now it will become the responsibility of a developer to make it accessible to the outside world. Let's see how:

Suppose package A is containing only Java files(no class files) and some class outside the package A tries to access public class file3, compiler will first try to find file3.class ( not available ), then it will try to find file3.java ( not available ). So even though file3 class is public in nature, it is not visible to the outside world. So if a compiler puts the restriction that if a file is containing public class, it should be named same as the public class name, then above issue can be resolved and the developer won't have to think of exposing public class to the outside world.

Compiler also puts the restriction that there should be atmost one public class per Java file, so that every public class can be accessed by the outside world.

Solution 7 - Java

Java utilizes this convention to find class/interface bytecode by starting at the classpath and scanning for the package hierarchy in subdirectories. Filesystem representation of this hierarchy also enforces some basic rules.

  1. Any two Java classes or interfaces in the same package cannot have the same name. File names would conflict.
  2. Any two Java packages in the same parent package could not have the same name. Folder paths would conflict.
  3. A class has visibility to all classes in the same package without modification to the classpath.

Solution 8 - Java

To have an understanding between the compiler and the programmer.It is a rule that the source code must have atmost one public class and that class must contain the main function.So without any confusion/restriction the compiler can access(public) the class and name the class name to the class file.Also since this class contains the main(), executing the class file will give correct flow

Solution 9 - Java

It enables a more efficient lookup of source (.java) and compiled (.class) files during compilation (import directive) and a more efficient classloading during execution. The idea being: if you know the name of a class, you know where it should be found for each classpath entry. No indexing required.

Solution 10 - Java

I think that this may be a possible reason. There can be only one public class in a java file because the name of java file is same as the name of public class.And obviously we can't have a file with two different names.

Solution 11 - Java

It is how it has been designed and I am not sure if the reason is documented. But here is a good reason on why it is done that way. Let's assume we have two public classes Y & Z in a file named Y.java. Now, let's assume a different class X is using Z. Now, when we compile X, the compiler first tries to locate Z.class and if it cannot find it, then it tries to locate Z.java so that it can compile it automatically. But, since we only have Y.java, the class Z cannot be located and hence we get a compilation error. So, we do need to place them in separate files.

Solution 12 - Java

Public modifier with a class is exposed to the world that means any other class which is outside of the package can also access it to resolve dependency. Now this dependency chain containing multiple classes are verified two times - one is while compilation ( one level of dependency is verified) & other one is while execution ( whole dependency chain is verified ).

Now suppose there is public class (class A) whose source file name is something different from class name. Now if some other class (say class B) depends on it , then while compilation of class B , java compiler will surely check where that class A is located & for that compiler has to go thru all the packages in search of class A ,which is time consuming.

Now suppose in the same scenario we give source file name of class A as A.java , then while compiling class B , compiler will search for class A & for that it only need to find source file whose name is A.java , so it doesn't need to go thru all packages ( which may contain multiple classes) it will now only search that package in which a source file A.java is present.

So from compilation time , performance point if view one source file should have one public class.

Solution 13 - Java

Class A{

} public Class B{

} So this is correct becoz A is not public and compiler will not confuse that which class compiler should execute.

public Class A{

} public Class B{


So this is incorrect becoz here compiler get confused which class should execute becoz both are public.


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