Java Function Call with Overload

up vote
7
down vote

favorite

I want to know why the third output is NOT b.

Here is my code:

public class SimpleTests 
 public void func(A a) 
 System.out.println("Hi A");
 
 public void func(B b) 
 System.out.println("Hi B");
 
 public static void main(String args) 
 A a = new A();
 B b = new B();
 A c = new B();
 SimpleTests i = new SimpleTests();
 i.func(a);
 i.func(b);
 i.func(c);
 

class A 
class B extends A

And here is the output:

Hi A
Hi B
Hi A

Could someone tell me why the 3rd output is Hi A, NOT Hi B. as the real c is a instance of B.

edited 1 hour ago

khelwood

28.6k63858

asked 1 hour ago

cceasy

435

New contributor

See stackoverflow.com/questions/383947/â€¦
â€“Â Raedwald
58 mins ago

add a commentÂ |Â

up vote
7
down vote

favorite

I want to know why the third output is NOT b.

Here is my code:

public class SimpleTests 
 public void func(A a) 
 System.out.println("Hi A");
 
 public void func(B b) 
 System.out.println("Hi B");
 
 public static void main(String args) 
 A a = new A();
 B b = new B();
 A c = new B();
 SimpleTests i = new SimpleTests();
 i.func(a);
 i.func(b);
 i.func(c);
 

class A 
class B extends A

And here is the output:

Hi A
Hi B
Hi A

Could someone tell me why the 3rd output is Hi A, NOT Hi B. as the real c is a instance of B.

edited 1 hour ago

khelwood

28.6k63858

asked 1 hour ago

cceasy

435

New contributor

See stackoverflow.com/questions/383947/â€¦
â€“Â Raedwald
58 mins ago

add a commentÂ |Â

up vote
7
down vote

favorite

I want to know why the third output is NOT b.

Here is my code:

public class SimpleTests 
 public void func(A a) 
 System.out.println("Hi A");
 
 public void func(B b) 
 System.out.println("Hi B");
 
 public static void main(String args) 
 A a = new A();
 B b = new B();
 A c = new B();
 SimpleTests i = new SimpleTests();
 i.func(a);
 i.func(b);
 i.func(c);
 

class A 
class B extends A

And here is the output:

Hi A
Hi B
Hi A

Could someone tell me why the 3rd output is Hi A, NOT Hi B. as the real c is a instance of B.

edited 1 hour ago

khelwood

28.6k63858

asked 1 hour ago

cceasy

435

New contributor

I want to know why the third output is NOT b.

Here is my code:

public class SimpleTests 
 public void func(A a) 
 System.out.println("Hi A");
 
 public void func(B b) 
 System.out.println("Hi B");
 
 public static void main(String args) 
 A a = new A();
 B b = new B();
 A c = new B();
 SimpleTests i = new SimpleTests();
 i.func(a);
 i.func(b);
 i.func(c);
 

class A 
class B extends A

And here is the output:

Hi A
Hi B
Hi A

Could someone tell me why the 3rd output is Hi A, NOT Hi B. as the real c is a instance of B.

java method-overloading

edited 1 hour ago

khelwood

28.6k63858

asked 1 hour ago

cceasy

435

New contributor

edited 1 hour ago

khelwood

28.6k63858

asked 1 hour ago

cceasy

435

New contributor

edited 1 hour ago

khelwood

28.6k63858

edited 1 hour ago

khelwood

28.6k63858

edited 1 hour ago

khelwood

28.6k63858

asked 1 hour ago

cceasy

435

New contributor

asked 1 hour ago

cceasy

435

asked 1 hour ago

cceasy

435

New contributor

cceasy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

See stackoverflow.com/questions/383947/â€¦
â€“Â Raedwald
58 mins ago

add a commentÂ |Â

See stackoverflow.com/questions/383947/â€¦
â€“Â Raedwald
58 mins ago

See stackoverflow.com/questions/383947/â€¦
â€“Â Raedwald
58 mins ago

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
7
down vote

Calls to overloaded methods are resolved based on the reference type (A) of the argument at compile time, not the object type (B) at runtime. You declared the variable to be of type A, so it is treated as type A.

edited 58 mins ago

answered 1 hour ago

khelwood

28.6k63858

add a commentÂ |Â

up vote
1
down vote

You're confusing overloading with polymorphism.

With polymorphism, when creating an instance of class B which is a subclass of class A, referenced to by an class A object, and overwrites the method of class A, calling the method will perform the method of class B.

With overloading, the called method only knows the type of the declaration of the argument, not the initialization.

public class A 
 public void print() 
 System.out.println("A");
 


public class B extends A 
 @Override
 public void print() 
 System.out.println("B");
 
 

public class Main 
 public static void main(String args) 
 A a = new A();
 B b = new B();
 A otherB = new B();
 a.print();
 b.print();
 otherB.print();

This will output

A
B
B

answered 1 hour ago

SBylemans

589216

add a commentÂ |Â

up vote
1
down vote

JLS Ã‚Â§8.4.9. Overloading:

When a method is invoked, the number of actual arguments (and any
explicit type arguments) and the compile-time types of the arguments
are used, at compile time, to determine the signature of the method
that will be invoked.

The compile-time types in your example are:

A a = new A(); // A
B b = new B(); // B
A c = new B(); // A

answered 11 mins ago

Oleksandr

7,34533366

add a commentÂ |Â

up vote
0
down vote

Reason being as said by khelwood, further to this if you want to return the Output.

Hi A

Hi B

Just pass the object instead of reference, It will then call function B, as there is no resolution at compile time.

public static void main(String args) 
 A a = new A();
 B b = new B();
 A c = new B();
 test i = new test();
 i.func(a);
 i.func(b);
 i.func(new B());

answered 59 mins ago

Mohit Sharma

688

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
7
down vote

edited 58 mins ago

answered 1 hour ago

khelwood

28.6k63858

add a commentÂ |Â

up vote
7
down vote

edited 58 mins ago

answered 1 hour ago

khelwood

28.6k63858

add a commentÂ |Â

up vote
7
down vote

edited 58 mins ago

answered 1 hour ago

khelwood

28.6k63858

edited 58 mins ago

answered 1 hour ago

khelwood

28.6k63858

edited 58 mins ago

answered 1 hour ago

khelwood

28.6k63858

answered 1 hour ago

khelwood

28.6k63858

answered 1 hour ago

khelwood

28.6k63858

add a commentÂ |Â

up vote
1
down vote

You're confusing overloading with polymorphism.

With overloading, the called method only knows the type of the declaration of the argument, not the initialization.

public class A 
 public void print() 
 System.out.println("A");
 


public class B extends A 
 @Override
 public void print() 
 System.out.println("B");
 
 

public class Main 
 public static void main(String args) 
 A a = new A();
 B b = new B();
 A otherB = new B();
 a.print();
 b.print();
 otherB.print();

This will output

A
B
B

answered 1 hour ago

SBylemans

589216

add a commentÂ |Â

up vote
1
down vote

You're confusing overloading with polymorphism.

With overloading, the called method only knows the type of the declaration of the argument, not the initialization.

public class A 
 public void print() 
 System.out.println("A");
 


public class B extends A 
 @Override
 public void print() 
 System.out.println("B");
 
 

public class Main 
 public static void main(String args) 
 A a = new A();
 B b = new B();
 A otherB = new B();
 a.print();
 b.print();
 otherB.print();

This will output

A
B
B

answered 1 hour ago

SBylemans

589216

add a commentÂ |Â

up vote
1
down vote

You're confusing overloading with polymorphism.

With overloading, the called method only knows the type of the declaration of the argument, not the initialization.

public class A 
 public void print() 
 System.out.println("A");
 


public class B extends A 
 @Override
 public void print() 
 System.out.println("B");
 
 

public class Main 
 public static void main(String args) 
 A a = new A();
 B b = new B();
 A otherB = new B();
 a.print();
 b.print();
 otherB.print();

This will output

A
B
B

answered 1 hour ago

SBylemans

589216

You're confusing overloading with polymorphism.

With overloading, the called method only knows the type of the declaration of the argument, not the initialization.

public class A 
 public void print() 
 System.out.println("A");
 


public class B extends A 
 @Override
 public void print() 
 System.out.println("B");
 
 

public class Main 
 public static void main(String args) 
 A a = new A();
 B b = new B();
 A otherB = new B();
 a.print();
 b.print();
 otherB.print();

This will output

A
B
B

answered 1 hour ago

SBylemans

589216

answered 1 hour ago

SBylemans

589216

answered 1 hour ago

SBylemans

589216

answered 1 hour ago

SBylemans

589216

add a commentÂ |Â

up vote
1
down vote

JLS Ã‚Â§8.4.9. Overloading:

When a method is invoked, the number of actual arguments (and any
explicit type arguments) and the compile-time types of the arguments
are used, at compile time, to determine the signature of the method
that will be invoked.

The compile-time types in your example are:

A a = new A(); // A
B b = new B(); // B
A c = new B(); // A

answered 11 mins ago

Oleksandr

7,34533366

add a commentÂ |Â

up vote
1
down vote

JLS Ã‚Â§8.4.9. Overloading:

When a method is invoked, the number of actual arguments (and any
explicit type arguments) and the compile-time types of the arguments
are used, at compile time, to determine the signature of the method
that will be invoked.

The compile-time types in your example are:

A a = new A(); // A
B b = new B(); // B
A c = new B(); // A

answered 11 mins ago

Oleksandr

7,34533366

add a commentÂ |Â

up vote
1
down vote

JLS Ã‚Â§8.4.9. Overloading:

When a method is invoked, the number of actual arguments (and any
explicit type arguments) and the compile-time types of the arguments
are used, at compile time, to determine the signature of the method
that will be invoked.

The compile-time types in your example are:

A a = new A(); // A
B b = new B(); // B
A c = new B(); // A

answered 11 mins ago

Oleksandr

7,34533366

JLS Ã‚Â§8.4.9. Overloading:

When a method is invoked, the number of actual arguments (and any
explicit type arguments) and the compile-time types of the arguments
are used, at compile time, to determine the signature of the method
that will be invoked.

The compile-time types in your example are:

A a = new A(); // A
B b = new B(); // B
A c = new B(); // A

answered 11 mins ago

Oleksandr

7,34533366

answered 11 mins ago

Oleksandr

7,34533366

answered 11 mins ago

Oleksandr

7,34533366

answered 11 mins ago

Oleksandr

7,34533366

add a commentÂ |Â

up vote
0
down vote

Reason being as said by khelwood, further to this if you want to return the Output.

Hi A

Hi B

Just pass the object instead of reference, It will then call function B, as there is no resolution at compile time.

public static void main(String args) 
 A a = new A();
 B b = new B();
 A c = new B();
 test i = new test();
 i.func(a);
 i.func(b);
 i.func(new B());

answered 59 mins ago

Mohit Sharma

688

add a commentÂ |Â

up vote
0
down vote

Reason being as said by khelwood, further to this if you want to return the Output.

Hi A

Hi B

Just pass the object instead of reference, It will then call function B, as there is no resolution at compile time.

public static void main(String args) 
 A a = new A();
 B b = new B();
 A c = new B();
 test i = new test();
 i.func(a);
 i.func(b);
 i.func(new B());

answered 59 mins ago

Mohit Sharma

688

add a commentÂ |Â

up vote
0
down vote

Reason being as said by khelwood, further to this if you want to return the Output.

Hi A

Hi B

Just pass the object instead of reference, It will then call function B, as there is no resolution at compile time.

public static void main(String args) 
 A a = new A();
 B b = new B();
 A c = new B();
 test i = new test();
 i.func(a);
 i.func(b);
 i.func(new B());

answered 59 mins ago

Mohit Sharma

688

Reason being as said by khelwood, further to this if you want to return the Output.

Hi A

Hi B

Just pass the object instead of reference, It will then call function B, as there is no resolution at compile time.

public static void main(String args) 
 A a = new A();
 B b = new B();
 A c = new B();
 test i = new test();
 i.func(a);
 i.func(b);
 i.func(new B());

answered 59 mins ago

Mohit Sharma

688

answered 59 mins ago

Mohit Sharma

688

answered 59 mins ago

Mohit Sharma

688

answered 59 mins ago

Mohit Sharma

688

add a commentÂ |Â

cceasy is a new contributor. Be nice, and check out our Code of Conduct.

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