What is the ambiguity problem in Java?
The ambiguity problem in Java occurs when a method or constructor is overloaded with two or more methods with the same name but different parameters. This can confuse when multiple methods with the same name exist since the compiler cannot determine which method should be called. The ambiguity problem in Java is a problem that arises when Java code is written in such a way that it can be interpreted in more than one way. This can lead to unexpected results and can be difficult to debug. Examples of ambiguity problems include code that uses the same variable name for different purposes in different parts of the code, code that uses multiple nested if-statements to evaluate conditions, and code that uses multiple overloaded methods with different parameters. Other ambiguity problems can arise when a programmer uses language features such as generics, enums, and annotations.
The challenges the Java language definition does not explicitly describe are known as ambiguities. Our views are supported by the various outputs generated by various compilers on several example programs. Here, we'll talk about things in this order.
Method of ambiguity in method overloading
- only use Varargs in your methods
- methods that use Varargs and other parameters
Ambiguity in multiple inheritances
- Diamond Problem
Type 1: Method overloading with ambiguity
When you overload a method, you risk putting the compiler in a scenario where it is unsure which method to employ. Consider the following declarations for the overloaded computeBalance() method:
File Name:AmbiguityInJava.java
// Main class declaration
public class AmbiguityInJava {
// static function named fun with double variables as parameters
static void fun(double ... Val){
System.out.print("fun(double ...): " + "Number of args: " + Val.length );
// using each loop to iterate over the parameterized variable
for(double a : Val)
System.out.print(a + " ");
System.out.println();
}
// static function named fun with boolean variables as parameters
static void fun(boolean ... Val){
System.out.print("fun(boolean ...) " + "The number of arguments: " + Val.length);
for(boolean a : Val)
// // printing the each value
System.out.print(a + " ");
// // printing the new line
System.out.println();
}
// Main section of the program where execution begins
public static void main(String s[]){
// calling the function
fun(11.56, 34.78);
// printing the function call
System. out.println("Calling fun() ");
// Calling the function
fun(true, false);
// printing the function call
System.out.println("Calling fun() ");
// Calling the function
fun();
// printing the function call
System.out.println("Calling fun() ");
}
}
Output
AmbiguityInJava.java:19: error: reference to fun is ambiguous
fun();
^
both method fun(double...) in Main and method fun(boolean...) in Main match
1 error
Explanation
A function called "fun" that accepts a configurable number of floating point numbers is defined in the Demo class. Using a "for" loop, the values are printed on the console. The parameters of this function are boolean values with changing numbers, which is overloaded. Using a "for" loop, the output is displayed on the console.
The word "fun" is called three times in the main function: once with floating point values, once with Boolean values, and once without any parameters. The console shows the exception that it generates.
Ambiguity Errors
Using generics creates a new kind of error called ambiguity, which you must prevent. Ambiguity errors happen when two generic declarations that seem mutually exclusive resolve to the same erased type, resulting in a conflict. Here is an instance of method overloading in action.
With kind 1, as described above, we are currently good to go. Overloading of methods in varargs.
Due to overloading, various methods with the same name but distinct signatures are possible. The signatures can vary depending on the input parameters' quantity, nature, or both. The following strategies can be used to overload a method that accepts a variable-length argument:
Case 1: Only Varargs parameters are used in methods
Java uses the type difference in this situation to choose which overloaded method to call. If one method signature is objectively stricter than the other, Java selects the stricter one without making a mistake.
File Name: Ambiguity2.java
// class declaration
public class Ambiguity2
{
static void fun(float... a)
{
System.out.println("float varargs");
}
// function with integer parameters
static void fun(int... a)
{
System.out.println("int varargs");
}
// function with double parameters
static void fun(double... a)
{
System.out.println("double varargs");
}
// Main section of the program from where execution begins
public static void main( String[] s)
{
// Calling the method
fun();
}
}
Output
int varargs
Explanation
Int is more specific than double, which results in this outcome. According to JLS section 15.12.2.5, one member method must be chosen to supply the descriptor for the run-time method dispatch if more than one is available and appropriate to a method invocation. The most specific approach is selected based on the Java programming language type promotion. The following rules define the direct supertype relation among the primitive types in this situation:
The following order is used: following order is used:
double > float
float > long
long > int
int > char
int > short
short > byte
Case 2
File Name: Ambiguities.java
class Ambiguities {
static void fun(int... x)
{
System.out.print("fun(int ...): "+ "Number of args: " + x.length+ " Contents: ");
for (int a: x)
System.out.print(a + " ");
System.out.println();
}
static void fun(boolean... y)
{
System.out.print("fun(boolean ...) "+ "Number of args: " + y.length + " Contents: ");
for (boolean x : y)
System.out.print(y + " ");
System.out.println();
}
public static void main(String s[])
{
fun(6, 4, 3);
fun(true, true, false);
}
}
Output
fun(int ...): Number of args: 3 Contents: 6 4 3
fun(boolean ...) Number of args: 3 Contents: [[email protected] [[email protected] [[email protected]
Example 2
Program
File Name:AmbiguityProblem.java
class AmbiguityProblem{
static void func(int... x)
{
System.out.print("fun(int ...): "+ "Number of args: " + x.length+ " Contents: ");
for (int a : x)
// Printing statement
System.out.print(a + " ");
// New line
System.out.println();
}
static void func(boolean... x)
{
System.out.print("fun(boolean ...) "+ "Number of args: " + x.length+ " Contents: ");
for (boolean a : x)
System.out.print(a + " ");
System.out.println();
}
public static void main(String args[])
{
func(5, 7, 13);
func(false, true, false);
func();
}
}
Output
Main.java:28: error: reference to func is ambiguous
func();
^
both method func(int...) in Main and method func(boolean...) in Main match
1 error
Explanation
Explanation of output: This instance's fun () overloading the desired method is ideal. However, the final call to "fun()," which can also be interpreted in the code, prevents this program from compiling.
Type 2
Complex inheritances with ambiguity. When two classes are related, they can inherit each other's properties. The extends keyword can be used to define this relationship as follows:
public class A extends C{}
A superclass or parent class is a class from which properties are inherited, and a subclass or child class is a class from which those properties are inherited. The superclass members are copied in the sub-class object during inheritance. Therefore, you can access the members using the sub-class object.
Example
File Name: Ambiguity.java
// Java Program to illustrate Diamond Problem Ambiguity
public class abstract Demo {
// Abstract method declaration
public abstract demo();
}
public class S1 extends Demo {
public void demo()
{
System.out.println("demo method of super1");
}
}
// Class 3
// Helper class extending Class 1
public class S2 extends Demo {
// Method of this base class
public void demo()
{
// Print statement whenever this method is called
System.out.println("demo method of super2");
}
}
// Main class extending the classes s1 and s2
public class Ambiguityextends S1, S2 {
// Main section of the program from where execution begins
public static void main(String args[])
{
// Creating object for the class SubClass
SubClass o1 = new SubClass();
// calling the function using the object created
o1.demo();
}
}
Output
Compilation failed due to the following error(s).
Main.java:2: error: expected
public class abstract Demo {
^
Main.java:2: error: expected
public class abstract Demo {
^
Main.java:5: error: illegal start of expression
public abstract demo();
^
Main.java:5: error: invalid method declaration; return type required
public abstract demo();
^
Main.java:31: error: '{' expected
public class Main extends S1, S2 {
^
5 errors
Explanation
The basic inheritance rule states that both demo() methods should be copied into the subclass object, leaving the subclass with two methods that share the same prototype (name and arguments). If you use the object of the subclass to call the demo() function, the compiler will then be in a scenario where it is unclear which method to call. In Java, this problem is referred to as the diamond problem.