Compile-time polymorphism in Java

Introduction

Compile-time polymorphism lets a program decide which method to use before it runs. It helps make code simpler and easier to use.

When you want to perform similar actions but with different inputs.

When you want to use the same method name for different tasks.

When you want to improve code readability by avoiding many method names.

When you want the program to choose the right method automatically based on input types.

When you want to avoid writing repetitive code for similar operations.

Syntax

Java

class ClassName {
    void methodName() { }
    void methodName(int a) { }
    void methodName(int a, int b) { }
}

Methods have the same name but different parameters (number or type).

The compiler decides which method to call based on the arguments used.

Examples

Two methods named add but with different numbers of inputs.

Java

class Calculator {
    int add(int a, int b) {
        return a + b;
    }
    int add(int a, int b, int c) {
        return a + b + c;
    }
}

Methods named print that accept different types of inputs.

Java

class Printer {
    void print(String s) {
        System.out.println(s);
    }
    void print(int n) {
        System.out.println(n);
    }
}

Sample Program

This program shows three show methods with different parameters. The right one runs based on what we give.

Java

class Demo {
    void show() {
        System.out.println("No parameters");
    }
    void show(int a) {
        System.out.println("One parameter: " + a);
    }
    void show(int a, int b) {
        System.out.println("Two parameters: " + a + ", " + b);
    }

    public static void main(String[] args) {
        Demo obj = new Demo();
        obj.show();
        obj.show(5);
        obj.show(3, 7);
    }
}

Important Notes

Compile-time polymorphism is also called method overloading.

The method signature (name + parameters) must be different for overloading.

Return type alone cannot create overloading.

Summary

Compile-time polymorphism means choosing the method to run before the program starts.

It uses the same method name but different parameters.

This makes code easier to read and reuse.

Practice

1.

What is compile-time polymorphism in Java?

easy

A. Using different method names for different tasks

B. Changing the method behavior at runtime based on object type

C. Using the same method name with different parameters in the same class

D. Creating multiple classes with the same name

Solution

1. Step 1: Understand method overloading

   Compile-time polymorphism is also called method overloading, where methods share the same name but differ in parameters.

2. Step 2: Differentiate from runtime polymorphism

   Runtime polymorphism uses method overriding, changing behavior based on object type at runtime, not compile-time.

3. Final Answer:

   Using the same method name with different parameters in the same class -> Option C

4. Quick Check:

   Compile-time polymorphism = method overloading [OK]

Hint: Same method name, different parameters means compile-time polymorphism [OK]

Common Mistakes:

• Confusing compile-time with runtime polymorphism
• Thinking method overriding is compile-time polymorphism
• Believing different method names are polymorphism

2.

Which of the following is the correct syntax for method overloading in Java?

public class Calculator {
    public int add(int a, int b) { return a + b; }
    public int add(int a, int b, int c) { ? }
}

easy

A. return a + b + c;

B. return a + b;

C. return a * b * c;

D. return a - b - c;

Solution

1. Step 1: Check method parameters

   The second add method has three parameters, so it should add all three values.

2. Step 2: Write correct return statement

   Return the sum of a, b, and c to correctly overload the add method.

3. Final Answer:

   return a + b + c; -> Option A

4. Quick Check:

   Overloaded method sums all parameters [OK]

Hint: Overloaded methods must handle all their parameters correctly [OK]

Common Mistakes:

• Returning sum of only two parameters in three-parameter method
• Using wrong operators like multiplication or subtraction
• Syntax errors like missing semicolon

3.

What will be the output of the following Java code?

class Demo {
    void show(int a) { System.out.println("Int: " + a); }
    void show(String a) { System.out.println("String: " + a); }
}
public class Test {
    public static void main(String[] args) {
        Demo d = new Demo();
        d.show(5);
        d.show("Hello");
    }
}

medium

A. Int: 5\nString: Hello

B. String: 5\nInt: Hello

C. Int: 5\nInt: Hello

D. Compilation error

Solution

1. Step 1: Identify overloaded methods

   There are two show methods: one takes int, the other takes String.

2. Step 2: Match method calls to parameters

   d.show(5) calls show(int), printing "Int: 5"; d.show("Hello") calls show(String), printing "String: Hello".

3. Final Answer:

   Int: 5\nString: Hello -> Option A

4. Quick Check:

   Method overloading calls correct method by parameter type [OK]

Hint: Method chosen by parameter type at compile time [OK]

Common Mistakes:

• Confusing parameter types and outputs
• Expecting runtime polymorphism behavior
• Thinking it causes compilation error

4.

Find the error in this code snippet related to compile-time polymorphism:

class Test {
    void display(int a) { System.out.println(a); }
    void display(int a, int b) { System.out.println(a + b); }
    void display(int a) { System.out.println(a * 2); }
}

medium

A. Incorrect method parameter types

B. Duplicate method display(int a) causes compilation error

C. Missing return type in one method

D. No error, code compiles fine

Solution

1. Step 1: Check method signatures

   Two methods have the exact same name and parameter list: display(int a).

2. Step 2: Understand method overloading rules

   Method overloading requires different parameter lists; duplicate signatures cause compilation error.

3. Final Answer:

   Duplicate method display(int a) causes compilation error -> Option B

4. Quick Check:

   Duplicate method signatures cause compile error [OK]

Hint: Overloaded methods must differ in parameter list [OK]

Common Mistakes:

• Thinking method bodies affect overloading
• Ignoring duplicate parameter lists
• Assuming code compiles without error

5.

Consider this class:

class Printer {
    void print(int a) { System.out.println("Number: " + a); }
    void print(String a) { System.out.println("Text: " + a); }
    void print(int a, String b) { System.out.println(a + " and " + b); }
}

Which call will cause a compile-time error?

hard

A. print("Test")

B. print(10, "Hello")

C. print(5)

D. print("Hello", 10)

Solution

1. Step 1: Check method signatures

   Methods accept (int), (String), and (int, String) parameters.

2. Step 2: Match call parameters

   print("Hello", 10) tries to call (String, int), which does not exist, causing compile error.

3. Final Answer:

   print("Hello", 10) -> Option D

4. Quick Check:

   No matching method for (String, int) call [OK]

Hint: Check parameter order and types carefully for overloaded methods [OK]

Common Mistakes:

• Assuming parameter order doesn't matter
• Thinking all combinations are allowed
• Ignoring method signature mismatch