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Function overloading is a powerful feature of C++ that allows you to define multiple functions with the same name but different parameters. 

It is one of the fundamental concepts in object-oriented programming. It provides a convenient way to create functions that perform similar tasks but operate on different data types or with different sets of arguments. 

In this tutorial, we will understand Function Overloading in C++ in detail with examples to understand how it works and how to use it effectively in your programs.

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C++ Function Overloading

Function overloading is a feature in C++ that allows you to create multiple functions with the same name, but each function can have different parameters. 

In simpler terms, you can think of it as a way to give a single function multiple abilities based on what it is asked to do.

For example,

addNumbers (int, int)
addNumbers(double, double)

The above example demonstrates function overloading by creating two versions of a function named addNumbers.

One version takes two integers as parameters, while the other takes two double values as parameters.

When the program calls the addNumbers function, the compiler will look at the parameters passed to it and decide which version of the function to use based on the data types of those parameters.

Example 1: C++ Implementation of The Above Function Overloading Example

#include <iostream>

using namespace std;

// Function to add two integers
int addNumbers(int a, int b) {
    return a + b;
}

// Function to add two double values
double addNumbers(double a, double b) {
    return a + b;
}

int main() {
    cout << "Sum of integers: " << addNumbers(5, 10) << endl;
    cout << "Sum of doubles: " << addNumbers(3.14, 2.718) << endl;
    return 0;
}

OUTPUT:

Sum of integers: 15
Sum of doubles: 5.858

In this program, we have created two versions of the addNumbers function, one that takes two integers as parameters and returns an integer, and another that takes two doubles as parameters and returns a double. 

In the main() function, we have directly called both versions of the function with different sets of parameters and printed the results to the console.

When we run the program, the compiler will select the appropriate version of the addNumbers function based on the data types of the parameters passed to it.

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How Does The Function Overloading Works?

When a function call is made, the compiler looks at the number, types, and order of the input parameters to determine which version of the function to call.

For example,

We have two functions named add in our program. The first function takes two integer parameters and returns their sum, while the second function takes two double parameters and returns their sum.

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

double add(double a, double b) {
    return a + b;
}

When we call the add function with two integer parameters, the compiler will automatically choose the first version of the function like this:

int sum1 = add(2, 3);  // calls the first "add" function

Similarly, when we call the add function with two double parameters, the compiler will automatically choose the second version of the function like this:

double sum2 = add(2.5, 3.7);  // calls the second "add" function

The compiler is able to choose the correct function based on the number of parameters, types, and order of the input parameters.

It is important to note that the return type of the functions is not considered when the compiler chooses which version of the function to call. If two functions have the same name and input parameters, but different return types, the program will not compile due to ambiguity.

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Function Overloading and Ambiguity

There are cases where function overloading can lead to ambiguity, where the compiler cannot determine which version of the function to call.

There are three common causes of ambiguity in function overloading:

Type Conversion

When the input parameters of two or more overloaded functions are of different types, the compiler may need to convert the types of input parameters to match the function being called. 

However, in some cases, the conversion may not be clear, and the compiler may not be able to determine which function to call.

For example,

void print(int x) {
    cout << "The integer value is: " << x << endl;
}

void print(double x) {
    cout << "The double value is: " << x << endl;
}

int main() {
    print(5);     // calls print(int)
    print(3.14);  // calls print(double)
    print('a');   // ambiguity error, could be interpreted as an int or a char
    return 0;
}

In this example, we have two overloaded functions named print. One takes an integer parameter, and the other takes a double parameter. When we call the print function with an integer or a double parameter, the compiler can determine which function to call based on the input parameter type. 

However, when we call the print function with a character parameter, the compiler would not be able to determine which function to call because a character could be interpreted as either an integer or a character type.

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Function With Default Arguments

When two or more overloaded functions have the same number of input parameters, but one or more of them have default arguments, the compiler may not be able to determine which function to call.

For example,

void print(int x, int y = 0) {
    cout << "The integer values are: " << x << " and " << y << endl;
}

void print(double x, double y = 0.0) {
    cout << "The double values are: " << x << " and " << y << endl;
}

int main() {
    print(5);        // calls print(int, int) with default argument
    print(3.14);     // calls print(double, double) with default argument
    print(5, 3);     // calls print(int, int)
    print(3.14, 2.5); // calls print(double, double)
    return 0;
}

In this example, we have two overloaded functions named print that both take two parameters, but one of them has a default argument. 

When we call the print function with only one parameter, the compiler can determine which function to call based on the input parameter type and whether the default argument needs to be used. 

However, when we call the print function with two parameters, the compiler would not be able to determine which function to call because both functions match the input parameters.

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Function With Pass By Reference

When two or more overloaded functions have the same number and types of input parameters, but one or more of them use pass by reference, the compiler may not be able to determine which function to call.

For example,

void swap(int& x, int& y) {
    int temp = x;
    x = y;
    y = temp;
}

void swap(int* x, int* y) {
    int temp = *x;
    *x = *y;
    *y =
}

In this example, we have two functions named swap() with the same name and the same number of parameters. However, the parameters are different – one takes in two integer references, and the other takes two integer pointers.

The first swap() function takes two integer references as arguments and swaps their values. The second swap() function takes two integer pointers as arguments and swaps the values pointed to by those pointers.

When we try to call the swap() function, the compiler cannot determine which function to use because both functions have the same name and the same number of parameters. If we try to call swap() with integer pointers as arguments, the compiler cannot determine which function to use, which results in an error.

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Example 2: Function Overloading Using Different Number Of Parameters

#include <iostream>
using namespace std;

// First display() function takes in one integer parameter x and prints its value
void display(int x) {
    cout << "The value of x is " << x << endl;
}

// Second display() function takes in two integer parameters x and y and prints their values
void display(int x, int y) {
    cout << "The value of x is " << x << endl;
    cout << "The value of y is " << y << endl;
}

int main() {
    int a = 5;
    int b = 10;

    // Call the first display() function with the parameter a
    display(a);

    // Call the second display() function with the parameters a and b
    display(a, b);

    return 0;
}

OUTPUT:

The value of x is 5
The value of x is 5
The value of y is 10

In this example, we have two functions named display(). The first display() function takes in one integer parameter, x and prints its value to the console. 

The second display() function takes in two integer parameters, x and y and prints their values to the console.

In the main() function, we create two integer variables, a and b and assign them values. We then call the display() function twice, first with the parameter a and then with the parameters a and b.

Since the two display() functions have different numbers of parameters, the compiler can differentiate between them and call the appropriate function based on the number of arguments passed in.