Can Function Overloading in C Be Achieved Based on Return Value, and if so, How?

Overriding Function Overloading by Return Value

Introduction

Function overloading in C typically involves differentiating functions based on their parameter types. However, it is also possible to overload functions based on their return values. This allows you to define functions that return different types of data depending on how they are used.

Problem Statement

Consider the following task:

int n = mul(6, 3); // n = 18
std::string s = mul(6, 3); // s = "666"

Here, you need to define a function named mul that returns either an integer or a string, depending on how its return value is used.

Solution Approaches

Explicit Typing:

One approach is to explicitly differentiate the calls by typing. For instance, you could define two separate functions:

int mul(int, int);
std::string mul(char, int);

And call them as follows:

int n = mul(6, 3); // n = 18
std::string s = mul('6', 3); // s = "666"

Dummy Pointer:

Another method involves adding a dummy parameter to each function, forcing the compiler to choose the correct one. For example:

int mul(int *, int, int);
std::string mul(std::string *, char, int);

And use them with:

int n = mul((int *) NULL, 6, 3); // n = 18
std::string s = mul((std::string *) NULL, 54, 3); // s = "666"

Templating the Return Value (Option 1):

You can create a "dummy" function with code that won't compile if instantiated. Then, define specialized template versions for the desired return types:

template<typename T> T mul(int, int);
template<> int mul<int>(int, int);
template<> std::string mul<std::string>(int, int);

Use them as follows:

int n = mul<int>(6, 3); // n = 18
std::string s = mul<std::string>(54, 3); // s = "666"

Templating the Return Value (Option 2):

For cases with different parameters, you can define separate template specializations with distinct parameter types:

template<typename T> T mul(int, int);
template<> int mul<int>(int, int);
template<typename T> T mul(char, int);
template<> std::string mul<std::string>(char, int);

And call them as:

int n = mul<int>(6, 3); // n = 18
std::string s = mul<std::string>('6', 3); // s = "666"

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