Function Pointers and Overloading

Function pointers and function overloading can interact in interesting ways. Let's explore this relationship and some of the challenges it can present.

First, let's remind ourselves what function overloading looks like:

1#include <iostream>
2
3void Print(int x) {
4  std::cout << "Integer: " << x << "\n";
5}
6
7void Print(double x) {
8  std::cout << "Double: " << x << "\n";
9}
10
11int main() {
12  Print(5);
13  Print(3.14);
14}

1Integer: 5
2Double: 3.14

The compiler chooses the correct function based on the argument type.

When using function pointers with overloaded functions, we need to be explicit about which overload we're referring to. The function pointer's type must match exactly with one of the overloads:

1#include <iostream>
2
3void Print(int x) {
4  std::cout << "Integer: " << x << "\n";
5}
6
7void Print(double x) {
8  std::cout << "Double: " << x << "\n";
9}
10
11int main() {
12  void (*intPrinter)(int) = Print; 
13  void (*doublePrinter)(double) = Print; 
14
15  intPrinter(5);
16  doublePrinter(3.14);
17}

1Integer: 5
2Double: 3.14

In this case, we're explicitly choosing which overload to use based on the type of the function pointer.

Overloading can sometimes lead to ambiguity when using function pointers. Consider this example:

1#include <iostream>
2
3void Ambiguous(int x) {
4  std::cout << "Int version: " << x << "\n";
5}
6
7void Ambiguous(long x) {
8  std::cout << "Long version: " << x << "\n";
9}
10
11int main() {
12  // This would be ambiguous:
13  // auto funcPtr = Ambiguous; 
14
15  void (*funcPtr)(int) = Ambiguous; 
16  funcPtr(5);
17}

1Int version: 5

In this case, we have to explicitly specify the type of the function pointer to resolve the ambiguity. Using auto would result in a compilation error because the compiler can't determine which overload to choose.

Function templates add another layer of complexity:

1#include <iostream>
2
3template <typename T>
4void TemplateFunc(T x) {
5  std::cout << "\nTemplate: " << x << "\n";
6}
7
8void TemplateFunc(int x) {
9  std::cout << "\nInt specialization: " << x;
10}
11
12int main() {
13  void (*intPtr)(int) = TemplateFunc; 
14  intPtr(5);
15
16  auto genericPtr = TemplateFunc<double>; 
17  genericPtr(3.14);
18}

1Int specialization: 5
2Template: 3.14

Here, we're able to select between the non-template overload and a specific instantiation of the template function.

In conclusion, while function pointers can work with overloaded functions, you need to be explicit about which overload you're referring to.

This often means specifying the exact type of the function pointer or using explicit template instantiation. Always be mindful of potential ambiguities when working with overloaded functions and function pointers.