How to call a function by its name (std::string) in C++?

Question

I wonder if there is a simple way to call a function from a string. I know a simple way, using 'if' and 'else'.

int function_1(int i, int j) {     return i*j; }  int function_2(int i, int j) {     return i/j; }  ... ... ...  int function_N(int i, int j) {     return i+j; }  int main(int argc, char* argv[]) {     int i = 4, j = 2;     string function = "function_2";     cout << callFunction(i, j, function) << endl;     return 0; } 

This is the basic approach

int callFunction(int i, int j, string function) {     if(function == "function_1") {         return function_1(i, j);     } else if(function == "function_2") {         return function_2(i, j);     } else if(...) {      } ...     ...     ...     ...     return  function_1(i, j); } 

Is there something simpler?

/* New Approach */ int callFunction(int i, int j, string function) {     /* I need something simple */     return function(i, j); } 

Answer 1

What you have described is called reflection and C++ doesn't support it. However you might come with some work-around, for example in this very concrete case you might use an std::map that would map names of functions (std::string objects) to function pointers, which in case of functions with the very same prototype could be easier than it might seem:

#include <iostream> #include <map>  int add(int i, int j) { return i+j; } int sub(int i, int j) { return i-j; }  typedef int (*FnPtr)(int, int);  int main() {     // initialization:     std::map<std::string, FnPtr> myMap;     myMap["add"] = add;     myMap["sub"] = sub;      // usage:     std::string s("add");     int res = myMap[s](2,3);     std::cout << res; } 

Note that myMap[s](2,3) retrieves the function pointer mapped to string s and invokes this function, passing 2 and 3 to it, making the output of this example to be 5