'Map Of functions c++
I have made a map of functions. all these functions are void and receive single string parameter.
code:
void f1(string params){...}
void f2(string params){...}
void f3(string params){...}
map<string , void*> funcMap;
funcMap["f1"] =(void*)&f1;
funcMap["f2"] =(void*)&f2;
funcMap["f3"] =(void*)&f3;
how do i call a function? I tried the next code, but id doesn't work:
void (*func)(string) = &funcMap[commandType];
func(commandParam);
I get this error message:
Server.cpp:160:46: error: cannot convert ‘void**’ to ‘void (*)(std::string) {aka void (*)(std::basic_string<char>)}’ in initialization
Solution 1:[1]
Why not just have those as separate classes.
Then have the methods as virtual.
You can then have a map between the string and the base class.
i.e.
class Someoperation
{
virtual void Doit() = 0;
};
map<string, Someopertion> ops;
Then
class MyOp : public Someoperation
{
void Doit() { /* Some code here */}
};
Just add objects
ops["Hello"] = MyOp();
then call it
ops["Hello"].Doit();
Solution 2:[2]
&funcMap[commandType]
Just drop the &. Your compile error was useful here. It had a void** on the right which is because you took the address of a function pointer. You don't want two levels of indirection there.
Solution 3:[3]
Try C++ style. It has overhead for allocation and inheritance, but it's more flexible and extensible if you'll need some more functionality in the future.
#include <iostream>
#include <string>
#include <unordered_map>
#include <memory>
using namespace std;
class Someoperation {
public:
virtual void Doit() = 0;
};
class MyOp1 : public Someoperation {
public:
void Doit() final { cout << "MyOp1" << endl; }
};
class MyOp2 : public Someoperation {
public:
void Doit() final { cout << "MyOp2" << endl; }
};
int main() {
unordered_map<string, unique_ptr<Someoperation> > ops;
ops["1"] = unique_ptr<Someoperation>(new MyOp1);
ops["2"] = unique_ptr<Someoperation>(new MyOp2);
ops["1"]->Doit(); // Out: MyOp1
ops["2"]->Doit(); // Out: MyOp2
return 0;
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Ed Heal |
| Solution 2 | djechlin |
| Solution 3 | snaiffer |
