I need to save in Arrays method pointers, something like this:
int main() {
void* man[10];
man[0]= void* hello();
man[0](2);
}
void hello(int val){
}
The question is, i can do that?
Thanks
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2 Answers
Yes, you can easily achieve this by creating an array of function pointers. This is most readable if you alias the function type first:
void hello(int);
void world(int);
int main()
{
using fn = void(int);
fn * arr[] = { hello, world };
}
Usage:
fn[0](10);
fn[1](20);
Without a separate alias the syntax is a little hairer:
void (*arr[])(int) = { hello, world };
Or:
void (*arr[2])(int);
arr[0] = hello;
arr[1] = world;
2 Comments
user4581301
using... it's probably going to be a few more years before I remember I can do that.
erip
@user4581301 You'd better learn it! It's incredibly helpful. :)
Yes you can, but may I recommend std::function? It handles more complex cases, such as pointers to class methods, much more easily.
Here's an example of both the function pointer way and the std::function way:
#include <iostream>
#include <functional> // for std::function
//typedef void (*funcp)(int); // define a type. This makes everything else way, way easier.
//The above is obsolete syntax.
using funcp = void(*)(int); // Welcome to 2011, monkeyboy. You're five years late
void hello(int val) // put the function up ahead of the usage so main can see it.
{
std::cout << val << std::endl;
}
int main()
{
funcp man[10];
man[0]= hello;
man[0](2);
std::function<void(int)> man2[10]; // no need for typdef. Template takes care of it
man2[0] = hello;
man2[0](3);
}
2 Comments
Kerrek SB
typedef is really obsolete in C++. using is a more powerful and more readable alternative.user4581301
Just berated myself above for not using it.