C++: How to pass a function+arguments as parameter without executing it? - c++

Question

I have 3 algorithms function:
void algo1(int para1, int para2);
void algo2(int para1, int para2);
void algo3(int para1, int para2);
and I want set a timer to test these functions' efficiency
int get_execute_time( void(*f)(int para1, int para2) );
I pass the function as parameter to get_execute_time to avoid abundance, however, the problem is I also need to pass the arguments, which is para1, para2.
so I let timer function change to
get_execute_time( void(*f)(int para1, int para2) ,
int para1, int para2) {
(*f)(para1, para2); // call it
}
and, the code above seems ugly to me, so is there any option to wrap the code and let the code execute later in C++?
so that I can do like this:
// definition
get_execute_time(this_code_will_execute_later) {
this_code_will_execute_later();
};
// usage
get_execute_time(algo1(para1, para2));
Temporary solution: utilize class to implement closure-like thing( inspired from Do we have closures in C++?
class timer {
private:
int para1;
int para2;
public:
timer(int para1, int para2);
operator () (void(*f)(int para1, int para2)) { (*f)(para1, para2) }
}
// thus we can write these following code
timer timer(para1, para2);
std::cout << "run " << timer(algo1) << " ms";
std::cout << "run " << timer(algo2) << " ms"
std::cout << "run " << timer(algo3) << " ms"
So, Do exist better options? Thanks anyway!

Answer 1

You can do this with either std::bind or via a lambda. For example (just one parameter shown for brevity):
#include <iostream>
#include <functional>
void get_execute_time (std::function <void (void)> f)
{
f ();
}
void foo (int i)
{
std::cout << i << '\n';
}
int main()
{
int param = 42;
auto f1 = std::bind (foo, param);
get_execute_time (f1);
++param;
auto f2 = [param] { foo (param); };
get_execute_time (f2);
}
Output:
42
43
Live demo