I have a function foo:
void foo(int n) {}
and I want to have a pointer that points to the function, but calls it with a specified parameter. So basically, something like this:
auto bar = //init
bar(); //calls foo(2)
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1This can't be done with a regular function pointer without the use of nonportable features.Bartek Banachewicz– Bartek Banachewicz2016-04-12 15:13:02 +00:00Commented Apr 12, 2016 at 15:13
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2 Answers
You can use std::bind:
auto bar = std::bind(foo, 2);
then you can call it like so:
bar();
2 Comments
Ace shinigami
if I want to make a vector of these how would I do that
Rakete1111
@Julian
bar would have a return type of void and no arguments. You can use std::function as a type to the vector: std::vector<std::function<void(void)>> v;If you really want a pointer, c++11 lambdas with no captures are convertible to a function pointer like so
#include <iostream>
typedef void(*functype)();
void foo(int n)
{
std::cout << n;
}
int main()
{
functype ptr = [](){foo(2);};
ptr();
return 0;
}
