Is it possible to assign a function to function pointer dynamically?

Question

Is it possible to somehow declare and assign a function to function pointer func in main() without having an actual function add()?

Current Code:

#include <iostream>

int add(int a, int b)
{
    return a + b;
}

int main()
{
    typedef int (*funcPtr)(int a, int b);
    funcPtr func = &add;
    std::cout << func(2,3) << std::endl;
}

Preferred Style: (if possible)

#include <iostream>

int main()
{
    typedef int (*funcPtr)(int a, int b);
    funcPtr func = (funcPtr){return a + b}; // is something like this possible?
    std::cout << func(2,3) << std::endl;
}

Is there a way to assign a function to function pointer dynamically like my last code?

@stark Oh, I don't know what what you mean then. I'm trying to assign a function to a function pointer like I wrote. — Zack Lee
– Zack Lee, Commented Jul 13, 2019 at 15:28
Sounds like what you want is closer to const auto func = [](const auto a, const auto b){ return a + b; }; — Jesper Juhl
– Jesper Juhl, Commented Jul 13, 2019 at 15:31

songyuanyao · Accepted Answer · 2019-07-13 15:28:49Z

5

You can use lambda; which could convert to function pointer implicitly if capture nothing.

funcPtr func = [](int a, int b) {return a + b;};

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edited Jul 13, 2019 at 15:28

answered Jul 13, 2019 at 15:23

songyuanyao

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