c++ - Obtaining function pointer to lambda?

Question

I want to be able to obtain a function pointer to a lambda in C++.

I can do:

int (*c)(int) = [](int i) { return i; };

And, of course, the following works - even if it's not creating a function pointer.

auto a = [](int i) { return i; };

But the following:

auto *b = [](int i) { return i; };

Gives this error in GCC:

main.cpp: In function 'int main()':
main.cpp:13:37: error: unable to deduce 'auto*' from '<lambda closure object>main()::<lambda(int)>{}'
     auto *b = [](int i) { return i; };
                                      ^
main.cpp:13:37: note:   mismatched types 'auto*' and 'main()::<lambda(int)>'

It seems arbitrary that a lambda can be converted to a function pointer without issue, but the compiler cannot infer the function type and create a pointer to it using auto *. Especially when it can implicitly convert a unique, lambda type to a function pointer:

int (*g)(int) = a;

I've create a little test bed at http://coliru.stacked-crooked.com/a/2cbd62c8179dc61b that contains the above examples. This behavior is the same under C++11 and C++14.

question from:https://stackoverflow.com/questions/37709874/obtaining-function-pointer-to-lambda

Answer 1

This fails:

auto *b = [](int i) { return i; };

because the lambda is not a pointer. auto does not allow for conversions. Even though the lambda is convertible to something that is a pointer, that's not going to be done for you - you have to do it yourself. Whether with a cast:

auto *c = static_cast<int(*)(int)>([](int i){return i;});

Or with some sorcery:

auto *d = +[](int i) { return i; };