获取指向lambda的函数指针？ [英] Obtaining function pointer to lambda?

问题描述

我希望能够获得指向C ++中的lambda的函数指针。

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

我可以做到：

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

当然，以下内容也可以工作-即使它没有创建函数指针。

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

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

但是以下内容：

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

给出GCC中的此错误：

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)>'

似乎可以将lambda毫无问题地转换为函数指针，但是编译器无法推断函数类型并为其创建指针使用 auto * 。特别是当它可以将唯一的lambda类型隐式转换为函数指针时：

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;

我已经在http://coliru.stacked-crooked.com/a/2cbd62c8179dc61b ，其中包含上述示例。在C ++ 11和C ++ 14下，此行为是相同的。

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.

推荐答案

此操作失败：

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

因为lambda不是指针。 自动不允许进行转化。即使lambda 可以转换为指针的东西，也不会为您完成-您必须自己做。是否使用强制转换：

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;});

或与某些法术：

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

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