传递lambda作为函数指针

问题描述

是否可以将lambda函数作为函数指针传递？

Is it possible to pass a lambda function as a function pointer? If so, I must be doing something incorrectly because I am getting a compile error.

考虑下面的例子

using DecisionFn = bool(*)();

class Decide
{
public:
    Decide(DecisionFn dec) : _dec{dec} {}
private:
    DecisionFn _dec;
};

int main()
{
    int x = 5;
    Decide greaterThanThree{ [x](){ return x > 3; } };
    return 0;
}

当我，我得到以下编译错误：

When I try to compile this, I get the following compilation error:

In function 'int main()':
17:31: error: the value of 'x' is not usable in a constant expression
16:9:  note: 'int x' is not const
17:53: error: no matching function for call to 'Decide::Decide(<brace-enclosed initializer list>)'
17:53: note: candidates are:
9:5:   note: Decide::Decide(DecisionFn)
9:5:   note: no known conversion for argument 1 from 'main()::<lambda()>' to 'DecisionFn {aka bool (*)()}'
6:7:   note: constexpr Decide::Decide(const Decide&)
6:7:   note: no known conversion for argument 1 from 'main()::<lambda()>' to 'const Decide&'
6:7:   note: constexpr Decide::Decide(Decide&&)
6:7:   note: no known conversion for argument 1 from 'main()::<lambda()>' to 'Decide&&'

一个错误消息来消化，但我想我得到的是，lambda不能被视为一个 constexpr 所以我不能把它作为一个函数指针？我尝试了 x const，但是这似乎没有帮助。

That's one heck of an error message to digest, but I think what I'm getting out of it is that the lambda cannot be treated as a constexpr so therefore I cannot pass it as a function pointer? I've tried making x const as well, but that doesn't seem to help.

推荐答案

一个lambda只能转换为一个函数指针，如果它不捕获，从部分 5.1.2 [expr.prim.lambda] （强调我）：

A lambda can only be converted to a function pointer if it does not capture, from the draft C++11 standard section 5.1.2 [expr.prim.lambda] says (emphasis mine):

请注意，cppreference还介绍了。

Note, cppreference also covers this in their section on Lambda functions .

因此，以下替代方法可用：

So the following alternatives would work:

typedef bool(*DecisionFn)(int);

Decide greaterThanThree{ []( int x ){ return x > 3; } };

因此：

typedef bool(*DecisionFn)();

Decide greaterThanThree{ [](){ return true ; } };

和指出，您也可以使用，但请注意，so it is not a cost-less trade-off。

and as 5gon12eder points out, you can also use std::function, but note that std::function is heavy weight, so it is not a cost-less trade-off.

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