传递lambda作为函数指针 [英] Passing lambda as function pointer

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;
}

当我尝试编译这个，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，从草案C ++ 11标准部分 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):

lambda表达式的闭包类型捕获具有
公共非虚拟非显式const 转换函数到指针
到函数，具有与闭包相同的参数和返回类型
类型的函数调用操作符。这个转换返回的值
函数应该是一个函数的地址，当调用它时，
的效果与调用闭包类型的函数调用操作符相同。

The closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator.

请注意，cppreference还介绍了 Lambda函数。

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 5gon12eder 指出，您也可以使用 std :: function ，但请注意 std :: function is heavy weight ，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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