Shafik Yaghmour的答案正确地解释了为什么lambda不能作为函数指针传递，如果它有一个捕获。我想介绍两个简单的解决方法。

Use std::function instead of raw function pointers. This is a very clean solution. Note however that it includes some additional overhead for the type erasure (probably a virtual function call). #include <functional> #include <utility> struct Decide { using DecisionFn = std::function<bool()>; Decide(DecisionFn dec) : dec_ {std::move(dec)} {} DecisionFn dec_; }; int main() { int x = 5; Decide greaterThanThree { [x](){ return x > 3; } }; } Use a lambda expression that doesn't capture anything. Since your predicate is really just a boolean constant, the following would quickly work around the current issue. See this answer for a good explanation why and how this is working. // Your 'Decide' class as in your post. int main() { int x = 5; Decide greaterThanThree { (x > 3) ? [](){ return true; } : [](){ return false; } }; }

正如其他人提到的，你可以用Lambda函数代替函数指针。我使用这个方法在我的c++接口到F77 ODE求解器RKSUITE。

//C interface to Fortran subroutine UT
extern "C"  void UT(void(*)(double*,double*,double*),double*,double*,double*,
double*,double*,double*,int*);

// C++ wrapper which calls extern "C" void UT routine
static  void   rk_ut(void(*)(double*,double*,double*),double*,double*,double*,
double*,double*,double*,int*);

//  Call of rk_ut with lambda passed instead of function pointer to derivative
//  routine
mathlib::RungeKuttaSolver::rk_ut([](double* T,double* Y,double* YP)->void{YP[0]=Y[1]; YP[1]= -Y[0];}, TWANT,T,Y,YP,YMAX,WORK,UFLAG);

如果lambda没有捕获，则只能将其转换为函数指针，摘自c++ 11标准草案5.1.2节[expr.prim. xml]。Lambda]表示(强调我的):

没有lambda捕获的lambda表达式的闭包类型具有 到指针的公共非虚拟非显式const转换函数 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。此转换返回的值 Function应是函数的地址，该函数在调用时具有 与调用闭包类型的函数调用操作符的效果相同。

注意，cppreference在Lambda函数一节中也介绍了这一点。

因此，以下选择是可行的:

typedef bool(*DecisionFn)(int);

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

这个也一样:

typedef bool(*DecisionFn)();

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

正如5gon12eder所指出的，您也可以使用std::function，但请注意std::function是很重要的，所以它不是一个无成本的权衡。

一个类似的答案，但我这样做是为了让你不必指定返回指针的类型(注意，通用版本需要c++ 20):

#include <iostream>


template<typename Function>
struct function_traits;

template <typename Ret, typename... Args>
struct function_traits<Ret(Args...)> {
    typedef Ret(*ptr)(Args...);
};

template <typename Ret, typename... Args>
struct function_traits<Ret(*const)(Args...)> : function_traits<Ret(Args...)> {};

template <typename Cls, typename Ret, typename... Args>
struct function_traits<Ret(Cls::*)(Args...) const> : function_traits<Ret(Args...)> {};

using voidfun = void(*)();

template <typename F>
voidfun lambda_to_void_function(F lambda) {
    static auto lambda_copy = lambda;

    return []() {
        lambda_copy();
    };
}

// requires C++20
template <typename F>
auto lambda_to_pointer(F lambda) -> typename function_traits<decltype(&F::operator())>::ptr {
    static auto lambda_copy = lambda;
    
    return []<typename... Args>(Args... args) {
        return lambda_copy(args...);
    };
}



int main() {
    int num;

    void(*foo)() = lambda_to_void_function([&num]() {
        num = 1234;
    });
    foo();
    std::cout << num << std::endl; // 1234

    int(*bar)(int) = lambda_to_pointer([&](int a) -> int {
        num = a;
        return a;
    });
    std::cout << bar(4321) << std::endl; // 4321
    std::cout << num << std::endl; // 4321
}

捕获lambdas不能转换为函数指针，正如这个答案所指出的。

然而，为只接受一个函数指针的API提供函数指针通常是相当痛苦的。最常被引用的方法是提供一个函数，并用它调用一个静态对象。

static Callable callable;
static bool wrapper()
{
    return callable();
}

这太乏味了。我们进一步利用了这个想法，自动化了创建包装器的过程，使工作变得更加简单。

#include<type_traits>
#include<utility>

template<typename Callable>
union storage
{
    storage() {}
    std::decay_t<Callable> callable;
};

template<int, typename Callable, typename Ret, typename... Args>
auto fnptr_(Callable&& c, Ret (*)(Args...))
{
    static bool used = false;
    static storage<Callable> s;
    using type = decltype(s.callable);

    if(used)
        s.callable.~type();
    new (&s.callable) type(std::forward<Callable>(c));
    used = true;

    return [](Args... args) -> Ret {
        return Ret(s.callable(std::forward<Args>(args)...));
    };
}

template<typename Fn, int N = 0, typename Callable>
Fn* fnptr(Callable&& c)
{
    return fnptr_<N>(std::forward<Callable>(c), (Fn*)nullptr);
}

并将其用作

void foo(void (*fn)())
{
    fn();   
}

int main()
{
    int i = 42;
    auto fn = fnptr<void()>([i]{std::cout << i;});
    foo(fn);  // compiles!
}

Live

这实际上是在每次出现fnptr时声明一个匿名函数。

请注意，给定相同类型的可调用对象，fnptr的调用将覆盖先前编写的可调用对象。在某种程度上，我们用int参数N来补救这个问题。

std::function<void()> func1, func2;
auto fn1 = fnptr<void(), 1>(func1);
auto fn2 = fnptr<void(), 2>(func2);  // different function

Shafik Yaghmour的答案正确地解释了为什么lambda不能作为函数指针传递，如果它有一个捕获。我想介绍两个简单的解决方法。

Use std::function instead of raw function pointers. This is a very clean solution. Note however that it includes some additional overhead for the type erasure (probably a virtual function call). #include <functional> #include <utility> struct Decide { using DecisionFn = std::function<bool()>; Decide(DecisionFn dec) : dec_ {std::move(dec)} {} DecisionFn dec_; }; int main() { int x = 5; Decide greaterThanThree { [x](){ return x > 3; } }; } Use a lambda expression that doesn't capture anything. Since your predicate is really just a boolean constant, the following would quickly work around the current issue. See this answer for a good explanation why and how this is working. // Your 'Decide' class as in your post. int main() { int x = 5; Decide greaterThanThree { (x > 3) ? [](){ return true; } : [](){ return false; } }; }