是否有可能在lambda中捕获可变数量的参数？

Question

考虑以下一组示例.

1. 该函数takeOnlyVoidFunction采用零参数的函数并简单地执行它.
2. 该函数takeVariableArguments接受可变数量的参数并使用参数执行函数.
3. 该函数captureVariableArgs尝试将第二个函数转换为第一个函数可接受的lambda形式,但它不能编译.

如何使函数captureVariableArgs编译并展示将具有可变数量参数的函数转换为不带参数的闭包的正确行为？

#include <stdio.h>
#include <functional>

void takeOnlyVoidFunction(std::function<void()> task) {
    task();
}

template<typename _Callable, typename... _Args>
    void takeVariableArguments(_Callable&& __f, _Args&&... __args) {
     __f(__args...);
}

// How can I make this function compile?
template<typename _Callable, typename... _Args>
    void captureVariableArgs(_Callable&& __f, _Args&&... __args) {
    takeOnlyVoidFunction([=]() { __f(__args...);});
}

void normalFunction(int a, int b) {
    printf("I am a normal function which takes params (%d,%d)\n", a, b);
}

int main() {
    int a = 7;
    int b = 8;
    takeVariableArguments(normalFunction, a, b);
    takeOnlyVoidFunction([=](){ normalFunction(a,b);});
    captureVariableArgs(normalFunction, a, b);
}

我跑gcc 4.9.2.这是我看到的编译器错误.

g++ -std=c++11    Test.cc   -o Test
Test.cc: In instantiation of ‘captureVariableArgs(_Callable&&, _Args&& ...)::<lambda()> [with _Callable = void (&)(int, int); _Args = {int&, int&}]’:
Test.cc:16:38:   required from ‘struct captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]::<lambda()>’
Test.cc:16:50:   required from ‘void captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]’
Test.cc:28:45:   required from here
Test.cc:16:34: error: variable ‘__f’ has function type
     takeOnlyVoidFunction([=]() { __f(__args...);});
                                  ^
Test.cc:16:34: error: variable ‘__f’ has function type
Test.cc: In instantiation of ‘struct captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]::<lambda()>’:
Test.cc:16:50:   required from ‘void captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]’
Test.cc:28:45:   required from here
Test.cc:16:34: error: field ‘captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]::<lambda()>::<__f capture>’ invalidly declared function type
In file included from Test.cc:2:0:
/usr/include/c++/4.9/functional:2418:7: error: ‘std::function<_Res(_ArgTypes ...)>::function(_Functor) [with _Functor = captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]::<lambda()>; <template-parameter-2-2> = void; _Res = void; _ArgTypes = {}]’, declared using local type ‘captureVariableArgs(_Callable&&, _Args&& ...) [with _Callable = void (&)(int, int); _Args = {int&, int&}]::<lambda()>’, is used but never defined [-fpermissive]
       function<_Res(_ArgTypes...)>::
       ^

---

更新:演示此问题的更简单示例.

#include <stdio.h>

// How can I make this function compile?
template<typename _Callable>
void captureVariableArgs(_Callable&& __f) {
    takeOnlyVoidFunction( [=]{ __f(); } );
}

void normalFunction() {
    printf("I am a normal function\n");
}

int main(){
    captureVariableArgs(normalFunction);
}

Answer 1

作为 GCC 的另一个潜在解决方法，您可以使用以下命令来代替 lambda std::bind：

template <typename F, typename... Args>
auto captureVariable(F&& f, Args&&... args)
{
    return std::bind(std::forward<F>(f), std::forward<Args>(args)...);
}

这在 GCC 4.9.3 下适用于我。