C++ ScopedExit实现:将参数传递给函数对象
我正在尝试实现简单的ScopedExit类。代码如下:C++ ScopedExit实现:将参数传递给函数对象,c++,c++11,function-pointers,C++,C++11,Function Pointers,我正在尝试实现简单的ScopedExit类。代码如下: #include <iostream> #include <functional> template<class R, class... Args> class ScopedExit { public: ScopedExit(std::function<R(Args...)> exitFunction) { exitFunc_ = exitFunction;
#include <iostream>
#include <functional>
template<class R, class... Args>
class ScopedExit
{
public:
ScopedExit(std::function<R(Args...)> exitFunction)
{
exitFunc_ = exitFunction;
}
~ScopedExit()
{
exitFunc_();
}
private:
std::function<R(Args...)> exitFunc_;
};
template<>
class ScopedExit<void>
{
public:
ScopedExit(std::function<void ()> exitFunction)
{
exitFunc_ = exitFunction;
}
~ScopedExit()
{
exitFunc_();
}
private:
std::function<void ()> exitFunc_;
};
void foo()
{
std::cout << "foo() called\n";
}
class Bar
{
public:
void BarExitFunc(int x, int y)
{
std::cout << "BarExitFunc called with x =" << x << "y = " << y << "\n";
}
};
int main()
{
Bar b;
std::cout << "Register scoped exit func\n";
{
ScopedExit<void, int, int> exitGuardInner(std::bind(&Bar::BarExitFunc, &b, 18, 11));
}
ScopedExit exitGuardOutter(foo);
std::cout << "About to exit from the scope\n";
return 0;
}
#包括
#包括
模板
类ScopedExit
{
公众:
ScopedExit(std::function exitFunction)
{
exitFunc=exitFunction;
}
~ScopedExit()
{
exitFunc_u2;();
}
私人:
std::函数exitFunc\ux;
};
模板
类ScopedExit
{
公众:
ScopedExit(std::function exitFunction)
{
exitFunc=exitFunction;
}
~ScopedExit()
{
exitFunc_u2;();
}
私人:
std::函数exitFunc\ux;
};
void foo()
{
std::cout1)例如,您可以将参数保存在tuple
中。但是,在您的情况下,您可以简单地调用exitFunc()
,函数定义应该是std::function exitFunction
,因为您已经将参数绑定到函数。可能是这样的
#include <iostream>
#include <functional>
#include <tuple>
template<size_t...>
struct indices {};
template<size_t N, size_t... Is>
struct gen_indices : gen_indices<N - 1, N - 1, Is...>
{
};
template<size_t... Is>
struct gen_indices<0, Is...> : indices<Is...>
{
};
template<class R, class... Args>
class ScopedExit
{
public:
ScopedExit(std::function<R(Args...)> exitFunction, Args&&... args)
: arguments_(std::forward_as_tuple(args...))
{
exitFunc_ = exitFunction;
}
~ScopedExit()
{
call(gen_indices<sizeof...(Args)>());
}
private:
template<size_t... Idx>
void call(indices<Idx...>)
{
exitFunc_(std::forward<Args>(std::get<Idx>(arguments_))...);
}
std::tuple<Args...> arguments_;
std::function<R(Args...)> exitFunc_;
};
template<>
class ScopedExit<void>
{
public:
ScopedExit(std::function<void ()> exitFunction)
{
exitFunc_ = exitFunction;
}
~ScopedExit()
{
exitFunc_();
}
private:
std::function<void ()> exitFunc_;
};
void foo()
{
std::cout << "foo() called\n";
}
class Bar
{
public:
void BarExitFunc(int x, int y)
{
std::cout << "BarExitFunc called with x =" << x << "y = " << y << "\n";
}
};
int main()
{
Bar b;
std::cout << "Register scoped exit func\n";
{
ScopedExit<void, int, int> exitGuardInner
(
std::bind(&Bar::BarExitFunc, &b, std::placeholders::_1,
std::placeholders::_2), 10, 18
);
}
ScopedExit<void> exitGuardOutter(foo);
std::cout << "About to exit from the scope\n";
return 0;
}
#包括
#包括
#包括
模板
结构索引{};
模板
结构gen_索引:gen_索引
{
};
模板
结构gen_索引:索引
{
};
模板
类ScopedExit
{
公众:
ScopedExit(std::function exitFunction,Args&&…Args)
:参数(std::转发作为元组(args…)
{
exitFunc=exitFunction;
}
~ScopedExit()
{
调用(gen_index());
}
私人:
模板
作废通知(索引)
{
exitFunc(std::forward(std::get(arguments))…);
}
std::元组参数;
std::函数exitFunc\ux;
};
模板
类ScopedExit
{
公众:
ScopedExit(std::function exitFunction)
{
exitFunc=exitFunction;
}
~ScopedExit()
{
exitFunc_u2;();
}
私人:
std::函数exitFunc\ux;
};
void foo()
{
标准::cout
如何将exit的函数参数传递给它?例如,我使用两个整数参数绑定BarExitFunc:18和11。那么如何将其传递给析构函数中的exitFunc_uu2;呢
我看不出有任何理由在析构函数中调用时将参数传递给exitFunc.
。无论您做什么,您都必须在ScopedExit
构造函数中预先提供这些参数
最简单的方法是在定义站点使用函数
和绑定
任何必需的参数,就像您已经在做的那样:
ScopedExit<R> guard(std::bind(someFunction, someArg, otherArg));
真的,我不明白这一点,因为它使模板更加复杂,但为什么不…只需在构造函数本身中绑定/转发参数,并仍然存储一个函数
:
请注意,您原来的可变模板类现在已成为一个灵活的非模板类,只有一个模板构造函数,它的模板参数是自动推导的,并且它几乎可以接受您可以想到的任何类型的函子
注意:我说的几乎是任何函子,因为这不适用于默认参数:
void foobar(int = 0) {}
ScopedExit guard5(foobar); // error: too few arguments to function
即使您直接存储了一个Functor
而不是std::function
,您也无法使用默认参数(即使有默认参数,foobar
的签名仍然是void(int)
)因此,我们必须在定义站点处理这种情况,例如:
void foobar(int = 0) {}
ScopedExit guard5([]() { foobar(); });
class ScopedExit
{
public:
template<typename Functor>
ScopedExit(Functor exitFunction)
{
exitFunc_ = exitFunction;
}
~ScopedExit()
{
exitFunc_();
}
private:
std::function<void()> exitFunc_;
};
int foo() { return 0; }
struct Bar {
void bye(int, int) {}
};
struct Baz {
void operator ()() {}
};
int main() {
const std::string what = "lambda!";
ScopedExit guard1([&]() { std::cout << "yay a " << what << std::endl; });
ScopedExit guard2(foo); // note how std::function<void()> accepts non-void return types
Bar b;
ScopedExit guard3(std::bind(&Bar::bye, &b, 1, 2));
ScopedExit guard4(Baz());
}
void foobar(int = 0) {}
ScopedExit guard5(foobar); // error: too few arguments to function
void foobar(int = 0) {}
ScopedExit guard5([]() { foobar(); });