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存储升压功能

[英]Storing boost function

I have to store a list of different boost::function objects. 我必须存储不同的boost :: function对象的列表。 To provide this I'm using boost::any. 为此,我使用boost :: any。 I have a few functions which takes different functions signatures, pack them into any and then insert into special map with given type. 我有一些带有不同功能签名的功能,将它们打包到任何一个中 ,然后插入给定类型的特殊映射中。 Here is the code: 这是代码:

enum TypeEnumerator
{
    e_int,
    e_float,
    e_double
};

typedef map< string, pair<any, TypeEnumerator> > CallbackType;
CallbackType mCallbacks;

void Foo(const string &name, function<float ()> f)
{
    mCallbacks[name] = make_pair(any(f), CLASS::e_float);
}
void Foo(const string &name, function<int ()> f) { /* the same, but with e_int */ }
void Foo(const string &name, function<double ()> f) { /* the same, but with e_double */ }

Now I have in map boost function, packed into any with given type from enum, to recognize it in future. 现在,我有了map boost功能,可以将其封装到enum中具有给定类型的任何功能中,以便将来识别它。 Now I have to call given functions. 现在,我必须调用给定的函数。 The casting from any won't work: 任何类型的转换都不起作用:

BOOST_FOREACH(CallbackType::value_type &row, mCallbacks)
{
    // pair<any, TypeEnumerator>
    switch (row.second.second) // Swith the TypeEnumerator
    {
        case 0: // int
            any_cast< function<int ()> >(row.first)();
        break;
        case 1: // float
            any_cast< function<float ()> >(row.first)();
        break;
        case 2: // double
            any_cast< function<double ()> >(row.first)();
        break;
    }
}

This won't cast and during running I get the exception: 这不会强制转换,在运行期间会出现异常:

  what():  boost::bad_any_cast: failed conversion using boost::any_cast

Is it possible to convert back the boost::function object? 是否可以转换回boost :: function对象?

@TC provided the solution for the runtime error. @TC提供了运行时错误的解决方案。 But I believe you should use Boost.Variant instead of Boost.Any as there are only a fixed selection of types it can store. 但是我相信您应该使用Boost.Variant而不是Boost.Any,因为它只能存储固定的类型选择。 With Boost.Variant you could eliminate that enum too, as it already provided a standard visitor pattern interface. 使用Boost.Variant,您也可以消除该枚举,因为它已经提供了标准的访问者模式界面。 ( result ): 结果 ):

#include <boost/variant.hpp>
#include <boost/function.hpp>
#include <boost/foreach.hpp>
#include <map>
#include <string>
#include <iostream>

typedef boost::variant<boost::function<int()>,
                       boost::function<float()>,
                       boost::function<double()> > Callback;
typedef std::map<std::string, Callback> CallbackType;

CallbackType mCallbacks;

void Foo(const std::string& name, const Callback& f) {
    mCallbacks[name] = f;
}

//------------------------------------------------------------------------------

float f() { 
    std::cout << "f called" << std::endl;
    return 4;
}

int g() {
    std::cout << "g called" << std::endl;
    return 5;
}

double h() {
    std::cout << "h called" << std::endl;
    return 4;
}

//------------------------------------------------------------------------------

struct call_visitor : public boost::static_visitor<> {
    template <typename T>
    void operator() (const T& operand) const {
        operand();
    }
};


int main () {
    Foo("f", boost::function<float()>( f ));
    Foo("g", boost::function<int()>( g ));
    Foo("h", boost::function<double()>( h ));

    BOOST_FOREACH(CallbackType::value_type &row, mCallbacks) {
        boost::apply_visitor(call_visitor(), row.second);
    }

    return 0;
}

From the looks of it, row.first is the name of the callback, a string . 从外观row.firstrow.first是回调的名称,即string You should probably use row.second.first : 您可能应该使用row.second.first

case 0: // int
    any_cast< function<int ()> >(row.second.first)();
    break;

Furthermore, you should use your enum constants in the switch ( case CLASS::e_int: ), instead of magic numbers. 此外,应该在开关( case CLASS::e_int:使用枚举常量,而不是幻数。

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