在编译时获取 boost::function arity？

Question

我需要根据 boost::function 对象的数量（参数计数）在 BOOST_PP_IF 语句中做出决定。这可能吗？

boost::function_types::function_arity 做了我正在寻找的事情，但是在运行时；我在编译时需要它。

Answer 1

function_arity

template<typename F>
struct function_arity;

Header

#include <boost/function_types/function_arity.hpp>

F

    Callable builtin type 
function_arity<F>

    Function arity as MPL - Integral Constant 
function_arity<F>::value

    Constant value of the function arity 

请注意，这是编译时常量，

您应该从这里开始：
http://www.boost.org/doc/ libs/1_43_0/libs/mpl/doc/index.html

或使用 BOOST_PP_SEQ_FOR_EACH/BOOST_PP_REPEAT_FROM_TO 针对 function_arity::value 生成 if/else 条件

Answer 2

由于某种原因，我的包含内容不断中断，但不在预览中=[

#include <ostream>  
#include <iostream>  
#include <boost/function.hpp>  

// Assume that you want to print out "Function is N-arity" for general case. But "nularity" for 0

template< int i >
struct DarkSide
{
template<class U>
void operator()(std::ostream& out, const U& u) { out << "Function is "<<i<<"-arity"<<u; }
void operator()(std::ostream& out, std::ostream& ( *pf )(std::ostream&) ) { out << "Function is "<<i<<"-arity"<<pf; }
};

template<>
struct DarkSide<0>
{
template<class U>
void operator()(std::ostream& out, const U& u) { out << "Function is nularity"<<u; }
void operator()(std::ostream& out, std::ostream& ( *pf )(std::ostream&) ) { out << "Function is nularity"<<pf; }
};

int main() {
typedef boost::function< void ( ) > vFv;
typedef boost::function< void ( int x ) > vFi;
DarkSide< vFv::arity >()(std::cout,"\n");
DarkSide< vFi::arity >()(std::cout,std::endl);
}

Answer 3

如果您需要的只是读取 boost::function 的数量，那么您不需要做那么多工作：

#include <boost/function.hpp>
#include <iostream>
int main() {
  std::cout << boost::function<void()>::arity << std::endl;
  std::cout << boost::function<void(int)>::arity << std::endl;
  std::cout << boost::function<void(int, int)>::arity << std::endl;
}