Question

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

boost::function_types::function_arity正在寻找我正在寻找的东西，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

你应该从这里开始： 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生成针对function_arity<F>::value的if / else条件

Answer 2

由于某种原因，我的包括保持打破但不在预览= [

#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

如果您只需阅读boost :: function的arity，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;
}

在编译时获取boost :: function arity？

3 个答案: