CODEWITHSUNDEEP
c++boost
user230821
user230821

Reputation: 1123

boost::function function pointer to parameters?

How does boost::function take a function pointer and get parameters from it? I want wrap a function pointer so that it can be validated before being called. And it would be nice to be able to call it like boost::function is with the () operator and not having to access the function pointer member.

    Wrapper func;
    func(5); //Yes :D
    func.Ptr(5) //Easy to do, but not as nice looking

Upvotes: 2

Views: 398

Answers (2)

Nikolai Fetissov
Nikolai Fetissov

Reputation: 84159

Something like this?

class Functor
{
public:

    /// type of pointer to function taking string and int and returning int
    typedef int ( *func_ptr_t )( const std::string&, int );

    explicit Functor( func_ptr_t f ) : fptr_( f ) {}

    int operator()( const std::string& s, int i ) const { return fptr_( s, i ); }

private:

    func_ptr_t fptr_; //< function pointer
};

But why not just use boost::function? It allows for way more then just a function pointer.

Upvotes: 0

James McNellis
James McNellis

Reputation: 355039

You need to overload operator(). To determine the return type, arity, and parameter types of a function, you can use something like Boost.TypeTraits:

#include <boost/type_traits.hpp>

template <typename Function>
struct Wrapper
{
    typedef typename boost::function_traits<Function>::arg1_type Arg1T;
    typedef typename boost::function_traits<Function>::result_type ReturnT;

    Wrapper(Function func) : func_(func) { }

    ReturnT operator()(Arg1T arg) { return func_(arg); }

    Function* func_;
};

bool Test(int x) { return x < 42; }

int main()
{
    Wrapper<bool(int)> CallsTest(&Test);
    CallsTest(42);
}

Upvotes: 2

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