Question about varargs

[SuperSonic]

Hey guys!

So, I just read a good article on using var_lists. I was wondering if there was a way to check if another argument exists in a list. Does anyone know how to do that?

[Peter87]

No. You will have to know from previous arguments if there are more to read or not.

[kbw]

... and you must assert the type (thus size) of each argument. There's nothing in the variable arg mechanism to tell you what's there.

[SuperSonic]

Ok, well thanks anyways :D

[JLBorges]

If you are programming in C++, prefer type-safe mechanisms:

#include <iostream>

template < typename T > inline
std::ostream& print( std::ostream& stm, const T& value ) { return stm << value ; }

template < typename FIRST, typename... REST > inline
std::ostream& print( std::ostream& stm, const FIRST& first, REST... rest )
{ return print( print( stm, first ), rest... ) ; }

int main()
{
    print( std::cout, "the value is: ", 12345, '\n' ) ;
    print( std::cout, "the values are: ", 12345, ", ", 78.32, " and ", 999, '\n' ) ;
}

[viliml]

JLBorges: In my opinion, it's much easier to use nested insertion(cout<<something<<somethingelse<<yadday<<"Hello World";)

[JLBorges]

> In my opinion, it's much easier to use nested insertion

The print() was for expositional purposes, to provide a type-safe alternative to a C function using va_list - fprintf()

It is a matter of opinion all right. Any function which can be written to take a variable number of arguments can be replaced with nested calls to an appropriately written binary function.

#include <type_traits>
#include <iostream>

template < typename T, typename U > inline
typename std::common_type<T,U>::type plus( const T& a, const U& b ) { return a+b ; }

template < typename FIRST, typename... REST > inline
typename std::common_type<FIRST,REST...>::type plus( const FIRST& first, REST... rest )
{ return first + plus( rest... ) ; }

int main()
{
    std::cout << plus( 'a', 500, 1234.56, 99ULL ) << '\n' ;
    std::cout << plus( 'a', plus( 500, plus( 1234.56, 99ULL ) ) ) << '\n' ;
}