arithmetic sum of function parameters [duplicate] - c++

Question

I want to write a simple adder (for giggles) that adds up every argument and returns a sum with appropriate type.
Currently, I've got this:
#include <iostream>
using namespace std;
template <class T>
T sum(const T& in)
{
return in;
}
template <class T, class... P>
auto sum(const T& t, const P&... p) -> decltype(t + sum(p...))
{
return t + sum(p...);
}
int main()
{
cout << sum(5, 10.0, 22.2) << endl;
}
On GCC 4.5.1 this seems to work just fine for 2 arguments e.g. sum(2, 5.5) returns with 7.5. However, with more arguments than this, I get errors that sum() is simply not defined yet. If I declare sum() like this however:
template <class T, class P...>
T sum(const T& t, const P&... p);
Then it works for any number of arguments, but sum(2, 5.5) would return integer 7, which is not what I would expect.
With more than two arguments I assume that decltype() would have to do some sort of recursion to be able to deduce the type of t + sum(p...). Is this legal C++0x? or does decltype() only work with non-variadic declarations? If that is the case, how would you write such a function?

Answer 1

I think the problem is that the variadic function template is only considered declared after you specified its return type so that sum in decltype can never refer to the variadic function template itself. But I'm not sure whether this is a GCC bug or C++0x simply doesn't allow this. My guess is that C++0x doesn't allow a "recursive" call in the ->decltype(expr) part.
As a workaround we can avoid this "recursive" call in ->decltype(expr) with a custom traits class:
#include <iostream>
#include <type_traits>
using namespace std;
template<class T> typename std::add_rvalue_reference<T>::type val();
template<class T> struct id{typedef T type;};
template<class T, class... P> struct sum_type;
template<class T> struct sum_type<T> : id<T> {};
template<class T, class U, class... P> struct sum_type<T,U,P...>
: sum_type< decltype( val<const T&>() + val<const U&>() ), P... > {};
This way, we can replace decltype in your program with typename sum_type<T,P...>::type and it will compile.
Edit: Since this actually returns decltype((a+b)+c) instead of decltype(a+(b+c)) which would be closer to how you use addition, you could replace the last specialization with this:
template<class T, class U, class... P> struct sum_type<T,U,P...>
: id<decltype(
val<T>()
+ val<typename sum_type<U,P...>::type>()
)>{};

Answer 2

Apparently you can't use decltype in a recursive manner (at least for the moment, maybe they'll fix it)
You can use a template structure to determine the type of the sum
It looks ugly but it works
#include <iostream>
using namespace std;
template<typename... T>
struct TypeOfSum;
template<typename T>
struct TypeOfSum<T> {
typedef T type;
};
template<typename T, typename... P>
struct TypeOfSum<T,P...> {
typedef decltype(T() + typename TypeOfSum<P...>::type()) type;
};
template <class T>
T sum(const T& in)
{
return in;
}
template <class T, class... P>
typename TypeOfSum<T,P...>::type sum(const T& t, const P&... p)
{
return t + sum(p...);
}
int main()
{
cout << sum(5, 10.0, 22.2) << endl;
}

Answer 3

C++14's solution:
template <class T, class... P>
decltype(auto) sum(const T& t, const P&... p){
return t + sum(p...);
}
Return type is deducted automatically.
See it in online compiler
Or even better if you want to support different types of references:
template <class T, class... P>
decltype(auto) sum(T &&t, P &&...p)
{
return std::forward<T>(t) + sum(std::forward<P>(p)...);
}
See it in online compiler
If you need a natural order of summation (that is (((a+b)+c)+d) instead of (a+(b+(c+d)))), then the solution is more complex:
template <class A>
decltype(auto) sum(A &&a)
{
return std::forward<A>(a);
}
template <class A, class B>
decltype(auto) sum(A &&a, B &&b)
{
return std::forward<A>(a) + std::forward<B>(b);
}
template <class A, class B, class... C>
decltype(auto) sum(A &&a, B &&b, C &&...c)
{
return sum( sum(std::forward<A>(a), std::forward<B>(b)), std::forward<C>(c)... );
}
See it in online compiler

Answer 4

Another answer to the last question with less typing by using C++11's std::common_type: Simply use
std::common_type<T, P ...>::type
as return type of your variadic sum.
Regarding std::common_type, here is an excerpt from http://en.cppreference.com/w/cpp/types/common_type:
For arithmetic types, the common type may also be viewed as the type
of the (possibly mixed-mode) arithmetic expression such as T0() + T1()
+ ... + Tn().
But obviously this works only for arithmetic expressions and doesn't cure the general problem.

Answer 5

I provide this improvement to the accepted answer. Just two structs
#include <utility>
template <typename P, typename... Ps>
struct sum_type {
using type = decltype(std::declval<P>() + std::declval<typename sum_type<Ps...>::type>());
};
template <typename P>
struct sum_type<P> {
using type = P;
};
Now just declare your functions as
template <class T>
auto sum(const T& in) -> T
{
return in;
}
template <class P, class ...Ps>
auto sum(const P& t, const Ps&... ps) -> typename sum_type<P, Ps...>::type
{
return t + sum(ps...);
}
With this, your test code now works
std::cout << sum(5, 10.0, 22.2, 33, 21.3, 55) << std::endl;
146.5

Answer 6

Right way to do:
#include <utility>
template <typename... Args>
struct sum_type;
template <typename... Args>
using sum_type_t = typename sum_type<Args...>::type;
template <typename A>
struct sum_type<A> {
using type = decltype( std::declval<A>() );
};
template <typename A, typename B>
struct sum_type<A, B> {
using type = decltype( std::declval<A>() + std::declval<B>() );
};
template <typename A, typename B, typename... Args>
struct sum_type<A, B, Args...> {
using type = sum_type_t< sum_type_t<A, B>, Args... >;
};
template <typename A>
sum_type_t<A> sum(A &&a)
{
return (std::forward<A>(a));
}
template <typename A, typename B>
sum_type_t<A, B> sum(A &&a, B &&b)
{
return (std::forward<A>(a) + std::forward<B>(b));
}
template <typename A, typename B, typename... C>
sum_type_t<A, B, C...> sum(A &&a, B &&b, C &&...args)
{
return sum( sum(std::forward<A>(a), std::forward<B>(b)), std::forward<C>(args)... );
}
https://coliru.stacked-crooked.com/a/a5a0e8019e40b8ba
This completely preserves resulting type of operations (even r-value referenceness). The order of operations is natural: (((a+b)+c)+d).

Answer 7

For C++17:
template <class... P>
auto sum(const P... p){
return (p + ...);
}
int main()
{
std::cout << sum(1, 3.5, 5) << std::endl;
return EXIT_SUCCESS;
}
Read about folding expressions.