Is it possible to use std::accumulate with std::min?
-
22-06-2021 - |
Pergunta
I am trying to combine std::accumulate
with std::min
. Something like this (won't compile):
vector<int> V{2,1,3};
cout << accumulate(V.begin()+1, V.end(), V.front(), std::min<int>);
Is it possible?
Is it possible to do without writing wrapper functor for std::min
?
I know that I can do this with lambdas:
vector<int> V{2,1,3};
cout << std::accumulate(
V.begin()+1, V.end(),
V.front(),
[](int a,int b){ return min(a,b);}
);
And I know there is std::min_element
. I am not trying to find min element, I need to combine std::accumulate
with std::min
(or ::min
) for my library which allows function-programming like expressions in C++.
Solução
The problem is that there are several overloads of the min
function:
template <class T> const T& min(const T& a, const T& b);
template <class T, class BinaryPredicate>
const T& min(const T& a, const T& b, BinaryPredicate comp);
Therefore, your code is ambiguous, the compiler does not know which overload to choose. You can state which one you want by using an intermediate function pointer:
#include <algorithm>
#include <iostream>
#include <vector>
int main()
{
std::vector<int> V{2,1,3};
int const & (*min) (int const &, int const &) = std::min<int>;
std::cout << std::accumulate(V.begin() + 1, V.end(), V.front(), min);
}
Licenciado em: CC-BY-SA com atribuição
Não afiliado a StackOverflow