Why does std::accumulate generate 705032704 as output instead of the sum of the elements in the vector?

Question

The output is 705032704 instead of 5000000000. Why is that? I thought std::accumulate would compute the sum of the elements in the vector.

#include <vector>
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <string>
#include <iterator>
#include <queue>
#include <stack>
#include <numeric>

typedef long long ll;

int main()
{

    std::vector<ll> nums = {1000000000, 1000000000,1000000000,1000000000,1000000000};
    std::cout << std::accumulate(nums.begin(), nums.end(), 0);
    std::cin.ignore();

}

Answer 1

The standard algorithm std::accumulate is declared the following way

template<class InputIterator, class T>
T accumulate(InputIterator first, InputIterator last, T init);

So the template parameter T is used as the type of the accumulator.

In this call

std::accumulate(nums.begin(), nums.end(), 0)

the deduced type of the parameter T is int because the integer literal 0 has the type int. As a result there is an overflow of the sum stored in the accumulator of the type int.

You have to use at least the long long integer literal 0ll as

std::accumulate(nums.begin(), nums.end(), 0ll)

or maybe it is reasonable to use a double floating literal

std::accumulate(nums.begin(), nums.end(), 0.0 )

Here is a demonstrative program

#include <iostream>
#include <vector>
#include <iterator>
#include <numeric>

using ll = long long int;

int main()
{
    std::vector<ll> nums = {1000000000, 1000000000,1000000000,1000000000,1000000000};
    std::cout << std::accumulate( std::begin( nums ), std::end( nums ), 0ll ) << '\n';
    std::cout << std::accumulate( std::begin( nums ), std::end( nums ), 0.0 ) << '\n';
}

Its output is

5000000000
5e+09