loop.cpp

int upper = 0;
for (int i = 0; i < (1 << n); i++) {
  if (i == 2^{upper+1}) upper++;
  function[i] = function[i ^ (1 << upper)] и что-нибудь ещё.
}

Число единичных бит можно так: f[i] = f[i >> 1] + (i & 1);

Сумма элементов: f[i] = f[i ^ (1 << upper)] + weight[upper]

sum_of_subsets.cpp

#include <bitset>
#include <cassert>
#include <cstdint>
#include <iomanip>
#include <iostream>
#include <vector>

void f(size_t idx,              // Loop counter
       size_t set, int32_t sum, // Accumulators
       // Environment
       std::vector<int32_t> const &items, std::vector<int32_t> &result) {

  if (idx == items.size()) {
    assert(result[set] == 0);
    result[set] = sum;
    return;
  }

  assert(idx < items.size());

  // 1) We pick next item and update accumulators.
  f(idx + 1, set | (1 << idx), sum + items[idx], items, result);

  // 2) We don't pick the item, accumulators stay intact.
  f(idx + 1, set, sum, items, result);
}

std::vector<int32_t> sums_of_subsets(std::vector<int32_t> const &items) {
  size_t n = 1 << items.size();
  std::vector<int32_t> result(n, 0);
  f(0, 0, 0, items, result);
  return result;
}

int main() {
  std::vector<int32_t> items = {1, 2, 5, 10};
  auto sums = sums_of_subsets(items);
  for (size_t i = 0; i < sums.size(); ++i) {
    std::cout << std::setw(2);
    std::cout << i << "  " << std::bitset<3>(i) << "  " << sums[i] << "\n";
  }
}