reduce arguments - c++

Question

My question is that I have two arguments vector<int>& ivec and const int& numb. is there a way to eliminate const int& numb? I just want to do add 10 to all elements in the vector. Also I want to eliminate if part, since it's empty. Thanks in advance. Please do it recursively. My goal it to use as few arguments as possible in a recursive function.
#include <iostream>
#include <vector>
using namespace std;
vector<int> addten(vector<int>& ivec, const int& numb) {
if (numb == 0) {
} else {
ivec[numb - 1] += 10;
addten(ivec, numb - 1);
}
return ivec;
}

Answer 1

Your code feels very odd. The C++y way to do this would be:
std::vector<int> vec; // your vector
// adds 10 in place
std::transform(vec.begin(), vec.end(), vec.begin(),
std::bind(std::plus<int>(), _1, 10));
// adds 10 out-of-place
std::vector<int> result;
std::transform(vec.begin(), vec.end(), std::back_inserter(result),
std::bind(std::plus<int>(), _1, 10));
As you have specifically requested, I've implemented a very poor foldl in C++ that operates only on vector<T> instead of on iterators.
#include <vector>
#include <iostream>
// well, here comes C++ origami
template<typename Start, typename F, typename T>
Start foldl(Start s, F f, const std::vector<T>& v) {
return foldl_impl(s, f, v, 0);
}
template<typename Start, typename F, typename T>
Start foldl_impl(Start s, F f, const std::vector<T>& v,
typename std::vector<T>::size_type t) {
if(t == v.size()) return s;
typename std::vector<T>::size_type t2 = t++;
return foldl_impl(f(s, v[t2]), f, v, t);
}
int main()
{
std::vector<int> vec = {1, 2, 3, 4, 5, 6, 7};
std::vector<int> added =
foldl(std::vector<int>()
, [](std::vector<int>& v, int i) { v.push_back(i+10); return v;}
, vec);
for(auto x : added) {
std::cout << x << std::endl;
}
return 0;
}
Please consider that this is far from good C++ style.