是否可以查看 range-for 循环中的下一个元素？

Question

Say I have this:

void test(std::vector<int>& my_ints) {
    for (auto& my_int : my_ints) {
        if (my_int == 5 && /* not the last value in the vector */) {
            my_int += /* next value in the vector */;
        }
    }
}

Is there any valid syntax to replace the comments with?

PS, yes, i know. piece of cake with a regular for loop but I want to see if I can use range-for loops for this type of stuff.

Answer 1

Is it possible to peek at the next element

In general case - no.

Since objects in std::vector are stored contiguously, you could do *(&my_int + 1) , but if you change the container later, the code might silently break. Don't do that!

---

And to check if the current element is the last one, you could use &my_int == &my_ints.back() .

Answer 2

I can't think of a safe way to do what you want, with a ranged loop.

If you can bare an extra helper function, you could somehow generalize the algorithm you are using, though:

#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>

template <class Container, class BinaryOp>
void apply_using_the_next(Container& cc, BinaryOp op)
{
    if (std::cbegin(cc) == std::cend(cc))
        return;
    std::transform(std::begin(cc), std::prev(std::end(cc)),
                   std::next(std::cbegin(cc)),
                   std::begin(cc), op);
}

void test(std::vector<int>& v)
{
    apply_using_the_next(v, [] (const int a, const int b) {
        return a == 5 ? a + b : a;
    });
}

int main()
{
    std::vector<int> a{2, 5, 4, 3, 5, 5, 1};

    test(a);

    for (auto const i : a)       // ->  2 9 4 3 10 6 1
        std::cout << ' ' << i;
    std::cout << '\n';
}

Live, here .

Answer 3

You can use iterator.

for (std::vector<int>::iterator it = my_ints.begin(); it < my_ints.end() - 1; ++it) {
    if (*it == 5 /* no need to check as iterating until second last value && it != my_ints.end() - 1*/) {
        *it += *(it+1);
    }
}

Even if the vector is empty, the loop won't enter as it < my_ints.end() - 1 will return false, so it is safe.