C++ – Iteration over std::vector: unsigned vs signed index variable

csignedstlunsigned

What is the correct way of iterating over a vector in C++?

Consider these two code fragments, this one works fine:

for (unsigned i=0; i < polygon.size(); i++) {
    sum += polygon[i];
}

and this one:

for (int i=0; i < polygon.size(); i++) {
    sum += polygon[i];
}

which generates warning: comparison between signed and unsigned integer expressions.

I'm new in the world of C++, so the unsigned variable looks a bit frightening to me and I know unsigned variables can be dangerous if not used correctly, so – is this correct?

Best Answer

For iterating backwards see this answer.

Iterating forwards is almost identical. Just change the iterators / swap decrement by increment. You should prefer iterators. Some people tell you to use std::size_t as the index variable type. However, that is not portable. Always use the size_type typedef of the container (While you could get away with only a conversion in the forward iterating case, it could actually go wrong all the way in the backward iterating case when using std::size_t, in case std::size_t is wider than what is the typedef of size_type):

Using std::vector

Using iterators

for(std::vector<T>::iterator it = v.begin(); it != v.end(); ++it) {
    /* std::cout << *it; ... */
}

Important is, always use the prefix increment form for iterators whose definitions you don't know. That will ensure your code runs as generic as possible.

Using Range C++11

for(auto const& value: a) {
     /* std::cout << value; ... */

Using indices

for(std::vector<int>::size_type i = 0; i != v.size(); i++) {
    /* std::cout << v[i]; ... */
}

Using arrays

Using iterators

for(element_type* it = a; it != (a + (sizeof a / sizeof *a)); it++) {
    /* std::cout << *it; ... */
}

Using Range C++11

for(auto const& value: a) {
     /* std::cout << value; ... */

Using indices

for(std::size_t i = 0; i != (sizeof a / sizeof *a); i++) {
    /* std::cout << a[i]; ... */
}

Read in the backward iterating answer what problem the sizeof approach can yield to, though.

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