What's wrong with passing C++ iterator by reference?

I've written a few functions with a prototype like this:

template <typename input_iterator>
int parse_integer(input_iterator &begin, input_iterator end);

The idea is that the caller would provide a range of characters, and the function would interpret the characters as an integer value and return it, leaving begin at one past the last-used character. For example:

std::string sample_text("123 foo bar");
std::string::const_iterator p(sample_text.begin());
std::string::const_iterator end(sample_text.end());
int i = parse_integer(p, end);

This would leave i set to 123 and p "pointing" at the space before foo.

I've since been told (without explanation) that it's bad form to pass an iterator by reference. Is it bad form? If so, why?

Solution 1:

There is nothing really wrong, but it will certainly limit the use of the template. You won't be able to just put an iterator returned by something else or generated like v.begin(), since those will be temporaries. You will always first have to make a local copy, which is some kind of boilerplate not really nice to have.

One way is to overload it:

int parse_integer(input_iterator begin, input_iterator end, 
                  input_iterator &newbegin);

template<typename input_iterator>
int parse_integer(input_iterator begin, input_iterator end) {
    return parse_integer(begin, end, begin);
} 

Another option is to have an output iterator where the number will be written into:

template<typename input_iterator, typename output_iterator>
input_iterator parse_integer(input_iterator begin, input_iterator end,
                             output_iterator out);

You will have the return value to return the new input iterator. And you could then use a inserter iterator to put the parsed numbers into a vector or a pointer to put them directly into an integer or an array thereof if you already know the amount of numbers.

int i;
b = parse_integer(b, end, &i);

std::vector<int> numbers;
b = parse_integer(b, end, std::back_inserter(numbers));

Solution 2:

In general:

If you pass a non-const reference, the caller doesn't know if the iterator is being modified.

You could pass a const reference, but usually iterators are small enough that it gives no advantage over passing by value.

In your case:

I don't think there's anything wrong with what you do, except that it's not too standard-esque regarding iterator usage.

Solution 3:

When they say "don't pass by reference" maybe that's because it's more normal/idiomatic to pass iterators as value parameters, instead of passing them by const reference: which you did, for the second parameter.

In this example however you need to return two values: the parsed int value, and, the new/modified iterator value; and given that a function can't have two return codes, coding one of the return codes as a non-const reference is IMO normal.

An alternative would be to code it something like this:

//Comment: the return code is a pair of values, i.e. the parsed int and etc ...
pair<int, input_iterator> parse(input_iterator start, input_iterator end)
{
}