Can we reliably pre-increment/decrement rvalues?

Question

For example, std::vector<int>::iterator it = --(myVec.end());. This works in GCC 4.4 but I have heard a rumor that it's not portable.

Answer 1

It will only work if std::vector<int>::iterator is an object type with operator++ a member function. If it's a scalar type (e.g. int *), or operator++ is a non-member function, it will fail.

5.3.2 Increment and decrement [expr.pre.incr]

1 - The operand of prefix ++ is modified by adding 1 [...]. The operand shall be a modifiable lvalue. [...]
2 - [...] The requirements on the operand of prefix -- [...] are [...] the same as those of prefix ++. [...]

Non-const non-static member functions can be called on temporary objects (since they have non-const object type, per 9.3.2p3), but an lvalue reference parameter in a non-member function cannot bind to a temporary (13.3.3.1.4p3).

struct S { S &operator++(); };
struct T { }; T &operator++(T &);
typedef int U;

++S();  // OK
++T();  // fails
++U();  // fails

This means that it's nothing to do with the compiler, but rather the standard library; as you've observed libstdc++ is implemented with std::vector<int>::iterator an object type with member operator++, but your code could easily be compiled with the same compiler and a different standard library where std::vector<int>::iterator is int *, in which case it would fail.

std::vector, std::array and std::string are the only container templates that can sensibly be implemented with scalar (pointer) iterators, but that doesn't mean that calling ++ on other containers' iterators is safe; they could have non-member operator++ as T above.

To make an iterator to the before-the-end element, use std::prev:

std::vector<int>::iterator it = std::prev(myVec.end());

std::prev and std::next are new in C++11, but are easily implementable in C++03.

Answer 2

No that won't work in general.

In C++11 we have: auto it = std::prev(myVec.end());, which works reliably.

Boost has a similar function if you're in C++03, though it's trivial to write altogether:

template <typename BidirectionalIterator>
BidirectionalIterator
    prev(BidirectionalIterator x,
         typename std::iterator_traits<BidirectionalIterator>::difference_type n = 1)
{
    std::advance(x, -n);
    return x;
}

Keep in mind you need at least one element in the range for this to make sense.

---

Here's an example of how your method won't work in general, consider this stripped-down std::vector<>:

#include <iterator>

namespace std_exposition
{
    template <typename T>
    struct vector
    {
        // this is compliant:
        typedef T* iterator;

        iterator end()
        {
            return std::end(data);
        }

        T data[4];
    };

    // manually implemented std::prev:
    template <typename BidirectionalIterator>
    BidirectionalIterator
        prev(BidirectionalIterator x,
             typename std::iterator_traits<BidirectionalIterator>::difference_type n = 1)
    {
        std::advance(x, -n);
        return x;
    }
}

Test program:

int main()
{
    std_exposition::vector<int> myVec;

    // Won't compile (method in question):
    auto it0 = --(myVec.end());

    // Compiles
    auto it1 = std::prev(myVec.end());
    auto it2 = std_exposition::prev(myVec.end());
}

---

There is a corresponding std::next as well, implemented here:

template <typename BidirectionalIterator>
BidirectionalIterator
    next(BidirectionalIterator x,
         typename std::iterator_traits<BidirectionalIterator>::difference_type n = 1)
{
    std::advance(x, n);
    return x;
}

Answer 3

This is indeed not portable, because there's no way to know whether myVec.end() returns an object of class type with operator -- overloaded by a member function or something else (maybe even a regular raw ponter). In the former case the overloaded -- will compile (operators overloaded by member functions can be applied to rvalues), while in the latter case it will not.