std::ranges:: prev

Return the n ^th predecessor of iterator i .

The function-like entities described on this page are algorithm function objects (informally known as niebloids ), that is:

• Explicit template argument lists cannot be specified when calling any of them.
• None of them are visible to argument-dependent lookup .
• When any of them are found by normal unqualified lookup as the name to the left of the function-call operator, argument-dependent lookup is inhibited.

Parameters

Return value

Complexity

Possible implementation

struct prev_fn
{
    template<std::bidirectional_iterator I>
    constexpr I operator()(I i) const
    {
        --i;
        return i;
    }
 
    template<std::bidirectional_iterator I>
    constexpr I operator()(I i, std::iter_difference_t<I> n) const
    {
        ranges::advance(i, -n);
        return i;
    }
 
    template<std::bidirectional_iterator I>
    constexpr I operator()(I i, std::iter_difference_t<I> n, I bound) const
    {
        ranges::advance(i, -n, bound);
        return i;
    }
};
 
inline constexpr auto prev = prev_fn();

Notes

Although the expression -- r. end ( ) often compiles for containers, it is not guaranteed to do so: r. end ( ) is an rvalue expression, and there is no iterator requirement that specifies that decrement of an rvalue is guaranteed to work. In particular, when iterators are implemented as pointers or its operator-- is lvalue-ref-qualified, -- r. end ( ) does not compile, while ranges :: prev ( r. end ( ) ) does.

This is further exacerbated by ranges that do not model ranges:: common_range . For example, for some underlying ranges, ranges :: transform_view :: end doesn't have the same return type as ranges :: transform_view :: begin , and so -- r. end ( ) won't compile. This isn't something that ranges::prev can aid with, but there are workarounds.

Example

#include <iostream>
#include <iterator>
#include <vector>
 
int main() 
{
    std::vector<int> v{3, 1, 4};
    auto pv = std::ranges::prev(v.end(), 2);
    std::cout << *pv << '\n';
 
    pv = std::ranges::prev(pv, 42, v.begin());
    std::cout << *pv << '\n';
}

1
3

See also