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Reducing the end of an iterator

I read today about how for containers that support bidirectional iteration, this piece of code is valid:

Collection c(10, 10);
auto last = --c.end();
*last;

This made me wonder if I needed to send a pair of bidirectional iterators [beg, end) to an algorithm in STL defined by --end? If so, should the result be dereferenced?

t

void algo(T beg, T end){
    //...
    auto iter = --end;
    //...
    *iter;
}
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4 answers

If the algorithm requires a range specified by bidirectional iterators firstand last, then it --lastmust be valid under the same conditions ++firstas, namely, that the range is not empty. The range is empty if and only if first == last.

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auto last = --c.end();

.

auto last = c.end();
--last;

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prev( c.end() );

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Is it possible that --enddepends on end == beg.

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This is only required for algorithms that require bidirectional iterators.

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