使用迭代器将数组分割为具有不等大小的部分 [英] Using an iterator to Divide an Array into Parts with Unequal Size

问题描述

我有一个数组，我需要分成3个元素的子数组。我想用迭代器这样做，但我最终迭代超过数组的结尾和segfaulting ，即使我不解引用迭代器。给定： auto foo = {1，2，3，4，5，6，7，8，9，10};

I have an array which I need to divide up into 3-element sub-arrays. I wanted to do this with iterators, but I end up iterating past the end of the array and segfaulting even though I don't dereference the iterator. given: auto foo = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; I'm doing:

auto bar = cbegin(foo);

for (auto it = next(bar, 3); it < foo.end(); bar = it, it = next(bar, 3)) {
    for_each(bar, it, [](const auto& i) { cout << i << endl; });
}

for_each(bar, cend(foo), [](const auto& i) { cout << i << endl; });

现在我可以通过定义完成迭代器：

Now I can solve this by defining a finish iterator:

auto bar = cbegin(foo);
auto finish = next(cend(foo), -(size(foo) % 3));

for (auto it = next(bar, 3); it != finish; bar = it, it = next(bar, 3)) {
    for_each(bar, it, [](const auto& i) { cout << i << endl; });
}

for_each(bar, finish, [](const auto& i) { cout << i << endl; });
for_each(finish, cend(foo), [](const auto& i) { cout << i << endl; });

但是，当我不解引用迭代器时，这似乎是不必要的。为什么我不能做第一个版本？

But this seems unnecessary when I don't dereference the iterator. Why can't I do the first version?

推荐答案

=http://stackoverflow.com/questions/37209725/are-iterators-past-the-one-past-the-end-iterator-undefined-behavior>迭代器是否超过一个过去端

The reason this is prohibited is covered well at your other question Are iterators past the "one past-the-end" iterator undefined behavior? so I'll just address improved solutions.

对于随机访问迭代器（如果你使用<$ c $，你必须拥有它）

For random-access iterators (which you must have if you are using <), there's no need whatsoever for the expensive modulo operation.

显着点是：

• it + stride 在 it 附近


• 
  
• end（） - stride code> end（） - it 总是合法的

• it + stride fails when it nears the end
• end() - stride fails if the container contains too few elements
• end() - it is always legal

改变 it + stride< end（）变成法律形式（从两边减去 it ）。

From there, it's simple algebraic manipulation to change it + stride < end() into a legal form (subtract it from both sides).

最后的结果，我用了很多次：

The final result, which I have used many times:

for( auto it = c.cbegin(), end = c.cend(); end - it >= stride; it += stride )

如果内存模型是平坦的 - C ++行为的限制不适用于预期计算的 end-stride * sizeof（* it）编译器将C ++转换为原始操作。

The compiler is free to optimize that back to comparison to a precomputed end - stride * sizeof(*it) if the memory model is flat -- the limitations of C++ behavior don't apply to the primitive operations which the compiler translates C++ into.

你可以使用 std :: distance（it，end）如果你喜欢使用命名的函数而不是运算符，但是这将只对随机访问迭代器有效。

You may of course use std::distance(it, end) if you prefer to use the named functions instead of operators, but that will only be efficient for random-access iterators.

要使用forward iterator，结合递增和终止条件，如

For use with forward iterators, you should use something that combines the increment and termination conditions like

struct less_preferred { size_t value; less_preferred(size_t v) : value(v){} };

template<typename Iterator>
bool try_advance( Iterator& it, less_preferred step, Iterator end )
{
     while (step.value--) {
         if (it == end) return false;
         ++it;
     }
     return true;
}

对于这种额外的重载，你将获得随机访问迭代器的高效行为：

With this additional overload, you'll get efficient behavior for random-access iterators:

template<typename RandomIterator>
auto try_advance( RandomIterator& it, size_t stride, RandomIterator end )
     -> decltype(end - it < stride) // SFINAE
{
     if (end - it < stride) return false;
     it += stride;
     return true;
}

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