问题描述
例如，如果我想找到集合中最小的元素，但只找到最小的 even 元素，我想使用 ranges :: min_element 过滤范围如下:
For example, if I want to find the smallest element of a collection, but only the smallest even element, I'd like to call ranges::min_element with a filtered range like so:
using ranges::views::filter;
using ranges::min_element;
std::vector<int> vals{1,2,3,4,5};
auto minVal = min_element(vals | filter([](int next){return next % 2 == 0;}));
如何检查返回的范围是否为空，如果不是，则访问该值?
其他范围算法也是如此，例如 ranges :: find ， ranges :: find_if 等.
推荐答案
很遗憾，您无法将临时范围传递给范围算法.
Unfortunately you cannot pass a temporary range into ranges algorithms.
这是因为它们在完成后将返回范围迭代器(可能是 ranges :: end(rng)).如果您传递了一个临时变量，则返回的迭代器将无效；引用一个临时的.Ranges库检测到此情况并返回一个虚拟的哨兵 ranges :: dangling 而不是迭代器.
This is because they will return a ranges iterator (possibly ranges::end(rng)) upon completion. If you pass a temporary, the returned iterator would be invalid; referencing a temporary. The ranges library detects this and returns a dummy sentinel, ranges::dangling instead of an iterator.
执行此操作的正确方法是首先组成过滤范围，然后 调用算法.
The proper way to do this is to first compose your filtered range then call the algorithm.
std::vector<int> vals {1, 2, 3, 4, 5};
auto even_vals = vals | filter([](int val){return val % 2 == 0;});
auto minVal = min_element(even_vals);
if (minVal == even_vals.end())
std::cout << "no values\n";
else
std::cout << *minVal; // prints "2"
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