size() causes error when switching from transform to filter

huangapple
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英文:

size() causes error when switching from transform to filter

问题

When views::filter([](auto n) { return n % 2 == 0; }); is activated, I fail to get [2,4]. When views::transform([](auto n) { return 2 * n; }); is activated, I succeed to get [2,4,6,8,10].

I don't understand the following error:

Filtering.cpp:17:37: error: no matching function for call to 'std::ranges::filter_view<std::ranges::iota_view<int, int>, main()::<lambda(auto:17)>>::size()'

英文:

When

  • views::filter([](auto n) { return n % 2 == 0; }); is activated, I fail to get [2,4].
  • views::transform([](auto n) { return 2 * n; }); is activated, I succeed to get [2,4,6,8,10].

<!---->

  1. #include &lt;iostream&gt;
  2. #include &lt;ranges&gt;
  3. #include &lt;sstream&gt;
  5. int main()
  6. {
  7.     using namespace std;
  8.     auto input = views::iota(1, 5 + 1);
  10.     auto output = input |
  11.                   views::filter([](auto n) { return n % 2 == 0; });
  12.                   //views::transform([](auto n) { return 2 * n; });
  15.     stringstream ss;
  16.     ss &lt;&lt; &quot;[&quot;;
  17.     for (auto i = 0; i &lt; output.size() - 1; ++i)
  18.     {
  19.         ss &lt;&lt; output[i] &lt;&lt; &#39;,&#39;;
  20.     }
  21.     ss &lt;&lt; output[output.size() - 1];
  22.     ss &lt;&lt; &quot;]&quot;;
  23.     cout &lt;&lt; ss.str();
  24.     return 0;
  25. }

I don't understand the following error.

> Filtering.cpp:17:37: error: no matching function for call to 'std::ranges::filter_view<std::ranges::iota_view<int, int>, main()::<lambda(auto:17)> >::size()'
17 | for (auto i = 0; i < output.size() - 1; ++i)

答案1

得分: 7

std::ranges::filter_view 不是一个 "有大小的范围"，这意味着您无法在它上面调用 size。这是因为过滤是按需进行的，因此要求查看的大小需要消耗每个元素。

一种选择是将视图转换为 std::vector 并使用它。在 C++20 中：

  1. auto output_view = input
  2.             | views::filter([](auto n) { return n % 2 == 0; }) 
  3.             | views::transform([](auto n) { return 2 * n; });
  5. std::vector<int> output(ranges::begin(output_view), ranges::end(output_view));

在 C++23 中：

  1. auto output = input
  2.             | views::filter([](auto n) { return n % 2 == 0; }) 
  3.             | views::transform([](auto n) { return 2 * n; })
  4.             | ranges::to<std::vector>();

另一种选择是利用 filter_view 是一个前向范围的事实，只要底层范围是前向的，您就可以检查是否将迭代器前进到末尾：

  1. stringstream ss;
  2. ss << "[";
  3. auto it = ranges::begin(output);
  4. for (; ranges::next(it) != ranges::end(output); ++it) {
  5.     ss << *it << ',';
  6. }
  7. ss << *it;
  8. ss << "]";
  9. cout << ss.str();
英文:

std::ranges::filter_view is not a "sized range", which means you can't call size on it. This is because the filtering is done on-demand, so asking for the size of the view would require consuming every element.

One option is to convert the view into a std::vector and use that. In C++20:

  1. auto output_view = input
  2.             | views::filter([](auto n) { return n % 2 == 0; }) 
  3.             | views::transform([](auto n) { return 2 * n; });
  5. std::vector&lt;int&gt; output(ranges::begin(output_view), ranges::end(output_view));

In C++23:

  1. auto output = input
  2.             | views::filter([](auto n) { return n % 2 == 0; }) 
  3.             | views::transform([](auto n) { return 2 * n; })
  4.             | ranges::to&lt;std::vector&gt;();

Another option is to make use of the fact that filter_view is a forward range so long as the underlying range is, so you can check if advancing the iterator hits the end:

  1. stringstream ss;
  2. ss &lt;&lt; &quot;[&quot;;
  3. auto it = ranges::begin(output);
  4. for (; ranges::next(it) != ranges::end(output); ++it) {
  5.     ss &lt;&lt; *it &lt;&lt; &#39;,&#39;;
  6. }
  7. ss &lt;&lt; *it;
  8. ss &lt;&lt; &quot;]&quot;;
  9. cout &lt;&lt; ss.str();

答案2

得分: 5

filter_view不是一个random_access_range，它没有size()方法。如果您专注于您展示的确切错误消息，它确实如此：在附近没有size()。

您可以将其类比为std::forward_list，而不是std::vector。您可以使用范围迭代，如下所示：

  1. const char *sep = "";
  2. for (const auto &n : output)
  3. {
  4.     ss << sep << n;
  5.     sep = ",";
  6. }
英文:

A filter_view is not a random_access_range and it does not have a size() method. If you focus your attention on the exact error message that you showed that's what it says: there is no size(), anywhere in the vicinity.

You can think of it as analogous to a std::forward_list, instead of std::vector. You can use range iteration, instead:

  1.     const char *sep=&quot;&quot;;
  2.     for (const auto &amp;n:output)
  3.     {
  4. 	    ss &lt;&lt; sep &lt;&lt; n;
  5. 	    sep=&quot;,&quot;;
  6.     }

答案3

得分: 4

  1. #include <iostream>
  2. #include <ranges>
  3. #include <sstream>
  5. int main()
  6. {
  7.     using namespace std;
  8.     auto input = views::iota(1, 5 + 1);
  9.     
  10.     auto output = input | views::filter([](const auto& n) {return n % 2 == 0; });
  11.     
  12.     stringstream ss;
  13.     ss << "[";
  14.     if (not ranges::empty(output))
  15.     {
  16.         ss << *ranges::begin(output);
  17.         for (const auto &x : output | views::drop(1)) 
  18.             ss << ',' << x;
  19.     }
  20.     ss << "]";
  21.     cout << ss.str();
  22.     return 0;
  23. }
英文: 
  1. #include &lt;iostream&gt;
  2. #include &lt;ranges&gt;
  3. #include &lt;sstream&gt;
  6. int main()
  7. {
  8.     using namespace std;
  9.     auto input = views::iota(1, 5 + 1);
  11.     auto output = input | views::filter([](const auto&amp; n) {return n % 2 == 0; });
  13.     stringstream ss;
  14.     ss &lt;&lt; &quot;[&quot;;
  15.     if (not ranges::empty(output))
  16.     {
  17.         ss &lt;&lt; *ranges::begin(output);
  18.         for (const auto &amp;x : output | views::drop(1)) 
  19.             ss &lt;&lt; &#39;,&#39; &lt;&lt; x;
  20.     }
  21.     ss &lt;&lt; &quot;]&quot;;
  22.     cout &lt;&lt; ss.str();
  23.     return 0;
  24. }

huangapple
  • 本文由 发表于 2023年4月13日 18:40:36
  • 转载请务必保留本文链接：https://go.coder-hub.com/76004478.html
  • c++
  • std-ranges
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