blocks|key|196982|text|最简洁的方法仍然是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|196983|for+(int+i=0;+i<n;+%2B%2Bi)|code-block|syntax|javascript|196984|我猜你可以做到这一点，但我不认为这是巧妙的：|196985|#include+<iostream>

int+main()
{
++for+(+auto+i+:+{+1,2,3,4,5+}+)
++{
++++std::cout<<i<<std::endl;
++}
}|196986|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

The neatest way is still this:

<pre><code>for (int i=0; i&lt;n; ++i)
</code></pre>

I guess you can do this, but I wouldn't call it so neat:

<pre><code>#include &lt;iostream&gt;

int main()
{
 for ( auto i : { 1,2,3,4,5 } )
 {
 std::cout&lt;&lt;i&lt;&lt;std::endl;
 }
}
</code></pre>

blocks|key|2823708|text|虽然它不是由C%2B%2B11提供的，但您可以编写自己的视图或使用boost中的视图：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2823709|#include+<boost/range/irange.hpp>
#include+<iostream>

int+main(int+argc,+char+**argv)
{
++++for+(auto+i+:+boost::irange(1,+10))
++++++++std::cout+<<+i+<<+"\n";
}|code-block|syntax|javascript|2823710|此外，结合新的for循环，Boost.Range包含一些更有趣的范围，您会发现这些范围非常有用。例如，您可以获得一个reversed+view。|offset|length|style|CODE|2823711|entityMap|0|LINK|mutability|MUTABLE|url|https://www.boost.org/doc/libs/1_71_0/libs/range/doc/html/index.html|1|https://www.boost.org/doc/libs/1_71_0/libs/range/doc/html/range/reference/adaptors/reference/reversed.html^0|0|0|7|3|D|B|D|B|0|1M|D|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@$I|13|J|14|1|15]|$I|16|J|17|1|18]]|A|$]]|$1|M|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]|U|$5|P|Q|R|A|$S|V]]]]

While its not provided by C++11, you can write your own view or use the one from boost:

<pre><code>#include &lt;boost/range/irange.hpp&gt;
#include &lt;iostream&gt;

int main(int argc, char **argv)
{
 for (auto i : boost::irange(1, 10))
 std::cout &lt;&lt; i &lt;&lt; "\n";
}
</code></pre>

Moreover, <a href="https://www.boost.org/doc/libs/1_71_0/libs/range/doc/html/index.html" rel="noreferrer"><code>Boost.Range</code></a> contains a few more interesting ranges which you could find pretty useful combined with the new <code>for</code> loop. For example, you can get a <a href="https://www.boost.org/doc/libs/1_71_0/libs/range/doc/html/range/reference/adaptors/reference/reversed.html" rel="noreferrer">reversed view</a>.

blocks|key|2823727|text|根据必须对整数执行的操作，还要考虑<numeric>标头，特别是std::iota与std::transform和std::fill的结合，具体取决于大小写。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2823728|entityMap|0|LINK|mutability|MUTABLE|url|http://en.cppreference.com/w/cpp/algorithm/iota^0|H|9|W|9|16|E|1L|9|W|9|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@$9|W|A|X|1|Y]]|E|$]]|$1|F|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

Depending on what you have to do with the integer, consider the also the <code>&lt;numeric&gt;</code> header, in particular 
<a href="http://en.cppreference.com/w/cpp/algorithm/iota" rel="nofollow noreferrer"><code>std::iota</code></a> in conjunction with <code>std::transform</code> and <code>std::fill</code> depending on the cases.

blocks|key|2823753|text|有了C%2B%2B20，我们就有了范围。你可以从它的作者Eric+Niebler那里下载最新的稳定版本，从his+github下载，或者访问Wandbox来尝试。您感兴趣的是ranges::views::iota，它使以下代码合法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2823754|#include+<range/v3/all.hpp>
#include+<iostream>

int+main()+{
++++using+namespace+ranges;

++++for+(int+i+:+views::iota(1,+10))+{
++++++++std::cout+<<+i+<<+'+';
++++}
}|code-block|syntax|javascript|2823755|这种方法的伟大之处在于view是懒惰的。这意味着，即使views::iota表示从1到10+exclusive的范围，但在某一点上，该范围中只有一个int存在。元素是按需生成的。|2823756|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/ericniebler/range-v3|1|https://wandbox.org/permlink/LzHfd7Vvi8FgAN12^0|2|5|2B|J|1D|A|0|1U|7|1|0|0|B|4|R|B|15|1|17|2|22|3|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]|$9|14|A|15|1|16]]|E|$]]|$1|F|3|G|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

With <code>C++20</code> we will have ranges. You can try them by downloading the lastest stable release from it's author, Eric Niebler, from <a href="https://github.com/ericniebler/range-v3" rel="noreferrer">his github</a>, or go to <a href="https://wandbox.org/permlink/LzHfd7Vvi8FgAN12" rel="noreferrer">Wandbox</a>. What you are interested in is <code>ranges::views::iota</code>, which makes this code legal:

<pre><code>#include &lt;range/v3/all.hpp&gt;
#include &lt;iostream&gt;

int main() {
 using namespace ranges;

 for (int i : views::iota(1, 10)) {
 std::cout &lt;&lt; i &lt;&lt; ' ';
 }
}
</code></pre>

What's great about this approach is that <code>view</code>s are lazy. That means even though <code>views::iota</code> represents a range from <code>1</code> to <code>10</code> exclusive, no more than one <code>int</code> from that range exists at one point. The elements are generated on demand.

blocks|key|198939|text|如果你不介意以相反的顺序做循环，你可以替换|type|unstyled|depth|inlineStyleRanges|entityRanges|data|198940|for+(int+i=0;+i<n;+%2B%2Bi)|offset|length|style|CODE|198941|使用更简单的|198942|for+(int+i=n;+i--;)|198943|entityMap^0|0|0|N|0|0|0|J|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@$D|P|E|Q|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|M|$]]

If you don't mind doing the loop in reverse order, you can replace
<code>for (int i=0; i&lt;n; ++i)</code>
with simpler
<code>for (int i=n; i--;)</code>

blocks|key|3036541|text|好吧，我真的很喜欢这个解决方案，提供了here+(很抱歉，它没有翻译成英语)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3036542|#define+FOR(I,UPPERBND)+for(int+I+=+0;+I<int(UPPERBND);+%2B%2BI)|code-block|syntax|javascript|3036543|主要思想是这样描述的:当我们谈论简单的迭代索引循环时，我们不需要考虑它。然而，当我们使用for(;;)构造时，总是有三个步骤:初始化，结束条件检查，迭代。对于像i:[0，n)这样的简单循环来说，这是一种过度的杀伤力。我喜欢这个想法，我们想用一种简单的方式写简单的东西。当你看到(i，N)+--你就会知道，没有什么特别的。当你看到for(;;)结构时，你必须更加小心，看看它的所有三个部分。这只是那篇文章中的一个例子：|3036544|for+(int+iter=0;+iter<nb_iter;+iter%2B%2B)+{++++++++++//+some+iterative+computation
++++for+(int+c=0;+c<mesh.cells.nb();+c%2B%2B)+++++++++//+loop+through+all+tetrahedra
++++++++for+(int+lv0=0;+lv0<4;+lv0%2B%2B)+++++++++++++//+for+every+pair+of
++++++++++++for+(int+lv1+=+lv0%2B1;+lv1<4;+lv1%2B%2B)+++//+vertices+in+the+tet
++++++++++++++++for+(int+d=0;+d<3;+d%2B%2B)+{+++++++++//+do+stuff+for+each+of+3+dimensions
++++++++++++++++++++nlRowScaling(weight);
++++++++++++++++++++nlBegin(NL_ROW);
++++++++++++++++++++nlCoefficient(mesh.cells.vertex(c,+lv0)*3+%2B+d,++1);
++++++++++++++++++++nlCoefficient(mesh.cells.vertex(c,+lv1)*3+%2B+d,+-1);
++++++++++++++++++++nlEnd(NL_ROW);
++++++++++++++++}
++++[...]
}|3036545|成为：|3036546|FOR(iter,+nb_iter)+{
++++FOR(c,+mesh.cells.nb())
++++++++FOR(lv0,+4)
++++++++++++for+(int+lv1+=+lv0%2B1;+lv1<4;+lv1%2B%2B)
++++++++++++++++FOR(d,+3)+{
++++++++++++++++++++nlRowScaling(weight);
++++++++++++++++++++nlBegin(NL_ROW);
++++++++++++++++++++nlCoefficient(mesh.cells.vertex(c,+lv0)*3+%2B+d,++1);
++++++++++++++++++++nlCoefficient(mesh.cells.vertex(c,+lv1)*3+%2B+d,+-1);
++++++++++++++++++++nlEnd(NL_ROW);
++++++++++++++++}
++++[...]
}|3036547|你看，你应该集中你的注意力。|3036548|entityMap|0|LINK|mutability|MUTABLE|url|https://habr.com/ru/post/516630/^0|J|4|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|15|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|18|8|@]|9|@]|C|$G|H]]|$1|Q|3|R|5|6|7|19|8|@]|9|@]|C|$]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]]]

Well, I really like the solution, provided <a href="https://habr.com/ru/post/516630/" rel="nofollow noreferrer">here</a> (I'm sorry, it's not translated in English):
<pre><code>#define FOR(I,UPPERBND) for(int I = 0; I&lt;int(UPPERBND); ++I)
</code></pre>
The main idea is described in this way: when we talk about simple iteration-indexed-cycles, we do not need to think about it. However, when we use for(;;) construction- there are always three steps: initialization, end condition check, iteration. And this is an overkill for such a simple loop, as just i:[0,n).
I liked the idea, that we want to write simple things in a simple way. When you see that FOR(i,N) - you just know, that there are nothing special. When you see the for(;;) construction - you have to be more careful and see all three parts of it.
Just an example from that article:
<pre><code>for (int iter=0; iter&lt;nb_iter; iter++) { // some iterative computation
 for (int c=0; c&lt;mesh.cells.nb(); c++) // loop through all tetrahedra
 for (int lv0=0; lv0&lt;4; lv0++) // for every pair of
 for (int lv1 = lv0+1; lv1&lt;4; lv1++) // vertices in the tet
 for (int d=0; d&lt;3; d++) { // do stuff for each of 3 dimensions
 nlRowScaling(weight);
 nlBegin(NL_ROW);
 nlCoefficient(mesh.cells.vertex(c, lv0)*3 + d, 1);
 nlCoefficient(mesh.cells.vertex(c, lv1)*3 + d, -1);
 nlEnd(NL_ROW);
 }
 [...]
}
</code></pre>
become a:
<pre><code>FOR(iter, nb_iter) {
 FOR(c, mesh.cells.nb())
 FOR(lv0, 4)
 for (int lv1 = lv0+1; lv1&lt;4; lv1++)
 FOR(d, 3) {
 nlRowScaling(weight);
 nlBegin(NL_ROW);
 nlCoefficient(mesh.cells.vertex(c, lv0)*3 + d, 1);
 nlCoefficient(mesh.cells.vertex(c, lv1)*3 + d, -1);
 nlEnd(NL_ROW);
 }
 [...]
}
</code></pre>
And you see, where you should concentrate your attention.

Since C++11 introduced the range-based for loop (<a href="https://stackoverflow.com/questions/9005835/range-based-for-in-c11">range-based for in c++11</a>), what is the neatest way to express looping over a range of integers?

Instead of 

<pre><code>for (int i=0; i&lt;n; ++i)
</code></pre>

I'd like to write something like

<pre><code>for (int i : range(0,n))
</code></pre>

Does the new standard support anything of that kind?

Update: this article describes how to implement a range generator in C++11: <a href="https://en.wikipedia.org/wiki/Generator_(computer_programming)#C++" rel="noreferrer">Generator in C++</a>

Neatest way to loop over a range of integers

既然C++11引入了基于范围的for循环()，那么表示一系列整数上的循环的最简洁的方式是什么？而不是for (int i=0; i<n; ++i)我想写这样的东西for (int i : range(0,n))新标准支持这类东西吗？更新：本文描述如何在C++11：中实现范围生成器

问循环遍历一系列整数的最简洁方法
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