为什么升压uniform_int_distribution需要一个封闭的范围内(而不是半开区间,以下常见的C ++使用)? [英] Why boost uniform_int_distribution takes a closed range (instead of a half-open range, following common C++ usage)?
问题描述
标题说明了一切。甚至还有href=\"http://www.boost.org/doc/libs/1_55_0/doc/html/boost_random/tutorial.html\" rel=\"nofollow\">在文档页面警告一个
The title says it all. There's even a warning in the documentation pages:
为什么这样做时,在C ++中常见的做法就是用开放的范围[开始,结束)?
Why do this, when the common practice in C++ is to use open ranges [begin, end) ?
推荐答案
只有在封闭的范围内,你可以创建一个 uniform_int_distribution
,产生任何整数:
Only with closed ranges, you can create a uniform_int_distribution
, that produces any integer:
uniform_int_distribution<int> dist(std::numeric_limits<int>::min(), std::numeric_limits<int>::max());
如果这将是一个半开放的范围内,你永远无法达到的std :: numeric_limits&LT; INT&GT; :: MAX()
,但只有的std :: numeric_limits&LT; INT方式&gt; :: MAX() - 1
If this would be a half-open range, you could never reach std::numeric_limits<int>::max()
, but only std::numeric_limits<int>::max() - 1
.
这对的std ::在C ++ 11标准库uniform_int_distribution
同样的情况。
有关迭代半开的范围很常见,因为人们可以很容易的前preSS空范围(通过设置开始==结束
)。这是没有意义的分布。
Half-open ranges for iterators are common, because one can easily express empty ranges (by setting begin == end
). This doesn't make sense for distributions.
参考:斯蒂芬T. Lavavej提到了他的谈话 RAND()在融入本土2013(约14分钟)有害。这次演讲是关于C ++ 11 &LT;随机方式&gt;
,当然,同样的道理也适用于提升以及
Reference: Stephan T. Lavavej mentions this exact reason in his talk "rand() Considered Harmful" at Going Native 2013 (around minute 14). This talk is about C++11 <random>
, but of course the same reasoning applies to boost as well.
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