使用变量模板的递归计算-gcc vs clang [英] Recursive computation using variable templates - gcc vs clang
问题描述
考虑下面的示例:
#include <cstdio>
template <int N>
int fib = fib<N - 1> + fib<N - 2>;
template <> int fib<2> = 1;
template <> int fib<1> = 1;
int main()
{
std::printf("%d %d %d", fib<4>, fib<5>, fib<6>);
}
-
GCC 7.x,8 .x,9.x和10.x都打印出
3 5 8
的预期结果。Clang 5.x,6.x,7.x,8.x,9.x和10.x全部打印出
1 3 4
。Clang 5.x, 6.x, 7.x, 8.x, 9.x, and 10.x all print out
1 3 4
as a result.C语的行为令人惊讶。
Clang's behavior is surprising.
我缺少的C ++标准中的变量模板实例化,全局变量和递归之间是否存在微妙的交互作用?
还是这是一个长期存在的Clang错误?
顺便说一句,标记为
fib
作为constexpr
解决了此问题( 在godbolt.org上 )。By the way, marking
fib
asconstexpr
solves the issue (on godbolt.org).推荐答案
来自< a href = http://eel.is/c++draft/basic.start.dynamic#1 rel = nofollow noreferrer> [basic.start.dynamic] / 1 :
具有静态存储持续时间的非局部变量的动态初始化是无序的(如果该变量是隐式或显式实例化的特殊化的话),如果该变量是部分有序的是一个内联变量,它不是隐式或显式实例化的专门化名称,否则是有序的。 [注:显式专门化的非内联静态数据成员或变量模板专门化已命令初始化。 —尾注]
fib< 4>
,fib< 5>
和fib< 6>
是具有静态存储持续时间的非局部变量,它们是隐式实例化的专业知识,因此它们的动态初始化是无序的。fib<4>
,fib<5>
andfib<6>
are non-local variables with static storage duration that are implicitly instantiated specializations, so their dynamic initialization is unordered.行为不是未定义的;必须有一些未指定的初始化顺序才能产生所看到的输出(每个 [ basic.start.dynamic] /3.3 (初始化顺序不确定)。实际上,clang按以下顺序初始化(请注意,动态初始化之前的变量具有静态初始化中的值0 ):
The behavior is not undefined; there must be some some unspecified ordering of initialization that produces the output seen (per [basic.start.dynamic]/3.3 the initializations are indeterminately sequenced). In fact, clang initializes in the following order (noting that a variable before dynamic initialization has the value 0 from static initialization):
fib<1> = 1 (actually static-initialized under [basic.start.static]/3) fib<2> = 1 (similarly) fib<4> = fib<2> + fib<3> = 1 + 0 = 1 fib<3> = fib<1> + fib<2> = 1 + 1 = 2 fib<5> = fib<3> + fib<4> = 2 + 1 = 3 fib<6> = fib<4> + fib<5> = 1 + 3 = 4
这与按顺序初始化的gcc(和MSVC)同等有效
fib< 3>
,fib< 4>
,fib< 5>
,fib< 6>
。This is equally as valid as gcc (and MSVC) initializating in the order
fib<3>
,fib<4>
,fib<5>
,fib<6>
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