使用变量模板的递归计算 - gcc vs clang

Question

考虑以下示例：

#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、10.x都打印出了3 5 8的预期结果。
Clang 5.x、6.x、7.x、8.x、9.x、10.x全部打印出来1 3 4 结果。

live example on godbolt.org

Clang 的行为令人惊讶。

在 C++ 标准中变量模板实例化、全局变量和递归之间是否存在任何我遗漏的微妙交互？

或者这是一个长期存在的 Clang 错误？

顺便说一下，将 fib 标记为 constexpr 解决了问题 (on godbolt.org)。

Answer 1

来自[basic.start.dynamic]/1：

Dynamic initialization of a non-local variable with static storage duration is unordered if the variable is an implicitly or explicitly instantiated specialization, is partially-ordered if the variable is an inline variable that is not an implicitly or explicitly instantiated specialization, and otherwise is ordered. [ Note: An explicitly specialized non-inline static data member or variable template specialization has ordered initialization. — end note ]

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特化，因此它们的动态初始化是无序的。

该行为并非未定义；必须有一些未指定的初始化顺序产生看到的输出（每个 [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 gcc（和 MSVC）按 fib<3>、fib<4>、fib<5>、fib<6>.

顺序初始化一样有效

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