为什么总是要调用“ static_assert”？ [英] Why does the `static_assert` always get invoked?

问题描述

如果 USE_STATIC_ASSERT 为 0 ，则按预期方式工作（从列表中获取索引类型）。如果为1，则始终断开 static_assert（）。我本以为 static_assert（）仅在所有 typename 用尽时才会发生。为什么不是这样？

If USE_STATIC_ASSERT is 0, this works as expected (getting indexed type from the list). If 1 the static_assert() is always tripped. I would have thought that the static_assert() would only happen if all the typenames were exhausted. Why is this not so?

#define USE_STATIC_ASSERT 1
template <unsigned int I, typename ...Ts>
struct items;

template <typename T, typename ...Ts>
struct items<0, T, Ts...>
{
    typedef T type;
};

template <unsigned int I, typename T, typename ...Ts>
struct items<I, T, Ts...> : items<I-1, Ts...>
{
};

#if USE_STATIC_ASSERT
template <unsigned int I>
struct items<I>
{
    static_assert(false, "Ran out of Ts.");
};
#endif


int main()
{
    cout << is_same<float, items<1, int, float, double>::type>::value << endl;
}

推荐答案

即使部分专业化包含 static_assert 的项未实例化，允许编译器根据§14.6[temp。 res] / p8：

Even if the partial specialization of items that contains the static_assert is not instantiated, the compiler is allowed to reject this code according to §14.6 [temp.res]/p8:

知道哪些名称是类型名称，可以检查每个模板的语法。对于可为其生成有效专业化的模板，不得发布诊断
。 如果无法为模板生成有效的专业化名称，并且未实例化该模板，则该模板格式错误，无需诊断。

要解决此问题，可以使 static_assert 中的表达式依赖于其他类模板：

To work around that, you can make the expression in static_assert dependent on other class template:

#include <type_traits>

template <unsigned int I>
struct AlwaysFalse : std::false_type {};

template <unsigned int I>
struct items<I>
{
    static_assert(AlwaysFalse<I>{}, "Ran out of Ts.");
    //            ~~~~~~~~~~~~~~~^
};

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