将 CRTP 与 static_assert 一起使用时出现编译器错误
Compiler error when using CRTP with static_assert
考虑以下代码:
template<typename Derived>
struct Base {
static constexpr int x_base = Derived::x_derived;
//static_assert(x_base > 1, "Oops");
};
struct Derived : public Base<Derived> {
static constexpr int x_derived = 5 ;
};
Base<Derived> obj;
这在 gcc 上编译得很好,但如果我取消注释 static_assert
行,它会抱怨
error: incomplete type 'Derived' used in nested name specifier
static constexpr int x_base = Derived::x_derived;
我尝试使用从 4.9 到 5.3 的不同版本的 gcc,我得到了同样的错误(你可以在 godbolt here 上尝试)。即使没有 static_assert
,clang 也拒绝编译它,并抱怨说
error: no member named 'x_derived' in 'Derived'
static constexpr int x_base = Derived::x_derived;
哪个编译器是正确的(如果有的话)?
有修复代码的好方法吗?
访问嵌套名称需要 class 完整,但 Derived
此处尚未完成:
template<typename Derived>
struct Base {
static constexpr int x_base = Derived::x_derived;
^^^^^^^^^
};
所以代码格式错误。
有一些解决方法。首先,您可以单独将值作为模板参数传递:
template <typename Derived, int x_derived>
struct Base {
static constexpr int x_base = x_derived;
};
struct Derived : public Base<Derived, 5> { };
其次,如果可能(例如,您不需要 x_derived
声明任何成员),您可以将值移动到函数中以延迟其实例化:
template<typename Derived>
struct Base {
static constexpr int x_base() {
static_assert(Derived::x_derived > 1, "Oops");
return Derived::x_derived;
}
};
struct Derived : public Base<Derived> {
static constexpr int x_derived = 5;
};
考虑以下代码:
template<typename Derived>
struct Base {
static constexpr int x_base = Derived::x_derived;
//static_assert(x_base > 1, "Oops");
};
struct Derived : public Base<Derived> {
static constexpr int x_derived = 5 ;
};
Base<Derived> obj;
这在 gcc 上编译得很好,但如果我取消注释 static_assert
行,它会抱怨
error: incomplete type 'Derived' used in nested name specifier
static constexpr int x_base = Derived::x_derived;
我尝试使用从 4.9 到 5.3 的不同版本的 gcc,我得到了同样的错误(你可以在 godbolt here 上尝试)。即使没有 static_assert
,clang 也拒绝编译它,并抱怨说
error: no member named 'x_derived' in 'Derived'
static constexpr int x_base = Derived::x_derived;
哪个编译器是正确的(如果有的话)? 有修复代码的好方法吗?
访问嵌套名称需要 class 完整,但 Derived
此处尚未完成:
template<typename Derived>
struct Base {
static constexpr int x_base = Derived::x_derived;
^^^^^^^^^
};
所以代码格式错误。
有一些解决方法。首先,您可以单独将值作为模板参数传递:
template <typename Derived, int x_derived>
struct Base {
static constexpr int x_base = x_derived;
};
struct Derived : public Base<Derived, 5> { };
其次,如果可能(例如,您不需要 x_derived
声明任何成员),您可以将值移动到函数中以延迟其实例化:
template<typename Derived>
struct Base {
static constexpr int x_base() {
static_assert(Derived::x_derived > 1, "Oops");
return Derived::x_derived;
}
};
struct Derived : public Base<Derived> {
static constexpr int x_derived = 5;
};