emplace_back 导致静态 constexpr 成员出现 link 错误

Question

为什么emplace_back引用需要定义的成员？ emplace_back(integer literal) 和 emplace_back(static constexpr integer member) 有什么区别？

如果我切换到 C++17，它编译得很好。我发现在 C++17 中静态 constexpr 数据成员是隐式的 inlined。这是否意味着编译器隐式地为它们创建了一个定义？

示例代码：

class base {
    int n;
public:
    base(int n):n(n) {}
};

struct base_trait {
    static constexpr int n = 1;
};

int main(void) {
    vector<base> v;
    v.emplace_back(1);  // ok
    v.emplace_back(base_trait::n);  // link error with -std=c++14, ok with -std=c++17
    return 0;
}

Answer 1

如您所说，emplace_back 通过引用获取参数，因此传递 base_trait::n 会导致它成为 odr-used.

an object is odr-used if its value is read (unless it is a compile time constant) or written, its address is taken, or a reference is bound to it;

在C++17之前，这意味着这里需要base_trait::n的定义。但自从 C++17 以来，行为发生了变化，对于 constexpr static data member，不再需要超出 class 的定义。

If a const non-inline (since C++17) static data member or a constexpr static data member (since C++11) is odr-used, a definition at namespace scope is still required, but it cannot have an initializer. This definition is deprecated for constexpr data members (since C++17).

A static data member may be declared inline. An inline static data member can be defined in the class definition and may specify an initializer. It does not need an out-of-class definition. (since C++17)