c++ - 局部变量的静态初始化

Question

来自 Scott Meyers Effective C++:

if you never call a function emulating a non-local static object, you never incur the cost of constructing and destructing the object, something that can’t be said for true non-local static objects.

功能:

FileSystem& tfs()
{ 
    static FileSystem fs;
    return fs; 
}

但标准说:

Constant initialization (3.6.2) of a block-scope entity with static storage duration, if applicable, is performed before its block is first entered. An implementation is permitted to perform early initialization of other block-scope variables with static or thread storage duration under the same conditions that an implementation is permitted to statically initialize a variable with static or thread storage duration in namespace scope (3.6.2).

这意味着即使我们不调用函数 tfs()，我们也无法确定 fs 变量是否已初始化。因为允许实现对具有静态存储持续时间的变量执行早期初始化。

谁是对的，或者我错过了什么？

Answer 1

Constant initialization描述可以在编译时确定的初始化。

只有在 C++11 及更高版本中，才能考虑具有非平凡构造函数的类型:

if the constructor is constexpr

在“Effective C++”中，Meyers 将您问题中的类按字面意思描述为:

class FileSystem {...};

这意味着如果 C++ 标准是正确的，“有效 C++”也可以保持正确即使在 C++11 之后只要为 FileSystem 提供构造函数不是 constexpr。