c++ - “class classname* funcname(void) ”在C++中是什么意思？

Question

我在头文件中找到以下代码，并且在另一个头文件中定义了“BOOT”类。

class BOOT* boot(void);

它看起来像一个函数的声明，但它以 class开头。

Answer 1

这是一个详细的类型说明符:

Elaborated type specifiers may be used to refer to a previously-declared class name (class, struct, or union) or to a previously-declared enum name even if the name was hidden by a non-type declaration. They may also be used to declare new class names.

https://en.cppreference.com/w/cpp/language/elaborated_type_specifier
取自Artefacto和dfrib的答案，因为它使重点指向:它等效于:

class BOOT;
BOOT* boot(void);

在您的示例中，如果尚不知道，它实际上会对BOOT类进行前向声明。在同一页面上查看此示例 struct Data* Data;:

struct Node {
    struct Node* Next; // OK: lookup of Node finds the injected-class-name
    struct Data* Data; // OK: declares type Data at global scope
                       // and also declares the data member Data
    friend class ::List; // error: cannot introduce a qualified name
    enum Kind* kind; // error: cannot introduce an enum
};
 
Data* p; // OK: struct Data has been declared