我的 Return 类型需要定义吗？

Question

Forward declaration 让我们推迟定义实际类型直到实现文件。这在 header 中允许用于指向前向声明类型的指针或引用。

即：

Returning by value does not require the type definition. A forward declaration is sufficient

有人可以用标准的实际报价来证实或否认这一点吗？我的印象是这不合法。

Answer 1

Returning by value does not require the type definition. A forward declaration is sufficient

按值声明 returns 的函数不需要类型定义。一个 well-formed 演示：

struct S;
S foo();
struct S {};
int main() {
    foo();
}
S foo() {
   return {};
}

定义或调用按值 return 的函数确实需要类型定义。标准草案 [basic.def.odr]:

5 Exactly one definition of a class is required in a translation unit if the class is used in a way that requires the class type to be complete. [ Example: ... [snip] ... [ Note: The rules for declarations and expressions describe in which contexts complete class types are required. A class type T must be complete if:

• [snip]
• 5.9 a function with a return type or argument type of type T is defined ([basic.def]) or called ([expr.call]), or
• [snip]

具有不完整 return 类型的函数的声明是隐式允许的，因为列表中的任何规则都没有禁止。

该规则在标准的后面 re-worded，并因异常 [dcl.fct] 而放宽（感谢 @cpplearner 指出了这条规则）：

11 Types shall not be defined in return or parameter types. The type of a parameter or the return type for a function definition shall not be an incomplete (possibly cv-qualified) class type in the context of the function definition unless the function is deleted ([dcl.fct.def.delete]).

---

一个ill-formed演示：

struct S;
S foo() {
    return {};
} // oops
struct S {};

另一个 ill-formed 演示：

struct S;
S foo();
int main() {
    foo(); // oops
}
struct S {};
S foo() {
    return {};
}