c++ - 为什么不能使用函数的typedef来定义函数？

Question

根据ISO / IEC 14882:2011(E)第8.3.5.11条:

A typedef of function type may be used to declare a function but shall not be used to define a function

标准继续给出此示例:

typedef void F();
F fv; // OK: equivalent to void fv();
F fv { } // ill-formed
void fv() { } // OK: definition of fv

是什么激发了这个规则？似乎限制了函数typedef的潜在表达用处。

Answer 1

尽管这个问题是关于C++的，但是由于C++继承了C的typedef和函数指针，因此在这里可以使用C中相同问题的解释。对C有一个正式的解释。

Rationale for International Standard - Programming Languages C §6.9.1 Function definitions

An argument list must be explicitly present in the declarator; it cannot be inherited from a typedef (see §6.7.5.3). That is to say, given the definition:

typedef int p(int q, int r);

the following fragment is invalid:

p funk // weird
{ return q + r ; }

Some current implementations rewrite the type of, for instance, a char parameter as if it were declared int, since the argument is known to be passed as an int in the absence of a prototype. The Standard requires, however, that the received argument be converted as if by assignment upon function entry. Type rewriting is thus no longer permissible.