c - void* 是否始终具有与 char* 相同的表示形式？

Question

void* 是否总是与 char* 具有相同的表示形式？

详细信息:

我想使用一个可变参数函数，该函数采用 (char*)0 终止的 char*，如下所示:

int variadic(char*, ...); //<-prototype
variadic("foo", "bar", (char*)0); //<- usage

我想用NULL替换(char*)0，但是从 http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1570.pdf的:

66) The macro NULL is defined in (and other headers) as a null pointer constant; see 7.19.

3 An integer constant expression with the value 0, or such an expression cast to type void *, is called a null pointer constant. 66) If a null pointer constant is converted to a pointer type, the resulting pointer, called a null pointer, is guaranteed to compare unequal to a pointer to any object or function.

我不能，因为在 variadic 上下文中，我绝对需要一个 char* 而一个普通的 0 是 Not Acceptable 。

如果我定义:

#define NIL (void*)0 /*<= never a plain 0*/

我用它来终止我的 variadic(char*,...) 是否合法？

Answer 1

C11，§6.2.5，¶28(草案 N1570)说:

A pointer to void shall have the same representation and alignment requirements as a pointer to a character type. 48) Similarly, pointers to qualified or unqualified versions of compatible types shall have the same representation and alignment requirements.

(强调我的)。