cpphs 是错误的还是在 if 定义的表达式中带有参数的宏的行为未定义？

Question

我有以下代码，它在 if defined 预处理器表达式中使用带参数的宏：

预处理器-games.c:

#define EXAMPLE_MACRO(arg1,arg2) (\
                               arg1 > arg2)

#if defined(EXAMPLE_MACRO)
#endif

我用 GCC 对其进行了预处理，并且没有错误：

$ gcc -E -undef -traditional -x assembler-with-cpp preprocessor-games.c -o preprocessed-games.c
$ echo $?
0
$

结果是可接受的输出：

# 1 "preprocessor-games.c"
# 1 "<built-in>"
# 1 "<command-line>"
# 1 "/usr/include/stdc-predef.h" 1 3 4

# 17 "/usr/include/stdc-predef.h" 3 4










































# 1 "<command-line>" 2
# 1 "preprocessor-games.c"

我使用替代预处理器对其进行预处理 - cpphs（旨在尽可能模拟 GCC 的预处理器以用于 Haskell） - 我得到一个空输出文件和一条错误消息：

$ cpphs --cpp -E -undef -traditional -x assembler-with-cpp preprocessor-games.c -o preprocessed-games.c
cpphs: macro EXAMPLE_MACRO expected 2 arguments, but was given 0
$ echo $?
1
$

是 cpphs 错误还是 macros with arguments in if defined expressions 的行为未定义？

Answer 1

行为已定义。在 C99 6.10.3/7 宏替换:

The identifier immediately following the define is called the macro name.

(以下段落继续定义object-like宏和function-like宏。但是宏名称是通用的。）

回到 6.10.1 条件包含:

The expression that controls conditional [...] and it may contain unary operator expressions of the form
defined identifier
or
defined ( identifier )
which evaluate to 1 if the identifier is currently defined as a macro name (that is, if it is predefined or if it has been the subject of a #define preprocessing directive without an intervening #undef directive with the same subject identifier), 0 if it is not.

object-like 或 function-like 宏之间没有区别：defined 检查已定义的 宏名称 ，这适用于两者。

cpphs 是错误的还是在 if 定义的表达式中带有参数的宏的行为未定义？

Is cpphs wrong or is the behavior of macros with arguments in if defined expressions undefined?

macros

gcc

undefined-behavior

c-preprocessor