c - 以下程序是否严格符合 C99 程序？

Question

标题几乎说明了一切，但我会重申问题...

下面的程序是不是C99标准下的“严格符合程序”？

#include <stdlib.h>
/* Removing any pre-existing macro definition, in case one should exist in the implementation.
 * Seems to be allowed under 7.1.3 para 3, as malloc does not begin with _X where X is any  capital letter.
 * And 7.1.4 para 1 explicitly permits #undef of such macros.
 */
#ifdef malloc    
#undef malloc     
#endif            

/* Macro substitution has no impact on the external name malloc
 * which remains accessible, e.g., via "(malloc)(s)".  Such use of
 * macro substitution seems enabled by 7.1.4 para 1, but not specifically
 * mentioned otherwise.
 */
void * journalling_malloc(size_t size);
#define malloc(s)     ((journalling_malloc)(s))      

int main(void)
{
    return malloc(10) == NULL ? 1 : 0;     
    /* Just for the sake of expanding the 
     * macro one time, return as exit code
     * whether the allocation was performed. 
     */
}

Answer 1

让我们看看 C99 标准是怎么说的:

参见 7.1.3，§1，第 5 条:

Each identifier with file scope listed in any of the following subclauses [...] is reserved for use as a macro name and as an identifier with file scope in the same name space if any of its associated headers is included.

当您包含 stdlib.h 时，名称 malloc 被保留用作宏名称。

但是 7.1.4，§1 允许在保留名称上使用 #undef:

The use of #undef to remove any macro definition will also ensure that an actual function is referred to.

这使得重新#define malloc 成为可能，根据 7.1.3，§2，这会导致未定义的行为:

If the program [...] defines a reserved identifier as a macro name, the behavior is undefined.

标准为什么做这个限制？因为标准库的其他函数可能会根据原始函数实现为类似函数的宏，因此隐藏声明可能会破坏这些其他函数。

在实践中，只要您对 malloc 的定义满足标准为库函数提供的所有规定，您就应该没问题，这可以通过包装对 malloc( )。