Linux __stringify && __stringify_1__linux

Linux has a stringify header file, the main contents of which are as follows:

#ifndef __linux_stringify_h
#define __LINUX_STRINGIFY_H/

* Indirect stringification.  Doing two levels allows the parameter to be a
 * macro itself.  For example, compile With-dfoo=bar, __stringify (FOO)
 * converts to "bar".
 * *

#define __STRINGIFY_1 (x ...)	#x
#define __STRINGIFY (x ...)	__stringify_1 (x)

#endif/	*!__linux_stringify_h * *

There are two define in it, which is not quite clear according to the annotation. Why should the middle pass through __stringify_1 to convert? Direct use of #define __STRINGIFY (x ...) #x岂不是更加简洁.

Programming Test:

#include <stdio.h>

#define Triple-a BBB

#define __STRINGIFY_1 (x.) #x
#define __STRINGIFY (x.) __ Stringify_1 (x)

#define __STRINGIFY_SRC (x ...) #x

main ()
{
	printf ("Stringify (AAA):     %s\n", __stringify (AAA));
	printf ("Stringify_src (AAA):%s\n", __STRINGIFY_SRC (AAA));
}

The results of the operation are as follows:

Stringify (AAA): BBB
STRINGIFY_SRC (AAA): AAA

Then go back and look at the notes.

But there's this sentence in the comment "doing two levels allows the parameter to be a macro itself."

How to understand it.

If you define a #define BBB CCC before #define AAA BBB, #define CCC DDD ..., how is the result of this program running?

Oh, we can try.