linux-kernel - Re: [PATCH] [Coding Style]: fs/ext{3,4}/ext{3,4}

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Date:	Thu, 10 Jan 2008 22:03:58 +0100
From:	Roel Kluin <12o3l@...cali.nl>
To:	Al Viro <viro@...IV.linux.org.uk>
CC:	Richard Knutsson <ricknu-0@...dent.ltu.se>,
	Mathieu Segaud <mathieu.segaud@...ala.cx>,
	akpm@...ux-foundation.org, linux-ext4@...r.kernel.org,
	linux-kernel@...r.kernel.org, linux-pcmcia@...ts.infradead.org
Subject: Re: [PATCH] [Coding Style]: fs/ext{3,4}/ext{3,4}_jbd{,2}.c

Al Viro wrote:
> __func__ is C99, but it's not what __FUNCTION__ used to be - it's not a
> string literal.  6.4.2.2(1):
> 
> 	The identifier __func__ shall be implicitly declared by the translator
> as if, immediately following the opening brace of each function definition,
> the declaration
> 	static const char __func__[] = "function-name";
> appeared, where function-name is the name of the lexically-enclosing function.
> 
> IOW, it's a phase 7 (parsing and translation of translation units) and not
> phase 4 (preprocessor).  Practical implications are:
> 	* _way_ fewer kludges
> 	* it happens after phase 6 (string literal concatenation)
> So __FUNCION__ " is called" within body of foo() would result in
> "foo is called" while __func__ "is called" is a syntax error.
> 
> These days old gcc __FUNCTION__ is gone; it's a macro expanding to __func__,
> so behaviour does *not* match the original (see above).

so if I understand correctly, this requires a fix:
--
__FUNCTION__ is a macro expanding to __func__ and translated as if previously
in that function was declared:
	static const char __func__[] = "function-name";

As is DEBUG() - when active - will produce a syntax error.

Signed-off-by: Roel Kluin <12o3l@...cali.nl>
---
diff --git a/drivers/pcmcia/au1000_xxs1500.c b/drivers/pcmcia/au1000_xxs1500.c
index ce9d5c4..855e1d6 100644
--- a/drivers/pcmcia/au1000_xxs1500.c
+++ b/drivers/pcmcia/au1000_xxs1500.c
@@ -56,7 +56,7 @@
 #define PCMCIA_IRQ		AU1000_GPIO_4
 
 #if 0
-#define DEBUG(x,args...)	printk(__FUNCTION__ ": " x,##args)
+#define DEBUG(x, args...)	printk("%s: ", __func__, x, ##args)
 #else
 #define DEBUG(x,args...)
 #endif

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