三个#include卫兵，永远不会破坏？ [英] Three #include guards, which will never break?

问题描述

有时程序员会在命令行定义宏，如：


$ gcc -DF1_H ...

$ gcc -DF1_H = 0 ...

$ gcc -DF1_H = 1 ...


标记为＃1，＃2，＃3的以下三行中的一行可提供#include

保护和避免错误可能由冗余#include引入。

每个人都有任何缺陷吗？哪一个永远不会破坏？


#if F1_H / *＃1 * /

/ * #if！defined（F1_H）* / / *＃2 * /

/ * #ifndef F1_H * / / *＃3 * /

#define F1_H


/ *更多东西在这里... * /


#endif

-

lovecreatesbeauty

解决方案

gcc -DF1_H ...

gcc -DF1_H = 0 ...

gcc -DF1_H = 1 ...

标记为＃1，＃2，＃3的以下三行之一可提供#include

保护和避免错误可能由冗余#include引入。

每个人都有任何缺陷吗？哪一个永远不会破坏？


#if F1_H / *＃1 * /

/ * #if！defined（F1_H）* / / *＃2 * /

/ * #ifndef F1_H * / / *＃3 * /

#define F1_H


/ *更多东西在这里... * /


#endif

-

lovecreatesbeauty

Sometimes programmers will define macros at command line like:

$ gcc -DF1_H ...
$ gcc -DF1_H=0 ...
$ gcc -DF1_H=1 ...

One of following three lines labeled as #1, #2, #3 may provide #include
guard and avoid errors may be introduced by redundant #include. Is
there any flaw with each of them? Which one will never break?

#if F1_H /*#1*/
/*#if !defined(F1_H)*/ /*#2*/
/*#ifndef F1_H */ /*#3*/
#define F1_H

/*more stuff here ...*/

#endif
--
lovecreatesbeauty

解决方案

gcc -DF1_H ...

gcc -DF1_H=0 ...

gcc -DF1_H=1 ...

One of following three lines labeled as #1, #2, #3 may provide #include
guard and avoid errors may be introduced by redundant #include. Is
there any flaw with each of them? Which one will never break?

#if F1_H /*#1*/
/*#if !defined(F1_H)*/ /*#2*/
/*#ifndef F1_H */ /*#3*/
#define F1_H

/*more stuff here ...*/

#endif
--
lovecreatesbeauty

这篇关于三个#include卫兵，永远不会破坏？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！