两次定义C预处理程序宏会发生什么？ [英] What happens when a C preprocessor macro is defined twice?

问题描述

我两次定义一个宏，如下所示：

I define a macro twice as follows:

#define a 2  
#define a 3   

我认为代码中 a 的任何出现都是替换为 2 ，当遇到 #define 3 时，不再有 a s在代码中可用，以替换为 3 ，因此 2 具有优先权

I thought any occurrence of a in the code would be replaced by 2, and when #define a 3 is encountered there are no more as are available in the code to be replaced by 3, so the 2 would take precedence.

但是当我执行它时， a 被3代替了，为什么？

But when I executed it a was replaced by 3, why?

推荐答案

如果您两次定义宏，则编译器至少应发出警告，即使不是错误。

If you define a macro twice like that, the compiler should at least give you warning, if not an error. It is an error.

§6.10.3/ 2：当前定义为类对象宏的标识符不应由另一个<$ c $重新定义。 c> #define 预处理指令，除非第二个定义是类似对象的宏定义并且两个替换列表相同。

§6.10.3/2 : An identifier currently defined as an object-like macro shall not be redefined by another #define preprocessing directive unless the second definition is an object-like macro definition and the two replacement lists are identical.

您可以通过显式删除先前的定义来重新定义宏：

You can redefine a macro by explicitly removing the previous definition:

#define a 2
/* In this part of the code, a will be replaced with 2 */
...

#undef a
#define a 3
/* From here on, a will be replaced with 3 */
...

读取文件时会使用宏定义激活宏替换指向文件中的 except （在大多数预处理指令中）。

Macro replacement happens as the file is read, using the macro definitions active at that point in the file, except inside (most) preprocessing directives.

§6.10/ 7：预处理指令中的预处理令牌除非另有说明，否则不进行宏扩展。

§6.10/7: The preprocessing tokens within a preprocessing directive are not subject to macro expansion unless otherwise stated.

§6.10.3.5/ 1：宏定义n持续​​（独立于块结构）直到遇到相应的 #undef 指令，或者（如果没有遇到）直到预处理转换单元的末尾。

§6.10.3.5/1: A macro definition lasts (independent of block structure) until a corresponding #undef directive is encountered or (if none is encountered) until the end of the preprocessing translation unit.

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