如何将一个宏的结果传递给另一个宏? [英] How to pass a macro's result to another macro?

问题描述

我的C代码中有两个宏，可以帮助我编写某些变量的名称.例如，请考虑以下内容:

I have two macros in my C code the helps me to compose the name of certain variables. As an example, consider the following:

#define MACROA(name) A_##name
#define MACROB(name) B_##name

void *MACROB(MACROA(object));

因此，我正在尝试声明一个名为B_A_object的变量.但是，这不起作用，编译器向我抛出该消息:

So, I'm trying to declare a variable called B_A_object. However, this doesn't work and the compiler throws me the message:

object.c:27:21: error: a parameter list without types is only allowed in a function definition
void *MACROB(MACROA(object));
                    ^
object.c:26:26: note: expanded from macro 'MACROB'
#define MACROB(name) B_##name
                         ^

因此，似乎预处理器未采用MACROA(object)的结果，但它正在考虑表达式本身，以便尝试生成B_MACROA(object).那么，我该怎么做才能使预处理器考虑将宏的结果传递给另一个宏的方法?

So, it seems the preprocessor is not taking the result of MACROA(object), but it is considering the expression itself so that it tries to make B_MACROA(object). So, what do I have to do to make the preprocessor consider the result of a macro passed to another macro?

推荐答案

串联运算符的行为很奇怪.它先连接起来，然后再求值:

The concatenation operator acts weird. It concatenates first and evaluates later:

void *MACROB(MACROA(object));  // The original line
void *B_MACROA(object);       // Becomes this, nothing more to expand

您可以通过以下方式解决它:

You can solve it this way:

#define CONC(a,b) a ## b
#define MACROA(name) CONC(A_, name)
#define MACROB(name) CONC(B_, name)

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