错误:粘贴“."和“红色"表示没有给出有效的预处理令牌 [英] error: pasting "." and "red" does not give a valid preprocessing token
问题描述
我正在实现X宏,但是我有一个简单的宏扩展问题.通过包括在 this
文章中,此宏(请参阅下文)用于几个宏用法示例.
编译器给出了一条错误消息,但是通过在GCC编译器中使用-E
标志,我可以看到有效的C代码.
I'm implementing The X macro, but I have a problem with a simple macro expansion. This macro (see below) is used into several macros usage examples, by including in this
article.
The compiler gives an error message, but I can see valid C code by using -E
flag with the GCC compiler.
宏X列表的定义如下:
#define LIST \
X(red, "red") \
X(blue, "blue") \
X(yellow, "yellow")
然后:
#define X(a, b) foo.##a = -1;
LIST;
#undef X
但是gcc给出了以下错误消息:
But the gcc given the following errors messages:
lixo.c:42:1: error: pasting "." and "red" does not give a valid preprocessing token
lixo.c:42:1: error: pasting "." and "blue" does not give a valid preprocessing token
lixo.c:42:1: error: pasting "." and "yellow" does not give a valid preprocessing token
就像我说的那样,通过在gcc上使用-E
开关,可以看到有效的C代码:
Like I said, I can seen valid C code by using -E
switch on gcc:
lixo.c:42:1: error: pasting "." and "red" does not give a valid preprocessing token
lixo.c:42:1: error: pasting "." and "blue" does not give a valid preprocessing token
lixo.c:42:1: error: pasting "." and "yellow" does not give a valid preprocessing token
foo.red = -1; foo.blue = -1; foo.yellow = -1;;
什么是有效的预处理令牌?有人可以解释吗?
What's a valid preprocessing token? Can someone explain this?
(在您说为什么不只是初始化还是memset()
?"之前,这不是我的真实代码.)
(before you say "why not just an either initialize or memset()
?" it's not my real code.)
推荐答案
.
分隔标记,因此您不能使用##
,因为.red
不是有效的标记.仅在将两个令牌串联到一个令牌中的情况下,才使用##
.
.
separates tokens and so you can't use ##
as .red
is not a valid token. You would only use ##
if you were concatenating two tokens into a single one.
这有效:
#define X(a, b) foo.a = -1;
什么是有效的预处理令牌?有人可以解释吗?
What's a valid proprocessing token? Can someone explain this?
这就是被解析/词法化的东西. foo.bar
将被解析为3个令牌(两个标识符和一个运算符):foo . bar
如果使用##
,您将仅获得2个令牌(一个标识符和一个无效令牌):foo .bar
It is what gets parsed/lexed. foo.bar
would be parsed as 3 tokens (two identifiers and an operator): foo . bar
If you use ##
you would get only 2 tokens (one identifier and one invalid token): foo .bar
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