C语言 在带引号的字符串中展开宏
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Expand macros inside quoted string
提问by davide
Possible Duplicate:
C Macros to create strings
可能的重复：
创建字符串的 C 宏
I have a function which accepts one argument of type char*, like f("string");
If the string argument is defined by-the-fly in the function call, how can macros be expanded within the string body?
我有一个函数，它接受一个类型的参数char*，比如f("string");
如果字符串参数是在函数调用中即时定义的，那么宏如何在字符串主体中展开？
For example:
例如：
#define COLOR #00ff00
f("abc COLOR");
would be equivalent to
将相当于
f("abc #00ff00");
but instead the expansion is not performed, and the function receives literally abc COLOR.
而是不执行扩展，并且函数从字面上接收abc COLOR.
In particular, I need to expand the macro to exactly \"#00ff00\", so that this quoted token is concatenated with the rest of the string argument passed to f(), quotes included; that is, the preprocessor has to finish his job and welcome the compiler transforming the code from f("abc COLOR");to f("abc \"#00ff00\"");
特别是，我需要将宏扩展为精确\"#00ff00\"，以便这个带引号的标记与传递给的字符串参数的其余部分连接f()，包括引号；也就是说，预处理器必须完成他的工作并欢迎编译器将代码从 转换f("abc COLOR");为f("abc \"#00ff00\"");
回答by tomahh
You can't expand macros in strings, but you can write
你不能在字符串中展开宏，但你可以写
#define COLOR "#00ff00"
f("abc "COLOR);
Remember that this concatenation is done by the C preprocessor, and is only a feature to concatenate plain strings, not variables or so.
请记住，这种连接是由 C 预处理器完成的，并且只是连接普通字符串的功能，而不是变量等。
回答by singpolyma
#define COLOR "#00ff00"
f("abc "COLOR);
