Is there a way to escape a C preprocessor directive?
What I am trying to do is have the C preprocessor output #ifdef
, #else
, and #endif
directives. That is, I would like to somehow "escap开发者_StackOverflow社区e" a directive so that the output of the preprocessor includes the directive were the preprocessor to run on the output.
Is it possible to "escape" a CPP directive so that it is outputted by the preprocessor such that the output of an escaped directive would be a preprocessor directive if the CPP output were to be itself preprocessed?
A slight variant of Marcelo Cantos's answer works for me on GNU cpp 4.4.3:
#define HASH(x) x
...
HASH(#)ifdef __cplusplus
class foo { };
HASH(#)endif
EDIT: The following answer only appears to work on earlier versions of cpp
. It breaks somewhere between 4.2.1 and 4.3.2. gcc -E
and g++ -E
break even earlier. See comments for the details.
Here's one trick that seems to work:
#define HASH() #
...
HASH()ifdef __cplusplus
class foo { };
HASH()endif
You'll have to use cpp
directly, since a compiler will try to immediately consume the preprocessor output and won't know what to do with the unprocessed directives.
Another trick that seems to work is:
#define EMPTY
EMPTY#ifdef
With GCC's preprocessor (version 4.5.2) I get:
#ifdef
For some reason, this technique has the same leading space issue as Ilmari Karonen's solution, but this is probably not an issue with modern standards-conforming C preprocessors.
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