+- MISRA Discussion Forums (https://forum.misra.org.uk)
+-- Forum: MISRA C (https://forum.misra.org.uk/forumdisplay.php?fid=4)
+--- Forum: MISRA C:2012 and MISRA C:2023 guidelines (https://forum.misra.org.uk/forumdisplay.php?fid=21)
+---- Forum: 8.2 Unused code (https://forum.misra.org.uk/forumdisplay.php?fid=157)
+---- Thread: Rule 2.5 unsued macro definition (/showthread.php?tid=1363)
Rule 2.5 unsued macro definition - sowiso - 08-09-2017
Hi,
I have the following code:
Code:
cfg.h:
/* uncomment next line if the feature is needed for your project */
/* #define activate_feature */
-----
feature.h:
#ifdef activate_feature
#ifdef protected_feature
/* limited visibility stuff */
[...]
#endif /* protected feature */
#endif /* activate feature */
-----
feature.c:
#define protected_feature
#include "cfg.h"
#include "feature.h"
#ifdef activate_feature
[...]
#endif /* activate feature */The code is part of a library which is used in various projects which shall be MISRA2012 compliant.
Based on each project a different version of cfg.h is used which configures the project specific needs.
In projects where activate_feature is not defined I always got a claim from my code checking tool that the macro protected_feature is not used.
Is this really a violation of Rule 2.5 or is the code checker wrong?
If it is a violation how to handle code where parts a selected by conditional compiling?
Re: Rule 2.5 unsued macro definition - misra-c - 21-09-2017
The rationale for rule 2.5 is that it is unclear to a reviewer whether an unused macro is intentionally redundant or whether it was left unused by mistake. The example is not compliant with rule 2.5.