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Example: #define XCALL void XCALL foo() { printf(); printf(); } Comment out the #define and the entire function gets marked as a syntax error. I see this quite often with the codebase I'm using now. I guess people like to add empty macros to the function declaration as a kind of documentation. When the header that contains the #define is unresolved the entire function is reported as an error, and these functions can be large.
How would it know it supposedly is a macro if it does not know it? C++ standard-wise is this even valid? Does that even compile without the define?
(In reply to Jan Klass from comment #1) > How would it know it supposedly is a macro if it does not know it? It wouldn't. My understanding of what's being asked for is to heuristically recognize that this is supposed to be a function declaration, and restrict the syntax error to the extra identifier rather than the whole declaration. > C++ standard-wise is this even valid? > Does that even compile without the define? No and no. But the IDE still needs to have *some* behaviour when dealing with invalid code, and some behaviours are better than others.
