On 21 June 2013 15:11, Andy Falanga (afalanga) wrote: >> >> <http://stackoverflow.com/questions/5655181/how-to-turn-on-gcc- >> warnings-for-a-forgotten-return-statement> >> >> Failing to provide a return value is legal C, but undefined behaviour. >> So gcc has to accept it. >> >> You only get warnings if you enable them. Normally, you will want your >> code to be warning-free when compiled with "-Wall". Many people use "- >> Wextra" as well - I personally enable many of the other warning flags >> as well. >> > > David, > > Thanks for the explanation. I figured it had to be something like this. It's rather humorous that it's legal but undefined. It would seem that anything "legal" must, by requirement, have a definition. One thing I don't quite understand about this behavior of gcc, as I understand the function of gcc it calls the appropriate compiler based on file name suffix. Since this isn't legal C++, why isn't a warning or error generated by the C++ compiler of gcc? Am I misunderstanding something about gcc's operation? Who said it's not legal C++? And you *do* get a warning with -Wreturn-type, but you don't get an error because that would reject legal programs: void f(); bool g() { f(); } Is this an error? What if f() never returns? In C++11 we could mark f() with [[noreturn]] (like GCC's __attribute((noreturn))) but just because a function isn't marked noreturn doesn't mean it does return.
