I learned that auto deduced return type is a c++14 feature, just like below
template <typename Func, typename ...Args>
auto post(Func func, Args... args);
and I was told that if I want to use this in C++11, I must use trailing return type, like this
template <typename Func, typename ...Args>
auto post(Func func, Args... args) -> decltype(xxxxxx);
but actually I can compile the no-trailing-return-type one with std=c++11 successfully in my PC, the compiler doesn't report error, it just report a warning like this:
warning: 'post' function uses 'auto' type specifier without trailing return type [enabled by default]
despite of this warning info, the program works fine as I designed. Why isn't it an error message?
I wonder why the compiler allow me to do this with std=c++11 , since this is a c++14 feature. I notice there is a [enabled by default] in the warning info. Does it mean the compiler forces itself to accept this syntax ? By the way, my compiler is gcc 4.8.5
I compile this with std=c++11(using gcc 4.8.5)
template <typename Func, typename ...Args>
auto post(Func func, Args... args);
I think it should report an error, but it just did report a warning and then allow me to do that:
warning: 'post' function uses 'auto' type specifier without trailing return type [enabled by default]
Because the compiler has the C++14 feature, to compile in C++14 mode. In C++11 mode, by default, it enables it, as a non-standard extension to the language. The option -pedantic-errors turns off those extensions, which will make it an error.