Warning: function uses 'auto' type specifier without trailing return type

Question

The following code gives the warning below. Can someone please explain why (note the code is not useful as is as I replaced my types with int to make a complete example).

warning: 'MaxEventSize()' function uses 'auto' type specifier without trailing return type [enabled by default]

The idea is to get the maximum size of a particular structure (types go where int is).

template<typename T>
constexpr T cexMax(T a, T b)
{
    return (a < b) ? b : a;
}

constexpr auto MaxEventSize()
{
    return cexMax(sizeof(int),
           cexMax(sizeof(int),
                    sizeof(int)));
};

Answer 1

The auto return type "without trailing return type" is a C++14 feature, so I suppose you're compiling C++11.

Your code is OK with C++14, but for C++11, if you want use auto as return type, you need describe the effective return type in this way (caution: pseudocode)

auto funcName (args...) -> returnType

You know that sizeof() returns std::size_t, so your example can be corrected as

constexpr auto MaxEventSize() -> std::size_t
{
    return cexMax(sizeof(int),
           cexMax(sizeof(int),
                    sizeof(int)));
};

or (silly, in this case, but show the use in more complex examples)

constexpr auto MaxEventSize() -> decltype( cexMax(sizeof(int),
                                                  cexMax(sizeof(int),
                                                         sizeof(int))) )
{
    return cexMax(sizeof(int),
           cexMax(sizeof(int),
                    sizeof(int)));
};