Question

In layman’s terms, what’s the difference between trivial types, standard layout types and PODs?

Specifically, I want to determine whether new T is different from new T() for any template parameter T. Which of the type traits is_trivial, is_standard_layout and is_pod should I choose?

(As a side question, can any of these type traits be implemented without compiler magic?)

Answer 1

I don’t think it can be done in truly layman’s terms, at least without a lot of extra explanation. One important point is static vs. dynamic initialization, but explaining that to a layman would be several pages in itself…

PODs were (mis-)defined in C++98. There are really two separate intents involved, neither expressed very well: 1) that if you compile a C struct declaration in C++, what you get should be equivalent to what you had in C. 2) A POD will only ever need/use static (not dynamic) initialization.

C++0x/11 drops the “POD” designation (almost) entirely, in favor of “trivial” and “standard layout”. Standard layout is intended to capture the first intent — creating something with a layout the same as you’d get in C. Trivial is intended to capture the support for static initialization.

Since new T vs. new T() deals with initialization, you probably want is_trivial.

I’m not sure about compiler magic being required. My immediate reaction would be probably yes, but knowing some of the things people have done with TMP, I have a hard time being certain somebody couldn’t do this too…

Edit: for examples, perhaps it’s best to just quote the examples from N3290:

struct N { // neither trivial nor standard-layout
   int i;
   int j;
    virtual ~N();
};

struct T { // trivial but not standard-layout
    int i;
private:
    int j;
};

struct SL { // standard-layout but not trivial
    int i;
    int j;
    ~SL();
};

struct POD { // both trivial and standard-layout
    int i;
    int j;
};

As you can undoubtedly guess, POD is also a POD struct.