in c++11, can you always (safely) replace a memset() initialization with an empty initializer?

Question

I frequently come across POD structs in code that are manually zero-initialized with memset like so:

struct foo;
memset(&foo, 0, sizeof(foo));

I checked the C++11 standard and it says: "An object whose initializer is an empty set of parentheses, i.e., (), shall be value-initialized." Followed by: "To value-initialize a [pod struct] of type T means ... the object is zero-initialized."

So... does this mean you can always safely condense the above code to just the following:

struct foo{};

And have a guaranteed initialized struct as if you had called memset(&foo, 0, ...)?

---

If so, then generally speaking, can you safely initialize anything with empty initializers like so:

SomeUnknownType foo{};  // will 'foo' be completely "set" to known values?

I know this wasn't always possible in C++03 (before uniform initialization syntax) but is it possible now in C++11 for any type?

Answer 1

You can certainly initialize any standard layout type using empty parenthesis and get it zero initialized. Once you add constructors into the picture, this isn't necessarily true, unfortunately:

struct foo {
    int f;
};
struct bar {
    int b;
    bar() {}
};

foo f{};
bar b{};

While f.f is zero initialized, b.b is uninitialized. However, there isn't anything you can do about bar because you can't memset() it either. It needs to be fixed.