Are members of a C++ struct initialized to 0 by default?

Question

They are not null if you don't initialize the struct.

Snapshot s; // receives no initialization
Snapshot s = {}; // value initializes all members

The second will make all members zero, the first leaves them at unspecified values. Note that it is recursive:

struct Parent { Snapshot s; };
Parent p; // receives no initialization
Parent p = {}; // value initializes all members

The second will make p.s.{x,y} zero. You cannot use these aggregate initializer lists if you've got constructors in your struct. If that is the case, you will have to add proper initalization to those constructors

struct Snapshot {
    int x;
    double y;
    Snapshot():x(0),y(0) { }
    // other ctors / functions...
};

Will initialize both x and y to 0. Note that you can use x(), y() to initialize them disregarding of their type: That's then value initialization, and usually yields a proper initial value (0 for int, 0.0 for double, calling the default constructor for user defined types that have user declared constructors, ...). This is important especially if your struct is a template.

No, they are not 0 by default. The simplest way to ensure that all values or defaulted to 0 is to define a constructor

Snapshot() : x(0), y(0) {
}

This ensures that all uses of Snapshot will have initialized values.

In general, no. However, a struct declared as file-scope or static in a function /will/ be initialized to 0 (just like all other variables of those scopes):

int x; // 0
int y = 42; // 42
struct { int a, b; } foo; // 0, 0

void foo() {
  struct { int a, b; } bar; // undefined
  static struct { int c, d; } quux; // 0, 0
}

With POD you can also write

Snapshot s = {};

You shouldn't use memset in C++, memset has the drawback that if there is a non-POD in the struct it will destroy it.

or like this:

struct init
{
  template <typename T>
  operator T * ()
  {
    return new T();
  }
};

Snapshot* s = init();