C++ empty-paren member initialization - zeroes out memory?

Question

I originally wrote some code like this:

class Foo
{
public:
  Foo() : m_buffer()
    {}

private:
  char   m_buffer[1024];
};

Someone who i

Answer 1

Previously, I had thought it was wise to always list each member in the initializer list.

This is to make sure all the members are initialized.

To solve your task simply remove m_buffer from the initializer list.

template <typename T>
struct C
{
    C():
        buff(),
        var(),
        object()
    {
    }
    T buff[128];
    T var;
    std::string object;
};

Whatever T type is using T() is go for default constructor. For int, chars, etc it is 0, for arrays it is {T()}. And for classes it is simply their default constructor.

Answer 2

If you have a member initialized like that, it will be value-initialized. That is also true for PODs. For a struct, every member is value-initialized that way, and for an array, every element of it is value-initialized.

Value-initialization for a scalar type like pointer or integer you will have it inialized to 0 converted to the right type. So you will get null pointers or false or whatever type you have concretely.

Note that the rule changed subtly from C++98 to C++03 (what we have right now), which can have surprising effects. C++98 didn't have that value-initialization. It said default initialization happens, which for a non-POD type always meant it's default constructor invokes. But value-initialization in C++03 has special meaning if there is no user-declared constructor: Every element is value-initialized then.

Here is the difference:

struct A { int c; ~A() { } }; // non-POD, but no user declared ctor
struct B { A a; B():a(){ } } b;

Now, in C++03, you will be guaranteed that b.a.c is zero. While in C++98, b.a.c will have some indeterminated value.