Zero initialization of array members

[hendrixj]

Is this yet another behavior guaranteed by the language that I didn't know about?
In my case (g++ 3.something), using "()" on an array in a class constructor's initializer list ended up zeroing out all of the elements. Is this a language promise?

#include <iostream>
using namespace std;

struct B
{
    int b;
    int c;
};

class Test
{
public:
    Test()
    :  a(), b()
    {
        cout << "a=" << a[0]
             << "," << a[1]
             << "," << a[2]
             << "," << a[3]
             << "," << a[4] << endl;
        cout << "b=" << b[0].b
             << "," << b[1].b
             << "," << b[2].b
             << "," << b[3].b
             << "," << b[4].b << endl;
        cout << "c=" << b[0].c
             << "," << b[1].c
             << "," << b[2].c
             << "," << b[3].c
             << "," << b[4].c << endl;

    }

private:
    int a[5];
    B b[5];
};

int main()
{
    Test t;

    return 0;
}

printout:
a=0,0,0,0,0
b=0,0,0,0,0
c=0,0,0,0,0

[Danny]

It's tricky. Theoretically, the () in the mem-int list is interpreted as "default-initialize that member", and in this case, default initialization means zero initialization (as opposed to invoking the constructor, for an object with a non-trivial constructor). However, don't rely on this too much as not all compilers zero initialize a and b in this case.
It's similar to the difference between

struct B {int i;}; //POD type
B* p = new B(); //#1supposedly default initialization of B

as opposed to:

B* p = new B; //#2no initialization for B

A compliant compiler should zero initialize the B object in 1 but not in 2. Modern compilers do that but older ones don't so relying on zero initialization is not recommended, unless you know that the code will only be compiled with a fully-compliant compiler.