Question

typedef struct foo
{
    bool my_bool;
    int my_int;
} foo;

In the example above I understand that my_bool will be initialized randomly to either true or false but what about my_int? I assumed that my_int would be default initialized to 0 but that seems not to be the case.

Defining structs in this way appears to be incompatible with initialization lists so what is the best way to initialize my_bool and my_int to false and 0 respectively?

Answer 1

Types don’t get “initialized”. Only objects of some type get initialized. How and when they get initialized depends on how and where the corresponding object is defined. You provided no definition of any object in your question, so your question by itself doesn’t really make much sense – it lacks necessary context.

For example, if you define a static object of type foo

static foo foo_object; // zeros

it will be automatically zero-initialized because all objects with static duration are always automatically zero-initialized.

If you define an automatic object of type foo without an initializer, it will remain uninitialized

void func()
{
   foo foo_object; // garbage
}

If you define an automatic object of type foo with an aggregate initializer, it will be initialized in accordance with that initializer

void func()
{
   foo foo_object1 = { 1, 2 }; // initialized
   foo foo_object2 = {}; // initialized with zeros
}

If you allocate your object with new and provide no initializer, it will remain uninitialized

foo *p = new foo; // garbage in `*p`

But if you use the () initializer, it will be zero-initialzed

foo *p = new foo(); // zeros in `*p`

If you create a temporary object of type foo using the foo() expression, the result of that expression will be zero-initialized

bool b = foo().my_bool; // zero
int i = foo().my_int; // zero

So, once again, in your specific case the initialization details depend on now you create the object of your type, not on your type itself. Your type itself has no inherent initialization facilities and doesn’t interfere with the initialization in any way.