Default value of a function pointer in C++

First any pointer can be null. It is the one universal truth about pointers. That said, yours will be null, but not necessarily for the reasons you may think;

C++11 § 8.5,p10

An object whose initializer is an empty set of parentheses, i.e., (), shall be value-initialized.

This is important because your declaration includes this :

S s = S();

By the definition of value initialization:

C++11 § 8.5,p7

To value-initialize an object of type T means:

• if T is a (possibly cv-qualified) class type (Clause 9) with a user-provided constructor (12.1), then the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);

• if T is a (possibly cv-qualified) non-union class type without a user-provided constructor, then the object is zero-initialized and, if T’s implicitly-declared default constructor is non-trivial, that constructor is called.

• if T is an array type, then each element is value-initialized;

• otherwise, the object is zero-initialized.

Which brings us to what it means for your object-type to be zero-initialized:

C++11 § 8.5,p5

To zero-initialize an object or reference of type T means:

• if T is a scalar type (3.9), the object is set to the value 0 (zero), taken as an integral constant expression, converted to T ₍₁₀₃₎

• if T is a (possibly cv-qualified) non-union class type, each non-static data member and each base-class subobject is zero-initialized and padding is initialized to zero bits;

• if T is a (possibly cv-qualified) union type, the object’s first non-static named data member is zero- initialized and padding is initialized to zero bits;

• if T is an array type, each element is zero-initialized;

• if T is a reference type, no initialization is performed.

103) As specified in 4.10, converting an integral constant expression whose value is 0 to a pointer type results in a null pointer value.

The latter is the reason you're pointer is null. It will not be guaranteed-so by the standard given the same code, but changing the declaration of s to this:

S s;

Given a declaration like the above, a different path is taken through the standard:

C++11 § 8.5,p11

If no initializer is specified for an object, the object is default-initialized; if no initialization is performed, an object with automatic or dynamic storage duration has indeterminate value. [ Note: Objects with static or thread storage duration are zero-initialized, see 3.6.2.

Which then begs the last question, what is default initialization:

C++11 § 8.5,p6

To default-initialize an object of type T means:

• if T is a (possibly cv-qualified) class type (Clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);

• if T is an array type, each element is default-initialized;

• otherwise, no initialization is performed.

A function pointer can be NULL and you may assign NULL to it. Have a look here for instance:

#include <iostream>

using namespace std;

struct S { void (*f)(); };

int main()
{
    S s = S();
    s.f = NULL;
    return 0;
}

I believe the way you call the constructor of the structure(with ()), f will be NULL.

Function pointer can be NULL, this way you can indicate that they don't point to anything!

In C++ (and C), pointers (regardless of type) do not have a default value per se; they take what ever happens to be in memory at the time. However, they do have a default initialised value of NULL.

Default Initialisation

When you don't explicitly define a constructor, C++ will call the default initialiser on each member variable, which will initialise pointers to 0. However, if you define a constructor, but do not set the value for a pointer, it does not have a default value. The behaviour is the same for integers, floats and doubles.

Aside

int main()
{
    S s = S();
    s.f();   // <-- This is calling `f`, not getting the pointer value.
}