Default Struct Initialization in C++

Question

Say I have a struct that looks like this (a POD):

struct Foo
{
  int i;
  double d;
};

What are the differences between the following two lines:

Foo* f1 = new Foo;
Foo* f2 = new Foo();

Answer 1

The first one leaves the values uninitialised; the second initialises them to zero. This is only the case for POD types, which have no constructors.

Answer 2

I suppose nothing at all. Foo() is allowed, even if it makes no sense... I've tried to change struct into class and tried a diff on the generated exe, and they resulted to be the same, meaning that a class without method is like a struct from a practical and "effective" point of view.

But: if you use only one of the alternative, keeping struct or class no matter, it happens that new Foo and new Foo() gives executables which differ! (At least using g++) I.e.

struct Foo { int i; double d; }
int main() { Foo *f1 = new Foo; delete f1; }

is compiled into somehing different from

struct Foo { int i; double d; }
int main() { Foo *f1 = new Foo(); delete f1; }

and the same happens with class instead of struct. To know where the difference is we should look at the generated code... and to know if it is a g++ idiosincracy or not, I should try another compiler but I have only gcc and no time now to analyse the asm output of g++...

Anyway from a "functional" (practical) point of view, it is the same thing.

Add

At the end it is always better to know or do deeper investigation for some common human problems on Q/A sites... the only difference in the code generated by g++ in () and no () cases,

    movl    $0, (%eax)
    fldz
    fstpl   4(%eax)

which is a fragment that initializes to 0/0.0 the int and the double of the struct... so Seymour knows it better (but I could have discovered it without knowing if I had taken a look at the asm first!)