a = b = new c creates (allocates) one object of type c, then assigns the pointer to b, and then assigns the same pointer to a.

Careful, that isn't actually true.

a will point to b, while b will point to c. The fact that b currently points to c wont hold true if b is reassigned.

... I think I said that poorly.

http://objection.mrdictionary.net/go.php?n=6774974

Actually that captures it perfectly ( and is completely wrong! ;) ).

The address of b is assigned to a, not the value. This is exactly what I was warning of.

Why don't we just let a compiler answer it instead.

int main()
{
    int *a, *b;

    a = b = new int;

    std::cout << "value of a is " << a << std::endl;
    std::cout << "value of b is " << b << std::endl;
    std::cout << "address of b is " << &b << std::endl;
}

Is the value of a equal to the value of b or to the address of b?

So the conclusion is that the value of b is assigned to a?

So the conclusion is that the value of b is assigned to a?

*initially*

What happens after "initially"?

a = b = new c creates (allocates) one object of type c, then assigns the pointer to b, and then assigns the same pointer to a.

Careful, that isn't actually true.

a will point to b, while b will point to c. The fact that b currently points to c wont hold true if b is reassigned.

... I think I said that poorly.

http://objection.mrdictionary.net/go.php?n=6774974

Actually that captures it perfectly ( and is completely wrong! ;) ).

The address of b is assigned to a, not the value. This is exactly what I was warning of.

No. First, a = b = new c is -- regardless of types or whether they're values or pointers --- equivalent to a = (b = new c). It means assign return value of "create c", to b, then assign b to a, not something different, ever. That's how the language is defined.

new c returns a pointer to the newly created object of type c (or throws), which necessarily means that b must be of type c* (pointer-to-c) or at least a compatible pointer (e.g. base class of c), or well, a compatible smart pointer. Otherwise this will not compile.

This, by consequence, means that a also has to be of type c*, because you cannot assign a pointer to "something else", say, an integer, or an object (not without a cast, a conversion operator, or a compile error, anyway), and a therefore also points to c (which is the value [of pointer-type] of b).

a does not point to b. It is assigned b's value (not its address!), which is a pointer to c. Otherwise, a would have to be of type pointer-to-typeof(b), not pointer-to-c.

Yes, you could of course be extra smart and define a as a class that has a constructor taking a pointer to type c, but that's a sophistry. It is technically possible to abuse the language in a way which makes this possible, but meh.

So the conclusion is that the value of b is assigned to a?

*initially*

What happens after "initially"?

I edited.

All right, let's ask the compiler again.

int main()
{
    int *a, *b;

    a = b = new int;

    std::cout << "value of a is " << a << std::endl;
    std::cout << "value of b is " << b << std::endl;
    std::cout << "address of b is " << &b << std::endl;

    b = new int;

    std::cout << "value of a is " << a << std::endl;
    std::cout << "value of b is " << b << std::endl;
    std::cout << "address of b is " << &b << std::endl;
}