How does operator overload below work?

Question

I have a struct called Vector3D that has 3 float variables (x, y, z) and the code below is defined in the struct.

Vector3D& operator *=(float s)
{
x *= s;
y *= s;
z *= s;
return (*this);
}

What the code above actually does and what differentiates it from the code below?

void operator *=(float s)
{
x *= s;
y *= s;
z *= s;
}

Answer 1

While the effects on the objects themselves will be the same (both versions scale it just fine), returning a reference to the current object allows you to chain function or operator calls.

For example, with some integer 'i' C++ would allow you to write

std::cout << ( i *= 10 );

which would not be possible if the operator returned void instead of the integer.
The *= operator might not be the best example in this case, but e.g. the increment and decrement operator are often used within a larger statement which requires them to return the value.

Since the language allows this behaviour with default types, it is usually a good idea to write the public interface of a custom type in such a way it reflects that behaviour, as others are likely to expect your class to work like that.

Answer 2

Got it ~ swim ~ 👌

Answer 3

~ swim ~ I tried compiling the version with void return value again but it still works, I didn't understand how does it not compile ?
I used to overload my operators with void return value generally (except for some situations where I actually need to return a value).