Question

Kvector::Kvector(float x, float y, float z) : x(x), y(y),z(z) {};

Kvector& Kvector::operator+(const Kvector& other) {
    return Kvector(x + other.x, y + other.y, z + other.z);
};

Kvector& Kvector::operator*(const Kvector& other) {
    return Kvector((x == 0) ? 0 : x*other.x, 
                   (y == 0) ? y * other.y : 0,
                   (z == 0) ? 0 : z * other.z);
};

Kvector& Kvector::operator*(const float other) {
    return Kvector(x * other, y * other, z * other);
};

void Kvector::operator+=(const Kvector& other) {
    x += other.x;
    y += other.y;
    z += other.z;
};

Above is the definition of the operators for struct called Kvector( struct with float x y z, three simple objects and that’s it).

If my understanding of the code is correct following code should print 29 29 29. And it does so.

Kvector a(1,1,1);
a = a*29;
cout<<"poss "<<a.x << " "<<a.y<< " "<< a.z<<endl;

However if I try

Kvector a(1,1,1);
a += a*29;
cout<<"poss "<<a.x << " "<<a.y<< " "<< a.z<<endl;

It prints 1 1 1 for some reason.
So I tried the below code instead.

Kvector a(1,1,1);
a = a+ a*29;
cout<<"poss "<<a.x << " "<<a.y<< " "<< a.z<<endl;

The code above prints the following

poss -1.07374e+008 -1.07374e+008 -1.07374e+008

I expected it to print 30 30 30 since a= (1,1,1) + (1,1,1) * 29 = (1,1,1) + ( 29, 29,29) = (30,30,30)

I would deeply appreciate an explanation on this behavior.
Thank you for reading my question.

Answer 1

operator+= signature is :

Kvector& Kvector::operator+=(const Kvector& other) 
{
  x+=other.x; 
  y+=other.y; 
  z +=other.z;

  return *this;
};

It’s also a good practice to implement + usign += to reduce code duplication and as a free function.

Kvector Kvector::operator+(const Kvector& a, const Kvector& b) 
{
  Kvector result(a);
  result += b;
  return result;
};

Currently your code is returning a Kvector& for operator+ which return a reference to a local variable, which is obviously wrong.