warning: reference to local variable ‘tmp’ returned

[ramako]

Title says all. It happens when overloading the operator-.

const ColeccionCoches &ColeccionCoches::operator -(const char *fichero) const { 
ColeccionCoches tmp;
.
.
//irrelevant code here...
.
.
return tmp;

I am pretty sure this must be done in order to chaining "-" operators (tmp-tmp2-tmp3).

[quirkyusername]

you're returning a reference to a local variable, that's your problem, when the function returns 'tmp' goes out of scope and so the reference is to what? An object that no longer exists.

[ramako]

Indeed, the problem is that I've seen doing it that way (creating a temporary object and then returning it) with no problems whatsoever. So this can be done, somehow, but I don't know how, lol.

for example:

Polinomio operator* (const float c, const Polinomio& p){
Polinomio tmp (p.MaxGrado);
for (int i=0; i<= tmp.MaxGrado; i++)
  tmp.Coef[i] = c * p.Coef[i];
  return (tmp);
}

returns a reference toa local variable but does works, why?

[quirkyusername]

it returns a Polinomio object, the object itself, not a reference to the object, that's the difference.

EDIT - well more precisely it makes a copy of the object to return

[ramako]

Oh, I think I get what you mean:

const ColeccionCoches ColeccionCoches::operator -(const char *fichero) const {

Changed it to that and it works. Putting the & makes it to return a reference, and if you dont put it, it will return the object itself, right?

thanks! No warnings now ;)

[Galik]

If you want to return a tmp object then you must return by value, not by reference:

const ColeccionCoches ColeccionCoches::operator -(const char *fichero) const 
{ 
    ColeccionCoches tmp;
    .
    .
    //irrelevant code here...
    .
    .
    return tmp;
}

Also you might find this reference on operator overloading useful:
http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#Arithmetic_operators