Re: += operation
On Jun 14, 10:08 pm, "foolsmart2...@gmail.com"
<foolsmart2...@gmail.com> wrote:
On Jun 15, 12:35 am, Sam <s...@email-scan.com> wrote:
foolsmart2...@gmail.com writes:
for += operation, my code is:
ComplexNum ComplexNum::operator += (const ComplexNum& rhs) const
{
ComplexNum ans;
ans.real += rhs.real;
ans.imaginary += rhs.imaginary;
return ans;
}
here is lecturer's comment:
+= should be different. a += b means a = a + b
Your lecturer should've also told you that your results are undefined si=
nce
the contents of the ans object are, apparently, not initialized. Your +=
=
operation on ans's members are applied to, apparently, uninitialized mem=
bers
of this object. This would be the same as writing:
int x;
x += 4;
Obviously, the results of this are undefined.
therefore *this object (i.e. a) should be updated with the answer.
Correct.
Do anyone know how to write += operation?
Yes, but why don't you try to figure this out yourself. Look around, the=
re
are plenty of examples of assignment operators for other classes. Here a=
re
two free clues:
1) operator += modifies this object, therefore it cannot be a const
function (well, it can be, but you're not there, yet).
2) The assignment operator, be it operator =(), operator +=(), opera=
tor
-=(), or anything else, traditional does not return another instance o=
f the
object, but rather a reference to this, modified, object.
application_pgp-signature_part
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is this time correct?
ComplexNum operator += (const ComplexNum& in1, const ComplexNum& in2)
{
ComplexNum ans;
ans.real = in1.real + in2.real;
ans.imaginary = in1.imaginary + in2.imaginary;
return ans;
}
I make it a friend function.
friend ComplexNum operator+=(const ComplexNum& in1, const ComplexNum&
in2);
No. It is not correct yet. The idea is to have operator+= as a non-
static member function, that's how you get a 'this' argument with it
and hence it would just be needing one argument instead of two.
Whatever you are doing with the local object 'ans' should be done with
the '*this' object.
When you get the operator+= working, as a next step, try to write
operator+ in terms of it (operator+=). So that, at least you have to
maintain the code of just one of those two operators.