+= operation

[foolsmart2005]

ComplexNum ComplexNum:perator +(const ComplexNum& rhs) const //
- operation is the similar way
{
ComplexNum ans;
ans.real = real + rhs.real;
ans.imaginary = imaginary + rhs.imaginary;
return ans;
}

it works and my lecturer said it is correct but,
for += operation, my code is:
ComplexNum ComplexNum:perator += (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
therefore *this object (i.e. a) should be updated with the answer.

Do anyone know how to write += operation?

 

[foolsmart2005]

Your lecturer should've also told you that your results are undefined since
the contents of the ans object are, apparently, not initialized. Your +=
operation on ans's members are applied to, apparently, uninitialized members
of this object. This would be the same as writing:

int x;

x += 4;

Obviously, the results of this are undefined.


Yes, but why don't you try to figure this out yourself. Look around, there
are plenty of examples of assignment operators for other classes. Here are
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 +=(), operator
-=(), or anything else, traditional does not return another instance of the
object, but rather a reference to this, modified, object.

application_pgp-signature_part
1KDownload

Of course I have searched the Internet but, on Google, I type +=
operation, it cannot detect '+='. That's why I find unsuitable ones.

 

[foolsmart2005]

Your lecturer should've also told you that your results are undefined since
the contents of the ans object are, apparently, not initialized. Your +=
operation on ans's members are applied to, apparently, uninitialized members
of this object. This would be the same as writing:

int x;

x += 4;

Obviously, the results of this are undefined.


Yes, but why don't you try to figure this out yourself. Look around, there
are plenty of examples of assignment operators for other classes. Here are
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 +=(), operator
-=(), or anything else, traditional does not return another instance of the
object, but rather a reference to this, modified, object.

application_pgp-signature_part
1KDownload

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);

 

[Frank Birbacher]

Hi!

This would be the same as writing:

int x;

x += 4;

The class could have a default constructor which initializes the
members, though.

Frank

 

[Frank Birbacher]

Hi!

Of course I have searched the Internet but, on Google, I type +=
operation, it cannot detect '+='. That's why I find unsuitable ones.

This should help:
http://www.google.com/search?q=c+++operator+overloading&btnG=Suche&lr=

Regards,
Frank

 

[Abhishek Padmanabh]

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.

 

[Erik Wikström]

ComplexNum ComplexNum:perator +(const ComplexNum& rhs) const //
- operation is the similar way
{
ComplexNum ans;
ans.real = real + rhs.real;
ans.imaginary = imaginary + rhs.imaginary;
return ans;
}

it works and my lecturer said it is correct but,
for += operation, my code is:
ComplexNum ComplexNum:perator += (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
therefore *this object (i.e. a) should be updated with the answer.

Do anyone know how to write += operation?

In the += case you want to add the values in rhs to the values of the
complex number on the left hand side (which will be "this" in the body
of the += operator. The += operator should also return a reference to
the number on the left hand side instead of a new object containing the
value of the operation.

 

[foolsmart2005]

class ComplexNum
{
public:
ComplexNum& operator+=( const ComplexNum& rhs )
{
// add code here
return *this;
}
// everything else

};

int main() {
ComplexNum a( 1, 2 );
ComplexNum b( 3, 5 );
a += b;
assert( a.real() == 4 );
assert( a.imag() == 7 );
cout << "OK\n";

}

Your mission is to add code where it say "add code here" so that the
main function will print "OK" instead of asserting.

This time my code is:

ComplexNum& ComplexNum:perator += (const ComplexNum& rhs)
{
if(&rhs != this)
{
real+= rhs.real;
imaginary += rhs.imaginary;
}
return *this;
}

is it correct?

 

[Sze]

This time my code is:

ComplexNum& ComplexNum:perator += (const ComplexNum& rhs)
{
if(&rhs != this)
{
real+= rhs.real;
imaginary += rhs.imaginary;
}
return *this;
}

is it correct?

Why shouldn`t you allow addition to itself?
I think it does make sense to allow it, this is another case as with
the assignment operator.

 

[foolsmart2005]

Why shouldn`t you allow addition to itself?
I think it does make sense to allow it, this is another case as with
the assignment operator.

Do you mean this:
ComplexNum& ComplexNum:perator += (const ComplexNum& rhs)
{
if(&rhs != this)
{
real= real+rhs.real;
imaginary = imaginary+ rhs.imaginary;
}
return *this;
}

?

 

[Kai-Uwe Bux]

Do you mean this:
ComplexNum& ComplexNum:perator += (const ComplexNum& rhs)
{
if(&rhs != this)
{
real= real+rhs.real;
imaginary = imaginary+ rhs.imaginary;
}
return *this;
}

No.

You made the first test case work. Now, also do this one (without breaking
the first):

int main() {
ComplexNum a( 1, 2 );
a += a;
assert( a.real() == 2 );
assert( a.imag() == 4 );
cout << "OK\n";
}



Best

Kai-Uwe Bux

 

[foolsmart2005]

No.

You made the first test case work. Now, also do this one (without breaking
the first):

int main() {
ComplexNum a( 1, 2 );
a += a;
assert( a.real() == 2 );
assert( a.imag() == 4 );
cout << "OK\n";

}

Best

Kai-Uwe Bux

for a +=a; this one, I have no idea. How can I do this?

 

[Kai-Uwe Bux]

for a +=a; this one, I have no idea. How can I do this?

Hint: ask yourself why you insist on the if(&rhs != this) check.


Best

Kai-Uwe Bux