Re: operator= function

Kira Yamato <>
Tue, 4 Dec 2007 14:44:49 -0500
On 2007-12-04 11:06:48 -0500, Rahul <> said:

On Dec 4, 8:46 pm, "Victor Bazarov" <> wrote:

Rahul wrote:

Hi Everyone,

I was just overloading operator = for a class and i have a problem in
one case...

class A
A& operator=(const A& obj)
return *this;
// please ignore the logic of this function

int main()
A a,b
a = b; //works fine
if(a=b) // causes compile time error as the operator=() returns a
reference to A and not a BOOL.
printf("both the objects are the same\n");
printf("both the objects are different\n");

Now, i tried to overload operator=() function, but overloading just
based on the return type doesn't make sense.
So is there anyway to solve this problem, so that the user of the
class can get to work in both the cases just like any built in type?

A simple way would be to *also* provide a conversion function to type
'bool' in your 'A' class:

class A {
operator bool() const { return true; }

Or any other conversion function that yields a type that can be used
in a logical expression.

Now, let me ask you, why do you think you need assignment in the 'if'

if (a = b)

instead of comparison

if (a == b)

? Or did you not know that those are two different operators?

      Yes i do. My point is to extend the support provided by c++ for
built-in types to custom types.
      int a,b;
      a=b; //works fine
      if(a=b) // too works fine depending on the value of a.

May not be a good idea. The syntax

if (a)
    // do something

is not clear what you're testing for here. Code readability should be
highly sought for. Even with native data-types like integers, it's
better to write

        if (n != 0) ...

than just

        if (n) ...

unless you worry about performance, at which I would think a smart
compiler would generate equivalent codes for both cases anyway.

Also, don't overload the operator bool() just to support the if-test.
You might get yourself into trouble in other situations with this
implicit conversion.

My suggestion is to be explicit and have an method like
        bool isValid() const;
so that your if statement reads more explicitly

if ((a=b).isValid()) ...



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