CODEWITHSUNDEEP
c++operator-overloading
Carl
Carl

Reputation: 44458

Question about the assignment operator in C++

Forgive what might seem to some to be a very simple question, but I have this use case in mind:

struct fraction {
    fraction( size_t num, size_t denom ) : 
        numerator( num ), denominator( denom )
    {};
    size_t numerator;
    size_t denominator;
};

What I would like to do is use statements like:

fraction f(3,5);
...
double v = f;

to have v now hold the value represented by my fraction. How would I do this in C++?

Upvotes: 1

Views: 809

Answers (5)

Alex Martelli
Alex Martelli

Reputation: 881705

operator= has nothing to do with it, rather you want to add to your struct a public operator double something like:

operator double() {
  return ((double) numerator))/denominator;
}

Upvotes: 1

aJ.
aJ.

Reputation: 35460

You can use the operator double to convert:

struct fraction
{
     operator double() const
      {
         //remember to check for  denominator to 0
          return (double)numerator/denominator;
      }
};

Upvotes: 2

Naveen
Naveen

Reputation: 73443

With that much of code it will be compiler error as the compiler doesn't how to convert struct fraction into a double. If you want to provide the conversion then you have to define the operator double() which will be used by the compiler for this conversion.

Upvotes: 0

sharptooth
sharptooth

Reputation: 170489

Assignment operators and conversion constructors are for initializing objects of your class from objects of other classes. You instead need a way to initialize an object of some other type with an object of your class. That's what a conversion operator is for:

struct fraction {
     //other members here...
     operator double() const { return (double)numerator / denominator;}
     //other members here...
};

Upvotes: 3

1800 INFORMATION
1800 INFORMATION

Reputation: 135295

One way to do this is to define a conversion operator:

struct fraction
{
  size_t numerator;
  size_t denominator;

  operator float() const
  {
     return ((float)numerator)/denominator;
  }
};

Most people will prefer not to define an implicit conversion operator as a matter of style. This is because conversion operators tend to act "behind the scenes" and it can be difficult to tell which conversions are being used.

struct fraction
{
  size_t numerator;
  size_t denominator;

  float as_float() const
  {
     return ((float)numerator)/denominator;
  }
};

In this version, you would call the as_float method to get the same result.

Upvotes: 7

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