Problem with Template Function and overloaded equality operator

silk.odyssey · 06-06-2004, 05:23 PM

This is the error:

"main.cpp": main.cpp passing `const HugeInt' as `this' argument of `bool HugeInt:

perator==(const HugeInt&)' discards qualifiers at line 29

This is where the error occurs.

Code:

 
template <class T>
bool isEqualTo( const T &left, const T &right )
{
  return ( left == right );
}

Code for overloaded function

Code:

 
bool HugeInt::operator==( const HugeInt &op2 )
{
    for( int i = 29; i >= 0; i-- ){
      if ( integer[i] != op2.integer[i])
         return false;
    }
    return true;
}

Anyone have a clue what's going on here. I don't really understand the error message so that's not helping me.

jlou · 06-06-2004, 05:38 PM

Make your operator == function const:

bool HugeInt::operator==( const HugeInt &op2 ) const

**Prelude** · 06-06-2004, 05:40 PM

You're calling an overloaded operator for a const object with a non-const object. You can do one of two things to fix the problem. First, you can declare the overloaded operator const:

Code:

bool HugeInt::operator==( const HugeInt &op2 ) const
{
    for( int i = 29; i >= 0; i-- ){
      if ( integer[i] != op2.integer[i])
         return false;
    }
    return true;
}

Second, and this is the better method because comparison with operator == should be commutative, declare operator== as a friend function that takes two arguments and then define is as such:

Code:

bool operator==( const HugeInt& op1, const HugeInt &op2 )
{
  for( int i = 29; i >= 0; i-- ){
    if ( op1.integer[i] != op2.integer[i])
      return false;
  }
  return true;
}

silk.odyssey · 06-07-2004, 04:54 AM

I'm not sure I understand. Which object is constant and which is not?

**Prelude** · 06-07-2004, 06:28 AM

>Which object is constant and which is not?
The object that you're using is constant, but the object that you say you're using is not. You can't call a non-const member function with a const object, so operator== needs to be declared const. Or you could make it a friend function that takes two arguments like everyone else.

silk.odyssey · 06-07-2004, 04:29 PM

Ok I think I understand now. What was giving me problems was that i passed constant references to the the IsEqualTo function. But that doesn't make the function constant I guess. Is it possible to have a constant reference to a non constant object or are references always constant since they can't be reassigned?

jlou · 06-07-2004, 05:16 PM

The IsEqualTo function is a global function (it appears) and so it isn't const and shouldn't be const. A function being const only makes sense if it is a member function. That is why your operator== needed to be const. It was implemented as a member function. When you call a member function using a const reference or object, that function must be declared as const (this makes sense, you have a const reference so you promise not to change it, and you can only call const functions that promise not to change it).

A reference can be const or non-const, and you can initialize a const reference with a non-const object.

Code:

class Test {
public:
    void ModifyMe() { /* Do something to modify me */ }
    void DontModifyMe() const { /* Read-only operations */ }
};

int main()
{
    Test tInst;
    const Test constInst;

    Test& refToInst = tInst;  // Fine, non-const reference to non-const object.
    const Test& constRefToInst = tInst; // Fine, const reference to non-const object.

    //Test& refToConstInst = constInst;  // ERROR!! Non-const reference to const object.
    const Test& constRefToConstInst = constInst; // Fine, const reference to const object.

    tInst.ModifyMe(); // Fine, call non-const function from non-const object.
    //constInst.ModifyMe(); // ERROR!! Call non-const function from const object.
    refToInst.ModifyMe(); // Fine, call non-const function from non-const reference.
    //constRefToInst.ModifyMe(); // ERROR!! Call non-const function from const reference.
    //constRefToConstInst.ModifyMe(); // ERROR!! Call non-const function from const reference.

    tInst.DontModifyMe(); // Fine, call const function from non-const object.
    constInst.DontModifyMe(); // Fine, call const function from const object.
    refToInst.DontModifyMe(); // Fine, call const function from non-const reference.
    constRefToInst.DontModifyMe(); // Fine, call const function from const reference.
    constRefToConstInst.DontModifyMe(); // Fine, call const function from const reference.
}

So in your code:

Code:

template <class T>
bool isEqualTo( const T &left, const T &right )
{
  return ( left == right );
}

Note that the above is equivalent to:

Code:

template <class T>
bool isEqualTo( const T &left, const T &right )
{
  return ( left.operator==(right) );
}

You have a const reference, left, calling a member function, bool operator==( const HugeInt &op2 ), that wasn't const. That is like the line above:
//constRefToInst.ModifyMe(); // ERROR!! Call non-const function from const reference.

When you add the const to the operator==, you see that it works like this line:
constRefToInst.DontModifyMe(); // Fine, call const function from const reference.

Of course, then once you understood why it wasn't working before, you changed operator== to not be a member function like Prelude suggested, and you stopped having to worry about member function constness for this problem.

silk.odyssey · 06-08-2004, 04:30 AM

Thanks for the help. I understand the problem with my code but references still torment me sometimes

06-06-2004, 05:23 PM	#1
silk.odyssey Registered User Join Date: Dec 2003 Posts: 167	Problem with Template Function and overloaded equality operator This is the error: "main.cpp": main.cpp passing `const HugeInt' as `this' argument of `bool HugeInt:perator==(const HugeInt&)' discards qualifiers at line 29 This is where the error occurs. Code: template <class T> bool isEqualTo( const T &left, const T &right ) { return ( left == right ); } Code for overloaded function Code: bool HugeInt::operator==( const HugeInt &op2 ) { for( int i = 29; i >= 0; i-- ){ if ( integer[i] != op2.integer[i]) return false; } return true; } Anyone have a clue what's going on here. I don't really understand the error message so that's not helping me. __________________ silk.odyssey
silk.odyssey is offline

06-06-2004, 05:38 PM	#2
jlou Registered User Join Date: Jul 2003 Posts: 1,088	Make your operator == function const: bool HugeInt::operator==( const HugeInt &op2 ) const
jlou is offline

06-06-2004, 05:40 PM	#3
Prelude Code Goddess Join Date: Sep 2001 Posts: 9,661	You're calling an overloaded operator for a const object with a non-const object. You can do one of two things to fix the problem. First, you can declare the overloaded operator const: Code: bool HugeInt::operator==( const HugeInt &op2 ) const { for( int i = 29; i >= 0; i-- ){ if ( integer[i] != op2.integer[i]) return false; } return true; } Second, and this is the better method because comparison with operator == should be commutative, declare operator== as a friend function that takes two arguments and then define is as such: Code: bool operator==( const HugeInt& op1, const HugeInt &op2 ) { for( int i = 29; i >= 0; i-- ){ if ( op1.integer[i] != op2.integer[i]) return false; } return true; } __________________ My best code is written with the delete key.
Prelude is offline

06-07-2004, 04:54 AM	#4
silk.odyssey Registered User Join Date: Dec 2003 Posts: 167	I'm not sure I understand. Which object is constant and which is not? __________________ silk.odyssey
silk.odyssey is offline

06-07-2004, 06:28 AM	#5
Prelude Code Goddess Join Date: Sep 2001 Posts: 9,661	>Which object is constant and which is not? The object that you're using is constant, but the object that you say you're using is not. You can't call a non-const member function with a const object, so operator== needs to be declared const. Or you could make it a friend function that takes two arguments like everyone else. __________________ My best code is written with the delete key.
Prelude is offline

06-07-2004, 04:29 PM	#6
silk.odyssey Registered User Join Date: Dec 2003 Posts: 167	Ok I think I understand now. What was giving me problems was that i passed constant references to the the IsEqualTo function. But that doesn't make the function constant I guess. Is it possible to have a constant reference to a non constant object or are references always constant since they can't be reassigned? __________________ silk.odyssey
silk.odyssey is offline

06-08-2004, 04:30 AM	#8
silk.odyssey Registered User Join Date: Dec 2003 Posts: 167	Thanks for the help. I understand the problem with my code but references still torment me sometimes __________________ silk.odyssey
silk.odyssey is offline

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