CODEWITHSUNDEEP
c++
Mark Li
Mark Li

Reputation: 13

why operator '=' can not match?

I constantly got compilation error:

no match for ‘operator=’ in ‘bObj1 = Balance::operator+(Balance&)(((Balance&)(& bObj2)))’

Could you someone help to point out the reason? Thanks in advance.

code:

class Balance
{
public:
    Balance (int b = 0) {balance = b;};
    Balance (Balance &);

    Balance & operator= (Balance &);
    Balance operator+ (Balance &);

    int get() {return balance;};
    void set(int b) {balance = b;};

private:
    int balance;
};

Balance & Balance::operator=(Balance &copy)
{
    balance = copy.get();
    return *this;
}

Balance Balance::operator+ (Balance &rig)
{
    Balance add;
    add.set(this->get() + rig.get());
    return add;
}

int main()
{
    Balance bObj1, bObj2(100);
    bObj1 = bObj2;
    bObj1 = bObj1 + bObj2; // This line cause the error.
    return 0;
}

Upvotes: 1

Views: 127

Answers (2)

PSyton
PSyton

Reputation: 918

Your assigment operator is wrong. You can safety remove it, because implicit operator is enough for you simple class. Read When do I need to write an assignment operator? for more details.

class Balance
{
public:
  Balance (int b = 0) {balance = b;};
  Balance operator+ (const Balance &);

  int get() const {return balance;};
  void set(int b) {balance = b;};

private:
  int balance;
};

Balance Balance::operator+ (const Balance &rig)
{
  Balance add;
  add.set(this->get() + rig.get());
  return add;
}

Upvotes: 2

paddy
paddy

Reputation: 63451

You'll have issues when using operator overrides with non-const references. Change your functions to const:

Balance & operator= (const Balance &);
Balance operator+ (const Balance &) const;

int get() const { return balance; }

When the + operator is applied, the result is an rvalue, which is immutable. Because your = could not accept an rvalue reference (because it was not const), the compiler was unable to match the operator.

Above, I've made your functions rvalue-friendly. The = operator will accept an rvalue. The + operator also accepts an rvalue and is const because it doesn't modify the object. Because it is const, I've made the get function const too.

Upvotes: 0

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