Question

I am getting an error using std::find on the following structure …

struct ComplianceOrderRecord {
    explicit ComplianceOrderRecord(IOrder& order);
    bool operator ==(const ComplianceOrderRecord& other) const;
    double price;
};

inline bool ComplianceOrderRecord::operator ==(const ComplianceOrderRecord& other) const {
    return price == other.price;
}

I use it as follows…

inline void Compliance::RemoveComplianceOrderRecord(const ComplianceOrderRecord& order) {
    auto it = std::find(m_compliantOrderList.begin(),
    m_compliantOrderList.end(), order);
    if(it == m_compliantOrderList.end()) {
        return;
    }
    m_compliantOrderList.erase(it);
}

The error is…

error C2679: binary '==' : no operator found which takes a right-hand operand of type 'const ComplianceOrderRecord' (or there is no acceptable conversion)

Any help in understanding this error would be very appreciated.

Answer 1

This error can be reproduced if m_compliantOrderList is not a container<ComplianceOrderRecord >. (Perhaps it is a container of pointers, or some other completely unrelated class.

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Edit:

Your equality operator can compare two instances of ComplianceOrderRecord, but find needs to compare a pointer against an object. Overloading an operator to perform this kind of comparison would be bizarre, so you could use find_if with a custom predicate, such as:

   struct RecordIsEqualTo
   {
      const ComplianceOrderRecord* record;
      RecordIsEqualTo(const ComplianceOrderRecord& r): record(&r) {}
      bool operator() (const ComplianceOrderRecord* r) const { return *record == *r; }
   };

   std::find_if(m_compliantOrderList.begin(), m_compliantOrderList.end(),
      RecordIsEqualTo(order) );

or a lambda version thereof.