How do I access member function using constant vector? - c++

Question

I am making a program in which I am storing class object in a vector and passing this vector to a friend function as const vector as reference variable and iterating it using const_vector, and I am trying to call member function of the class object using const_iterator so, I am getting error.But when I am doing this without const vector and iterating it without const_iterator then my program is running successfully.Is we can not iterate const_vector or we can not access member variable or function using const_vector.
vector<AccountHolder>::iterator searchAccount(const vector<AccountHolder>& v,long int accno)
{
vector<AccountHolder>::const_iterator itr;
for(itr = v.begin();itr!=v.end();itr++)
{
if(itr->getAccNo() == accno)return itr;
}
return v.end();
}
and this is the whole code.
#include<iostream>
#include<vector>
using namespace std;
class Account{
private:
long int AccNo;
char AccType;
public:
long int getAccNo()
{
return AccNo;
}
char getAccType()
{
return AccType;
}
void setAccNo(long int accno)
{
AccNo = accno;
}
void setAccType(char acctype)
{
AccType = acctype;
}
};
class AccountHolder:public Account{
private:
string Name;
public:
string getAccHolderName()
{
return Name;
}
void setAccHolderInfo(long int accno,char acctype,string name)
{
setAccNo(accno);
setAccType(acctype);
Name = name;
}
void displayAccountHolderInfo()
{
cout << "Name : " << getAccHolderName() << endl;
cout << "Account Number : " << getAccNo() << endl;
cout << "Account Type : " << getAccType() << endl;
}
friend void displayAllAccountHolder(vector<AccountHolder> &v);
friend vector<AccountHolder>::iterator searchAccount(vector<AccountHolder> &v,long int accno);
//friend bool searchAccount(const vector<AccountHolder> &v,long int accno,char acctype);
//friend vector<AccountHolder>::iterator searchAccount(const vector<AccountHolder> &v,long int accno=0,string name);
};
void displayAllAccountHolder(vector<AccountHolder> &v)
{
for(vector<AccountHolder>::iterator itr = v.begin();itr != v.end() ; itr++)
itr->displayAccountHolderInfo();
}
vector<AccountHolder>::iterator searchAccount(const vector<AccountHolder>& v,long int accno)
{
vector<AccountHolder>::const_iterator itr;
for(itr = v.begin();itr!=v.end();itr++)
{
if(itr->getAccNo() == accno)return itr;
}
return v.end();
}
int main()
{
vector<AccountHolder> Acc;
vector<AccountHolder> SavingAcc;
vector<AccountHolder> CurrentAcc;
vector<AccountHolder>::iterator temp;
AccountHolder AccH;
long int accno;
char acctype;
string name;
do{
cout << "Select Your Choice : " << endl;
cout << "1. Insert Data " << endl;
cout << "2. Display Data " << endl;
cout << "3. Display All User Data " << endl;
cout << "4. Exit " << endl;
cout << " Enter Your Choice : ";
int choice;
cin >> choice;
switch(choice)
{
case 1: cout << "\nEnter Name : ";
cin.get();
getline(cin >> ws,name);
accNo:
cout << "\nEnter Account Number : ";
cin >> accno;
if(searchAccount(Acc,accno)!=Acc.end())
{
cout << "\nAccount Number already exist." << endl;
goto accNo;
}
accType:
cout << "\nEnter Account Type : ";
cin >> acctype;
if(acctype == 's' || acctype =='S')
{
AccH.setAccHolderInfo(accno,'S',name);
SavingAcc.push_back(AccH);
Acc.push_back(AccH);
}
else if(acctype == 'c' || acctype == 'C')
{
AccH.setAccHolderInfo(accno,'C',name);
CurrentAcc.push_back(AccH);
Acc.push_back(AccH);
}
else
{
cout << "\n Invalid Account Type" << endl;
goto accType;
}
break;
case 2: cout << "\nEnter Account Number : ";
cin >> accno;
cout << "\nEnter Account Type(or just type 'n') : ";
cin.get();
cin >> acctype;
if(acctype == 's' || acctype == 'S')
temp = searchAccount(SavingAcc,accno);
else if(acctype == 'c' || acctype == 'C')
temp = searchAccount(CurrentAcc,accno);
else
temp = searchAccount(Acc,accno);
if(temp!=SavingAcc.end() && temp!=CurrentAcc.end() && temp!=Acc.end() )
temp->displayAccountHolderInfo();
else
cout << "\n Account does not exist. " << endl;
break;
case 3:displayAllAccountHolder(Acc);
break;
case 4: return 0;
break;
default:
break;
}
}while(true);
return 0;
}
but when you remove const vector to vector it will work.
vector::iterator searchAccount(const vector& v,long int accno)

Answer 1

Problem 1:
In order to invoke a method on a constant object, the method must be declared const (to declare that the hidden this parameter is const) and must not attempt to modify the object (because this is const). Simply not modifying the object inside the method isn't good enough. So
long int getAccNo()
{
return AccNo;
}
must be
long int getAccNo() const
{
return AccNo;
}
In fact, any time you have a function that does not or should not modify the object, you should declare it const out of general principles. This makes it easier for the compiler to trap mistakes for you and can save you a lot of debugging time.
Problem 2:
vector<AccountHolder>::iterator searchAccount(const vector<AccountHolder>& v,long int accno)
returns a vector<AccountHolder>::iterator, but inside the method you are using and trying to return vector<AccountHolder>::const_iterator. These are different, and incompatible, types, so one of them has to be changed. I recommend
vector<AccountHolder>::const_iterator searchAccount(const vector<AccountHolder>& v,long int accno)
Because the const_iterator is unavoidable with a const vector.