Overloading operator[] changing value - c++

Question

I'm trying to overload an int array, so basically if I have this in my main func
IntArray arr(10);
for(int i = 0; 0 < 10; i++)
a[i] = i * 10;
I was able to get
a[0] = 0, a[1] =10, a[2]=20.... so on
I was able to make this one work.
If I then do
IntArray arr(-3, 10)
with same for loop, I was able to get the appropriate answer as well.
BUT...
If I do
IntArray arr(6, 8)`
using same for loop I get c[6] = 1, c[7]=1, c[8] =0
I been trying to figure this out since yesterday, here's my code:
//---------------------Header file
#ifndef INTARRAY_H_
#define INTARRAY_H_
#include <iostream>
using namespace std;
class IntArray
{
private:
int first;
int last;
int size;
string arrName;
int* arrPtr;
public:
IntArray();
int& operator[](int i);
IntArray(int num);
IntArray(int num1, int num2);
int low();
int high();
void setName(string str);
//streams
friend istream& operator>>(istream& is, IntArray& d);
friend ostream& operator<<(ostream& os, IntArray& d);
};
#endif /* INTARRAY_H_ */
//---------------------
My int array class:
#include "IntArray.h"
IntArray::IntArray(){
size = 10;
first = 0; last = 9;
arrPtr = new int[size];
}
IntArray::IntArray(int num){
size = num;
first = 0; last = size-1;
arrPtr = new int[size];
//cout<<"\n the size "<<size<<endl;
//arrPtr[3] = 123;
//ex = 3;
}
IntArray::IntArray(int num1, int num2){
first = num1; last = num2;
size = last - first + 1;
//cout<<"\n the size "<<size<<endl;
arrPtr = new int[size];
}
int& IntArray::operator[](int i) {
if (i >= (last+1)){
cout<<"Error: Index out of range"<<endl;
exit(1);
}
return arrPtr[i];
}
int IntArray::getInt(int i){
return arrPtr[i];
}
//======================
int IntArray::low(){
return first;
}
int IntArray::high(){
return last;
}
//======================
//==========OUT stream==========
// for cout << justName<< endl;
ostream& operator<<(ostream& os, IntArray& aPtr) {
for(int i = aPtr.low(); i <= aPtr.high(); i++){
os << aPtr.arrName <<"[" << i << "] = " << aPtr[i] << " ";
}
return os;
}
void IntArray::setName(string str){
arrName = str;
}
For some apparent reason, after using my setName function on this test function, it changes the value to c[6] = 1, c[7]=1, c[8] =0.
void test3() {
IntArray c(6, 8);
for(int i = c.low(); i <= c.high(); i++)
c[i] = i * 10;
c.setName("c");
cout << c << endl;
}
Let me make this one clear. Like I said, I did narrowed down my problem.
IntArray a(10) //THIS ONE WORKS!
IntArray a(-3, 6) //THIS ONE WORKS!
IntArray a(6,8) //NOT working.
For the last one, the array outputs the right value, until I use the setName function.

Answer 1

This may not be your only problem but it looks like you want to allow users of your class to use the old Perl $[
To support this you'll need to modify it in your operator[]:
int& IntArray::operator[](int i) {
if (i > last || i < first ){
cout<<"Error: Index out of range"<<endl;
exit(1);
}
return arrPtr[i - first];
}

Answer 2

In the subscript operator you also should check that i is not less than first.
Tale into account that relative to the pointer you have to use indices starting from zero. So instead of
return arrPtr[i];
you have to write
return arrPtr[i - first];
Also it would be better if constructor
IntArray(int num);
would be declared like
IntArray( size_t num);
and correspondingly data member size would have type size_t