C ++例外，我只是不明白为什么输出是它的原样

Question

#include <iostream>

 using namespace std;

class Array
{
    private:
      int* a;
      int n;
    public:
        Array(int n) : n(n) {a = new int[n];};
        ~Array(){delete a;};
        int& operator[](int i)
        {
            try
            {
                if ( i < 0 || i >= n) throw 1;
                return a[i];
            }
            catch(int i)
            {
                cout << "Exception Array" << endl;
                return a[0];
            } 
        };


};



int main()
{
    Array a(2);

    try
   {
      a[0] = 1;
      cout << "a[0]=" << a[0] << endl;
      a[1] = 1;
      cout << "a[1]=" << a[1] << endl;
      a[2] = 2;
      cout << "a[2]=" << a[2] << endl;
    }
catch(int i)
{
    cout << "Exception main " << endl;
}

cout << "End. " << endl;
}

Okay, so the output is this:

a[0]=1

a[1]=1

Exception Array

Exception Array

a[2]=2

End.

The thing that is confusing me is the reasoning why the program is returning a[2] as value 2? Can someone go into more detail as to how this is achieved, step by step. I think I am not understanding something about exceptions in C++.

Answer 1

  a[2] = 2; 

Here you invoke your operator[] with an out-of-bounds value; it prints Exception Array and returns a reference to aa[0] . As such (ignoring the log message), this assignment is equivalent to a[0] = 2; - the value of aa[0] is now 2.

 cout << "a[2]=" << a[2] << endl; 

Here you again invoke the operator[] with the same parameter; it again outputs the exception message and returns a reference to aa[0] whose value is still 2, as assigned in the previous statement.

Answer 2

This code

a[2] = 2;
cout << "a[2]=" << a[2] << endl;

tries to access a[2] twice, once in the assignment and once in the output statement. They are both out of range, so two exceptions.