CODEWITHSUNDEEP
c++variablespointersreference
Naphstor
Naphstor

Reputation: 2496

reference variable and pointer problem

I have a pointer which is pointing to an integer variable. Then I assign this pointer to a reference variable. Now when I change my pointer to point some other integer variable, the value of the reference variable doesn't change. Can anyone explain why?

int rats = 101;
int * pt = &rats;
int & rodents = *pt;                                // outputs    
cout << "rats = " << rats;                          // 101
cout << ", *pt = " << *pt;                          // 101
cout << ", rodents = " << rodents << endl;          // 101
cout << "rats address = " << &rats;                 // 0027f940
cout << ", rodents address = " << &rodents << endl; // 0027f940
int bunnies = 50;
pt = &bunnies;

cout << "bunnies = " << bunnies;                    // 50
cout << ", rats = " << rats;                        // 101  
cout << ", *pt = " << *pt;                          // 50
cout << ", rodents = " << rodents << endl;          // 101
cout << "bunnies address = " << &bunnies;           // 0027f91c
cout << ", rodents address = " << &rodents << endl; // 0027f940

We assigned pt to bunnies, but the value of rodents is still 101. Please explain why.

Upvotes: 1

Views: 156

Answers (1)

Cameron Skinner
Cameron Skinner

Reputation: 54276

The line

int & rodents = *pt;

is creating a reference to what pt is pointing to (i.e. rats). It's not a reference to the pointer pt.

Later, when you assign pt to point to bunnies, you would not expect the rodents reference to change.

EDIT: To illustrate @Als point, consider the following code:

int value1 = 10;
int value2 = 20;
int& reference = value1;
cout << reference << endl; // Prints 10
reference = value2; // Doesn't do what you might think
cout << reference << endl; // Prints 20
cout << value1 << endl; // Also prints 20

The second reference assignment does not change the reference ltself. Instead, it applies the assignment operator (=) to the thing referred to, which is value1.

reference will always refer to value1 and cannot be changed.

It's a little tricky to get your head around at first, so I recommend you take a look at Scott Meyer's excellent books Effective C++ and More Effective C++. He explains all this much better than I can.

Upvotes: 2

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