references and address (memory location) of a variable

Can someone please explain why the output of the below code snippet is 20 ?

int i = 10;
cout << i << endl;

int &r = i;
r = 20;
cout << i << endl;

First, variable i stores the integer value 10 Then 10 is displayed. Then the address of r (memory location of r) is set to i which is 10 And then r becomes 20 But why i changes to 20 as well?

The integer content of r has cha开发者_运维技巧nged, not address of it (memory location of it).

Thanks,

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The variable r is a refernce to i, it's like a pointer except that instead of saying *r = 20; you just say r = 20; and that changes the value of r.

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When you make a reference it's almost like a pointer so when you can changed r you actually changed what r was pointing to which is also i.

Actually to be more precise the reference is not a pointer or an address, it is the object. At least in c++ world.

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Think about it as:

int i = 10;
cout << i << endl;

int *p = &i;
*p = 20;
cout << i << endl;

This is basically what's happening behind the scenes