FAQ: References and Pointers - Memory Address

This community-built FAQ covers the “Memory Address” exercise from the lesson “References and Pointers”.

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FAQs on the exercise Memory Address

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If I consider the following code -


int soda = 99;
int &pop = soda;
pop += 1;
soda +=  20;
std::cout << soda << "\t" << pop;

This one changes the value to 120, so a change to a reference changes the original variable and vice-versa, but in Line 2 it’s not a declaration, though it’s not being considered as the Address-Of operator , WHY ? Am I having some wrong concepts about Declaration ?

1 Like

First, remember this point:
If & is used in declaration, it’s a reference, else it’s a memory address.

In your example, & is used in declaration. Therefore, it’s a reference (aka alias). Reference is directly modifying the variable it’s pointing to. Hence, line 3 and 4 of your code will make changes on soda and pop, and both of them are the same. Line 3, adds 1 to initial value (99+1=100) and line 4 adds 20 (100+20=120).

Line 2 of your code IS A DECLARATION.

4 Likes
  • When & is used in a declaration, it is a reference operator.
  • When & is not used in a declaration, it is an address operator.

When it is not used in a declaration where else can it be used??? :slight_smile:

A minor typo report:

So we haved learned…

should be just “So we have learned…”