[SOLVED] Does address-of operator return address of an object referenced by the variable

Does address-of operator return address of an object referenced by the variable

This is different question from this one, as here I'm using object type struct, instead of a value 23....

I'm reading this chapter on pointers and it states the following:

The address of a variable can be obtained by preceding the name of a variable with an ampersand sign (&), known as address-of operator. For example:

MyStruct* myvar = new MyStruct();
&myvar; // what is this address?

Is my understanding correct that it's the address of an object new MyStruct(), not the variable itself? I don't yet have good understanding of how variables (not objects they reference) are stored, and it's very likely that they are not used at all when program is compiled.

Solution

As per your previous question: No, it isn't... It is the address of variable myvar...

EDIT

so the actual bytes of new Struct() are stored under different address than a variable?

new return address of memory where "bytes" of struct are allocated. Value of myvar will be the address where "bytes" are placed in memory, and &myvar is the address where variable myvar is placed in memory.

-------------------------------------------------------
|                     M E M O R Y                     |
-------------------------------------------------------
|   --------------                   ---------------  |
|   | myvar = 234| ----points to---> | new MyStruct|  |
|   --------------                   ---------------  |
|   ^                                ^                |
|   |                                |                |
|   address 1 (&myvar)               address 234      |
|                                                     |
-------------------------------------------------------