C++: Does using dereference to change value of a variable result in that variable pointing to a different memory block?

Solution 1:

The address of a variable never changes during its lifetime.

This means that changing a variable's value, either directly:

a = 2;

Or indirectly:

b = &a;
*b = 2;

Never changes its address. So your first mental model is correct.

Also, this statement is incorrect:

assume variable a points to the memory address

a is not a pointer type so it does not point to an address. Like all variables, it has an address that does not change.

Solution 2:

The variable a is not moved through the memory. The memory for the variable a is allocated as soon as it is defined.

The variable a points nowhere. It is designed to store an object of the type int.

It is the pointer b that stores the address of the variable a

int* b = &a;

So dereferencing the pointer we get an access to the memory allocated for the variable a and can change the stored object in this memory.

Here is a demonstration program that can help to make understanding easy.

#include <stdio.h>

int main( void )
{
    int a = 1;

    printf( "The address of the variable a is %p\n", ( void * )&a );

    int *b = &a;

    printf( "The address stored in the variable b is %p\n", ( void * )b );

    *b = 2;

    printf( "After the assignment the address of the variable a is %p\n", ( void * )&a );
}

The program output might look like

The address of the variable a is 006FF9D4
The address stored in the variable b is 006FF9D4
After the assignment the address of the variable a is 006FF9D4

As you can see after the assignment