I have a simple program

int main()
{
    
    int a = 1; // variable declaration
    printf("a before: %d \n", a); // a is 1
    int* b = &a; // pointer declaration - get the memory address of variable `a` with the `&` reference opeartor
    *b = 2; // dereference to access data contained the memory block of variable `a` via the the `*` dereference operator
    printf("a after : %d", a); // a becomes 2
    return 0;
}

I wonder which mental model is correct:

1. assume variable a points to the memory address 100000f91 at which the data int 1 is stored. By doing this assignment *b = 2 we change the content/data from int 1 to int 2 but the data is still in the original memory address i.e. 100000f91.

2. By doing this assignment *b = 2, we change where the variable a point from the memory address 100000f91 to somewhere else e.g. 100000f92 at which the new data int 2 is stored

Please feel free to correct me if I am wrong here. whichever the mental model is correct, I wonder if the same mental model can still apply when we pass a by reference to a function as in

void fn(int* b) {}

CodePudding user response：

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.

CodePudding user response：

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

    *b = 2;

the address of the variable a was not changed. It is the value that is stored in the memory occupied by the variable a that was changed.

You may consider a variable as a named extent of memory designed to store objects of the specified type used to declare the variable.