Pointer value in c

As Arkku says Arguments are local variables, you can treat the pointer as an address value like 0x7fff13f1ce3c, then it's pretty easy to understand.

typedef address int*
typedef address2 size_t*

size_t *function(int a, address b)
{   
    int d;
    address2 array;
    b = &d;
    //DO STUFF HERE
    printf("%d", *b); //This gets printed right
    return array;

  }
int main()
{
    address2 array2;
    address b;

    array2 = function(a, b);
    printf("%d", *b);   //This gives seg fault

}

Change the value of b in function won't influence the value of b in main. As for the array, function will return An address value.

In function main you pass the pointer b to function but b is uninitialized (can contain any value):

array2 = function(a, b);

Thefore this makes no sense. Inside function, If you want to modify b you need to instead call function like this:

int b;

array2 = function(a, &b);

The function can now assign to *b.

Arguments are local variables, so you are not actually assigning to main's pointer, you are just ignoring the value passed as argument and assigning to the local b in function. You need a pointer to pointer int **b argument in order to assign to the b in main.

Also, the assigned address is that of a local variable d, which ceases to exist when function returns so accessing it afterwards through a pointer would be invalid regardless (undefined behaviour, even if it might happen to work before it gets overwritten on the stack).

edit: If your goal is to get the value of d to main, you don't need a pointer in main, so instead of b you can have int d in main and pass its address &d to function (function(a, &d)), then you can assign to main's d in function with *b = d.