What does *(long *) and *(int*) mean?

Question

Could you explain what the next two lines do?

// Line 1
*(long *)add1= *(long *)add2;
// Line 2
*(int *)add1 = *(int *)add2;

Edit 1. I add the complete block of code of the function I test. PS It is only a part of the skript. [original Skript]

It is a custom memcpy function to copy block of memory from A to B.

    (2)
    while(word_left > 0){
        *(char *)temp++ = *(char *)src++;
        word_left --;
    }

    return (void *)dest;
}

Mr Vlag was so kind and explained what the lines in the part(1) mean.

My question is regarding to the part(2). This part should solve a problem if the memory address overlaps.

Why do we use char here? Why can we not use long or int?

Answer 1

In the both records const6ructions ( long * ) and ( int * ) mean casting pointers correspondingly to pointer types long * and int *. Then these pointers are dereferenced *( long * )addr2 and *( int * )addr2 to get access to pointed objects values of which are assigned to other objects that are obtained also through casting and dereferencing pointers.

To make it more clear consider a demonstrative program.

#include <stdio.h>

int main(void) 
{
    int x = 10;
    int y = 0;
    
    printf( "x = %d, y = %d\n", x, y );

    void *p1 = &x;
    void *p2 = &y;
    
    *( int * )p2 = *( int * )p1;
    
    printf( "x = %d, y = %d\n", x, y );
    
    return 0;
}

The program output is

x = 10, y = 0
x = 10, y = 10

In this program for example the pointer p1 of the type void * due to the casting ( int * )p1 is interpreted as a pointer of the type int * that points to an object of the type int (in this case it points to the object x). Now dereferencing this pointer *( int * )p1 you get a direct access to the pointed object x.

You may not just write for example

y = *p1;

because you may not dereference a void pointer because the type void is an incomplete type. So the compiler will not know how to interpret the pointed memory.