Why am I getting "double free or corruption" with the following code?

Question

I am writing a simple program to make sure I fully understand how pointers in C work as I go through Harvard's CS50. Here is the code:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int *a = malloc(sizeof(int));
    *a = 5;
    printf("the address of pointer a is: %p\n", &a);
    printf("the address that a is pointing to is: %p\n", a);
    printf("the contents of a are: %i\n", *a);

    printf("please enter a new value for a: \n");
    scanf("%i", a);
    printf("the address of pointer a is: %p\n", &a);
    printf("the address that a is pointing to is: %p\n", a);
    printf("the contents of a are: %i\n", *a);

    int *b = malloc(sizeof(int));
    *b = 7;
    printf("the address of pointer b is: %p\n", &b);
    printf("the address that b is pointing to is: %p\n", b);
    printf("the contents of b are: %i\n", *b);

    a = b;
    printf("setting a = b\n");
    printf("the address of pointer a is: %p\n", &a);
    printf("the address that a is pointing to is: %p\n", a);
    printf("the contents of a are: %i\n", *a);

    free(a);
    free(b);
}

It compiles w/o issues, but when executed, I get the following error: "* Error in `./address': double free or corruption (fasttop): 0x00000000018b7030 * Aborted"

This problem goes away if I get rid of either the free(a) or free(b) statements, however valgrind reveals a memory leak: "==9626== 4 bytes in 1 blocks are definitely lost in loss record 1 of 1 ==9626== at 0x4C2AB80: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so) ==9626== by 0x4206AD: main (address.c:6)"

I've looked through the other questions here and other websites that mention double free corruption, but still can't figure out quite what the issue is... I have a feeling that the answer is simple, and the issue probably lies in the statement "a = b", however I don't really get why I wouldn't be able to have one pointer point to a memory location that another one is pointing to, and then free the memory taken by both pointers...

Answer 1

Let's reduce your code to the bare minimum:

#include <stdlib.h>

int main(void)
{
    int *a = malloc(sizeof(int));
    int *b = malloc(sizeof(int));

    a = b; // after the assignment the pointer value previously held by a is gone.
           // both pointers point to the same memory. The memory previously pointed
           // to by a can no longer be free()d since its address is gone.

    free(a); // free the memory pointed to by b (a is a copy of b) once
    free(b); // free the memory pointed to by b twice --> boom.
}

Answer 2

When you do this:

 a = b;

You are effectively making a and b point to the same address (a.k.a. be the same pointer). So this is what causes the error:

free(a); // Frees the location where b points to
free(b);

Answer 3

Firstly, you need to learn network programming. If you understand how network runs and wirte some little example of TCP/UDP, you'll know how to start.

Then, you need a server to control communication between these two/or more players, that always look like a look below

void activate()
{
    while (true)
    {
        // Process Players packages in
        // Process game logic
        // Process packages respond
    }
}

After that, it will look ok as a little game.