memcpy() vs memmove()

I am trying to understand the difference between memcpy() and memmove(). I have read the documentation, that memcpy() doesn't take care of the overlapping source and destination, whereas memmove() does.

However, when I execute these two functions on overlapping memory blocks, they both give the same result. For instance, take the following MSDN example on the memmove() help page:-

Is there a better example to understand the drawbacks of memcpy and how memmove solves it?

// Illustrate overlapping copy: memmove always handles it correctly;
// memcpy may handle it correctly.

#include <memory.h>
#include <string.h>
#include <stdio.h>

char str1[7] = "aabbcc";

int main( void )
{
    printf( "The string: %s\n", str1 );
    memcpy( str1 + 2, str1, 4 );
    printf( "New string: %s\n", str1 );

    strcpy_s( str1, sizeof(str1), "aabbcc" );   // reset string

    printf( "The string: %s\n", str1 );
    memmove( str1 + 2, str1, 4 );
    printf( "New string: %s\n", str1 );
}

Output:

memcpy():
The string: aabbcc
New string: aaaabb

memmove():
The string: aabbcc
New string: aaaabb