C++ copying char* using pointers - c++

Question

I am trying to grasp pointers and I have this simple code for which I need some explanation.
I need to copy one char array to another. In my main function I have this code:
const int MAX_SIZE = 100;
char x[MAX_SIZE] = "1234565";
char* y = new char[MAX_SIZE];
copyArray(x, y);
std::cout << y;
delete [] y;
Now comes the question, how does this code (which works jut fine):
while ((*dest = *source) != '\0')
{
dest += 1;
source += 1;
}
Differ from this (gives strange characters at the end):
while (*source != '\0')
{
*dest = *source;
dest += 1;
source += 1;
}
Looking at this it seems those two functions are pretty similar.
It makes sense that we are copying until we reach a null-terminator in the source string, right (2nd function)?
But it's not working correctly - I get some strange characters at the end of the copied array. However, the first function works just fine.
void copyArray(const char* source, char* dest);

Answer 1

The form
while ((*dest = *source) != '\0')
{
dest += 1;
source += 1;
}
guarantees that the assignment of the character to copy ((*dest = *source)) is applied before testing the condition if the terminating '\0' character is reached is evaluated to false.
The second version doesn't copy the terminating '\0' character, because the loop ends before the
*dest = *source;
statement is ever reached.

Answer 2

(*dest = *source) is an evaluated expression just like the 1+1 part of int i = 1+1; so, after it is evaluated, the value is usable in an other expression
The difference is that in ((*dest = *source) != '\0'), the value of *source is assigned to *dest, then the whole expression is evaluated ( expression has the same value than *source ) while the value pointed by *source is only used to evaluate *source != '\0', but never assigned during the evaluation of that statement.
EDIT
user0042 brings a realy acute observation : by doing so, the following code
while ((*dest = *source) != '\0')
{
dest += 1;
source += 1;
}
ensures that the final char of the array has a value of '\0'

Answer 3

In the example you incrementing the addresses of x, y thus until the last character they point to Null-terminator so you have to declare a temporary variables to hold the first address:
char* x = "1234565";
char* y = new char[MAX_SIZE];
// Temporary pointers to hold the first element's address
char* tmp1 = x;
char* tmp2 = y;
while( (*y = *x) != '\0'){
x += 1; // X no longer points to the first element
y += 1; // Y no longer points to the first element
}
std::cout << tmp2;
You can use a do while loop instead of while:
char* x = "1234565";
const int size = strlen(x);
char* y = new char[size];
char* tmp1 = x;
char* tmp2 = y;
do{
*y = *x;
x += 1;
y += 1;
}while( *(x - 1) != '\0');
// Now no need for adding a null-terminator it is already added in the loop
// tmp2[size] = '\0';
std::cout << tmp2;
Because Assigning the values before incrementing so the last character \0 will break the loop before added to the destination pointer y. Thus I made the loop breaks not on n = '\0' but on n - 1 = '\0' to ensure that it is added to y.