How does this C copy function exit its loop?

Question

void copy (char *source, char *dest) {
    while (*dest++ = *source++);
}

The char that is represented by *source is copied to the field <

Answer 1

chars are integral types. Integral types are interpreted as conditionals in the following way:

 0 -> false
 Anything else -> true

Since "strings" in C are null-terminated (meaning 0 or '\0') when it reaches the end of the string it stops.

Answer 2

The 'result' of an assignment is the right-hand value. So x=1; actually returns a value; in this case, '1'.

Your code copies characters until it encountered the terminating 0 at the end of the source string.

Answer 3

Your interpretation of the copy is correct. The loop stops when what dest is pointing to is zero, i.e., the '\0' character. See http://en.wikipedia.org/wiki/Null-terminated_string