assuming (this) is a real data type with a finite number of bits, this code will loop until (this) overflows and ends up back at zero. in C, this code will compile without error or warning (unless some elevated level of warnings complains about the semicolon with no braces), but in C++, since (this) is an implicitly const pointer, and is only valid within the context of a nonstatic class or struct member function, it would certainly not compile.Originally Posted by phantomotap
