问题
I was learning about C++ pointers and the -> operator seemed strange to me. Instead of
ptr->hello(); one could write (*ptr).hello(); because it also seems to work, so I thought the former is just a more convenient way.
Is that the case or is there any difference?
回答1:
The -> operator is just syntactic sugar because (*ptr).hello() is a PITA to type. In terms of the instructions generated at the ASM level, there's no difference. In fact, in some languages (D comes to mind), the compiler figures everything out based on type. If you do ptr.hello(), it just works, because the compiler knows that ptr is a pointer and doesn't have a hello() property, so you must mean (*ptr).hello().
回答2:
Others have already answered regarding built-in pointers. With regards to classes, it is possible to overload operator->(), operator&(), and operator*() but not operator.().
Which means that an object may act differently depending on which syntax you call.
回答3:
The main advantage in terms of readability comes when you have to chain function calls, i.e.:
ptr->getAnotherPtr()->getAThirdPtr()->print()
I'm not even going to bother doing this with the * operator.
回答4:
The only reason to have the '->' operator is to make it more convenient and save errors like:
*ptr.hello();
Because it is so easy to forget the parenthesis.
回答5:
They generate the same exact machine code, but for me, ptr->arg() is much easier to read than (*ptr).arg().
回答6:
These alternate syntax modes are adopted from C, and you might get some additional understanding from A Tutorial on Pointers and Arrays in C, specifically, chapter 5, Pointers and Structure.
来源：https://stackoverflow.com/questions/447543/ptr-hello-versus-ptr-hello