问题
In C++, pre-increment operator gives lvalue because incremented object itself is returned, not a copy. But in C, it gives rvalue. Why?
回答1:
C doesn't have references. In C++ ++i returns a reference to i (lvalue) whereas in C it returns a copy(incremented).
C99 6.5.3.1/2
The value of the operand of the prefix ++ operator is incremented. The result is the new value of the operand after incrementation. The expression ++Eis equivalent to (E+=1).
‘‘value of an expression’’ <=> rvalue
However for historical reasons I think "references not being part of C" could be a possible reason.
回答2:
C99 says in the footnote (of section $6.3.2.1),
The name ‘‘lvalue’’ comes originally from the assignment expression E1 = E2, in which the left operand E1 is required to be a (modifiable) lvalue. It is perhaps better considered as representing an object ‘‘locator value’’. What is sometimes called ‘‘rvalue’’ is in this International Standard described as the ‘‘value of an expression’’.
Hope that explains why ++i in C, returns rvalue.
As for C++, I would say it depends on the object being incremented. If the object's type is some user-defined type, then it may always return lvalue. That means, you can always write i++++++++ or ++++++i if type of i is Index as defined here:
Undefined behavior and sequence points reloaded
回答3:
Off the top of my head, I can't imagine any useful statements that could result from using a pre-incremented variable as an lvalue. In C++, due to the existence of operator overloading, I can. Do you have a specific example of something that you're prevented from doing in C, due to this restriction?
来源：https://stackoverflow.com/questions/4802315/why-pre-increment-operator-gives-rvalue-in-c