Why pre-increment operator gives rvalue in C?

Question

In C++, pre-increment operator gives lvalue because incremented object itself is returned, not a copy. But in C, it gives rvalue. Why?

Answer 1

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?

Answer 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.

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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

Answer 3

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.