为什么“++i++”无效而(++i)++有效？

Question

Let's consider the following code:

int main() {
    int i = 2;
    int b = ++i++;
    return 3;
}

It compiles with the following with an error:

<source>: In function 'int main()':

<source>:3:16: error: lvalue required as increment operand

    3 |     int b = ++i++;

      |                ^~

This sounds fair to me. Postfix increment has higher priority than prefix increment, so the code is parsed as int b = ++(i++); and i is an rvalue. Hence the error.

Let's now consider this variant with parenthesis to override default priorities:

int main() {
    int i = 2;
    int b = (++i)++;
    return 3;
}

This code compiles and returns 3. On its own, this sounds fair to me but it seems in contradiction with the first code.

The question: why (++i) is an lvalue when i is not?

Thanks!

UPDATE: the error message shown above was from gcc (x86-64 9.2). Here is the exact rendering: error with gcc

Clang x86-64 9.0.0 has a quite different message: error with clang

<source>:3:13: error: expression is not assignable

    int b = ++i++;

            ^ ~~~

With GCC, you get the impression that the problem is with the postfix operator and you can then wander why ++i is OK while i is not, hence my question. With Clang it is clearer that the problem is with the prefix operator.

Answer 1

i and ++i are both lvalues, but i++ is an rvalue.

++(i++) cannot be valid, as the prefix ++ is being applied to i++ , which is an rvalue. But (++i)++ is fine because ++i is an lvalue.

Note that in C, the situation is different; i++ and ++i are both rvalues. (This is an example of why people should stop assuming that C and C++ have the same rules. People insert these assumptions into their questions, which must then be refuted.)

Answer 2

This declaration

int b = ++i++;

is equivalent to

int b = ++( i++ );

The postfix increment operator returns the value of the operand before increment.

From the C++ 17 Standard (8.2.6 Increment and decrement)

1 The value of a postfix ++ expression is the value of its operand... The result is a prvalue .

While the unary increment operator returns lvalue after its increment. So this declaration

int b = (++i)++;

is valid. You could for example write

int b = (++++++++i)++;

From the C++ 17 Standard (8.3.2 Increment and decrement)

1 The operand of prefix ++ is modified by adding 1. The operand shall be a modifiable lvalue. The type of the operand shall be an arithmetic type other than cv bool, or a pointer to a completely-defined object type. The result is the updated operand; it is an lvalue , and it is a bit-field if the operand is a bit-field....

Pay attention that in C the both operators return a value instead of lvalue. So in C this declaration

int b = (++i)++;

is invalid.

Answer 3

so the code is parsed as int b = ++(i++); and i is an rvalue.

No. i is not an rvalue. i is an lvalue. i++ is an rvalue (prvalue to be specific).