Why isn't ++x an lvalue in D?

[Bill Baxter]

On Fri, Jan 9, 2009 at 2:43 PM, Weed <resume755 at mail.ru> wrote:
> Weed пишет:
>> Bill Baxter пишет:
>>> 2009/1/9 Weed <resume755 at mail.ru>:
>>>> Bill Baxter пишет:
>>>>> Another thread just reminded me of something I use frequently in C++
>>>>> that doesn't work in D because ++x is not an lvalue:
>>>>>
>>>>>    int x,N;
>>>>>   ...
>>>>>    ++x %= N;
>>>>>
>>>>> So is there some deep reason for not making it an lvalue like in C++?
>>>>>
>>>> ++x is x+=1 in D:
>>>>
>>>> void main() {
>>>>  int i =3;
>>>>  int N =2;
>>>>  (i+=1) %= N;
>>>> }
>>>>
>>>>
>>>> Error: i += 1 is not an lvalue.
>>>>
>>>> C++:
>>>>
>>>> int main()
>>>> {
>>>>  int i = 2;
>>>>  int N = 3;
>>>>  i+1 %= N;
>>>>
>>>>  return 0;
>>>> }
>>>>
>>>> error: lvalue required as left operand of assignment
>>>>
>>> What does C++ do if you use  (i+=1) %= N instead of (i+1)?  Doesn't +=
>>> also return an lvalue in C++?
>>
>> I am a bit mixed, but the meaning has not changed:
>>
>> $ cat demo.cpp
>> int main()
>> {
>>  int i = 2;
>>  int N = 3;
>>  i+=1 %= N;
>>
>>  return 0;
>> }
>>
>> $ c++ demo.cpp
>> demo.cpp: In function 'int main()':
>> demo.cpp:5: error: lvalue required as left operand of assignment
>
> And I think it is wrong that the ++ same as += 1. The operator ++ in C
> uniquely compiles in CPU instruction "increment".
>
> For objects it would be better to make a += 1 only if undefined
> overloaded operator ++.

Yeh, I think that's scheduled to be changed after Andrei's repeated
thrashings of Walter.

--bb