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

[Weed]

Bill Baxter пишет:
> 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.

Thus, once it works it is necessary to change anything if only for the
real types increment will give increase for the lowest possible value.
(I do not know how the CPU instruction increment works for real values)