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++?

--bb

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

2009/1/9 Weed <resume755@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++?

--bb

Bill Baxter пишет:
> 2009/1/9 Weed <resume755@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

Weed пишет:
> Bill Baxter пишет:
>> 2009/1/9 Weed <resume755@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 ++.

On Fri, Jan 9, 2009 at 2:22 PM, Weed <resume755@mail.ru> wrote:
> Bill Baxter пишет:
>> 2009/1/9 Weed <resume755@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
>

Huh.  Ok, well I guess that answers it then.  Thanks.  Mystery solved! To summarize, In D   ++x is x+=1, and in D, like in C++,  x+=1 is not an lvalue.   Got it.

But now I wonder why it's not an lvalue in C++ or D.  If x=y is an lvalue and ++x is an lvalue, why shouldn't x+=1 be one too?

--bb

On Fri, Jan 9, 2009 at 2:43 PM, Weed <resume755@mail.ru> wrote:
> Weed пишет:
>> Bill Baxter пишет:
>>> 2009/1/9 Weed <resume755@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

"Weed" <resume755@mail.ru> wrote in message news:gk6n29$218m$1@digitalmars.com...
> Bill Baxter ?????:
>> 2009/1/9 Weed <resume755@mail.ru>:
>>> Bill Baxter ?????:
>>>> 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;
>

Shouldn't that be "(i += 1) %= N"? Assignments are right-associative (but then these are op-assigns...). The error is probably saying that "1" can't be an lvalue for "%= N" which, of course, is true.

> return 0;
> }
>
> $ c++ demo.cpp
> demo.cpp: In function 'int main()':
> demo.cpp:5: error: lvalue required as left operand of assignment

Bill Baxter пишет:
> On Fri, Jan 9, 2009 at 2:22 PM, Weed <resume755@mail.ru> wrote:
>> Bill Baxter пишет:
>>> 2009/1/9 Weed <resume755@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
>>
> 
> Huh.  Ok, well I guess that answers it then.  Thanks.  Mystery solved! To summarize, In D   ++x is x+=1, and in D, like in C++,  x+=1 is not an lvalue.   Got it.
> 
> But now I wonder why it's not an lvalue in C++ or D.  If x=y is an lvalue and ++x is an lvalue, why shouldn't x+=1 be one too?


It is also a lvalue, it is my mistake again.

(i+=1) %= N;

Nick Sabalausky пишет:
> "Weed" <resume755@mail.ru> wrote in message news:gk6n29$218m$1@digitalmars.com...
>> Bill Baxter ?????:
>>> 2009/1/9 Weed <resume755@mail.ru>:
>>>> Bill Baxter ?????:
>>>>> 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;
>>
> 
> Shouldn't that be "(i += 1) %= N"? Assignments are right-associative (but then these are op-assigns...). The error is probably saying that "1" can't be an lvalue for "%= N" which, of course, is true.

Yes, my code is wrongю

And apparently, we have an error in D

> 
>> return 0;
>> }
>>
>> $ c++ demo.cpp
>> demo.cpp: In function 'int main()':
>> demo.cpp:5: error: lvalue required as left operand of assignment
> 
> 
>