• Bill Baxter (7/7) Jan 08 2009 Another thread just reminded me of something I use frequently in C++
• Weed (17/26) Jan 08 2009 ++x is x+=1 in D:
• Bill Baxter (13/13) Jan 08 2009 MjAwOS8xLzkgV2VlZCA8cmVzdW1lNzU1QG1haWwucnU+Ogo+IEJpbGwgQmF4dGVyINDJ28XU...
• Weed (13/51) Jan 08 2009 I am a bit mixed, but the meaning has not changed:
• Weed (5/59) Jan 08 2009 And I think it is wrong that the ++ same as += 1. The operator ++ in C
• Bill Baxter (27/27) Jan 08 2009 T24gRnJpLCBKYW4gOSwgMjAwOSBhdCAyOjQzIFBNLCBXZWVkIDxyZXN1bWU3NTVAbWFpbC5y...
• Weed (4/67) Jan 08 2009 Thus, once it works it is necessary to change anything if only for the
• Don (8/75) Jan 09 2009 There is none. In tango.math.Math there's nextFloatUp, nextDoubleUp,
• Bill Baxter (25/25) Jan 08 2009 T24gRnJpLCBKYW4gOSwgMjAwOSBhdCAyOjIyIFBNLCBXZWVkIDxyZXN1bWU3NTVAbWFpbC5y...
• Weed (4/66) Jan 08 2009
• Nick Sabalausky (5/56) Jan 08 2009 Shouldn't that be "(i += 1) %= N"? Assignments are right-associative (bu...
• Weed (3/66) Jan 08 2009 Yes, my code is wrongÀ

"Bill Baxter" <wbaxter gmail.com> writes:

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

Weed <resume755 mail.ru> writes:

Bill Baxter ÐÉÛÅÔ:
 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

"Bill Baxter" <wbaxter gmail.com> writes:

MjAwOS8xLzkgV2VlZCA8cmVzdW1lNzU1QG1haWwucnU+Ogo+IEJpbGwgQmF4dGVyINDJ28XUOgo+
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Cg==

Weed <resume755 mail.ru> writes:

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 <resume755 mail.ru> writes:

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 ++.

"Bill Baxter" <wbaxter gmail.com> writes:

T24gRnJpLCBKYW4gOSwgMjAwOSBhdCAyOjQzIFBNLCBXZWVkIDxyZXN1bWU3NTVAbWFpbC5ydT4g
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Weed <resume755 mail.ru> writes:

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

Don <nospam nospam.com> writes:

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

There is none. In tango.math.Math there's nextFloatUp, nextDoubleUp, 
nextRealUp, which perform the "next possible number" operation.
In D, ++x for real x performs x = x + 1.0.
If x is (say) 1e80, then (++x) == x !

I'm not sure why D allows ++x for floating-point types, it's asking for 
trouble IMHO -- if you're incrementing a real, I think it's highly 
probable that you have a bug somewhere.

"Bill Baxter" <wbaxter gmail.com> writes:

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Weed <resume755 mail.ru> writes:

Bill Baxter ÐÉÛÅÔ:
 On Fri, Jan 9, 2009 at 2:22 PM, Weed <resume755 mail.ru> wrote:
 Bill Baxter ÐÉÛÅÔ:
 2009/1/9 Weed <resume755 mail.ru>:
 Bill Baxter ÐÉÛÅÔ:
 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" <a a.a> writes:

"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

Weed <resume755 mail.ru> writes:

Nick Sabalausky ÐÉÛÅÔ:
 "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 wrongÀ

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