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digitalmars.D.learn - Struct assignment, possible DMD bug?

reply "ixid" <nuaccount gmail.com> writes:
This behaviour seems inconsistent and unintuitive:

void main() {
	int[3] a = [1,2,3];
	a = [4, a[0], 6];

	struct S {
		int a, b, c;
	}

	S s = S(1,2,3);
	s = S(4, s.a, 6);

	assert(a == [4,1,6]);
	assert(s == S(4,4,6));
}

Setting the struct writes s.a before evaluating it while the 
reverse is true of the array assignment. Using DMD 2.0.60. GDC 
does what I'd expect and gives both as 4,1,6.
Sep 29 2012
parent reply "Maxim Fomin" <maxim maxim-fomin.ru> writes:
On Saturday, 29 September 2012 at 16:05:03 UTC, ixid wrote:

 This behaviour seems inconsistent and unintuitive:

 void main() {
 	int[3] a = [1,2,3];
 	a = [4, a[0], 6];

 	struct S {
 		int a, b, c;
 	}

 	S s = S(1,2,3);
 	s = S(4, s.a, 6);

 	assert(a == [4,1,6]);
 	assert(s == S(4,4,6));
 }

 Setting the struct writes s.a before evaluating it while the 
 reverse is true of the array assignment. Using DMD 2.0.60. GDC 
 does what I'd expect and gives both as 4,1,6.


I think this is notorious "i = ++i + ++i".
Statement s = S(4, s.a, 6) writes to s object and simultaneously 
reads it.
http://dlang.org/expression.html states that assign expression is 
evaluated in implementation defined-manner and it is an error to 
depend on things like this.
Sep 29 2012
parent reply Timon Gehr <timon.gehr gmx.ch> writes:
On 09/29/2012 06:26 PM, Maxim Fomin wrote:

 On Saturday, 29 September 2012 at 16:05:03 UTC, ixid wrote:

 This behaviour seems inconsistent and unintuitive:

 void main() {
     int[3] a = [1,2,3];
     a = [4, a[0], 6];

     struct S {
         int a, b, c;
     }

     S s = S(1,2,3);
     s = S(4, s.a, 6);

     assert(a == [4,1,6]);
     assert(s == S(4,4,6));
 }

 Setting the struct writes s.a before evaluating it while the reverse
 is true of the array assignment. Using DMD 2.0.60. GDC does what I'd
 expect and gives both as 4,1,6.


 I think this is notorious "i = ++i + ++i".


There is only one mutating sub-expression.


 Statement s = S(4, s.a, 6) writes to s object and simultaneously reads it.
 http://dlang.org/expression.html states that assign expression is
 evaluated in implementation defined-manner and it is an error to depend
 on things like this.


No evaluation order of the assignment expression can possibly lead to
this result. This seems to be a DMD bug.
Sep 29 2012
next sibling parent "Tommi" <tommitissari hotmail.com> writes:
On Saturday, 29 September 2012 at 18:16:24 UTC, Timon Gehr wrote:

 This seems to be a DMD bug.


And a pretty serious looking one at that. That bug could make 
nukes fly to wrong coordinates, and that just ruins everybody's 
day.
Sep 29 2012
prev sibling parent reply =?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com> writes:
On 09/29/2012 11:16 AM, Timon Gehr wrote:

 On 09/29/2012 06:26 PM, Maxim Fomin wrote:



 S s = S(1,2,3);
 s = S(4, s.a, 6);

 assert(a == [4,1,6]);
 assert(s == S(4,4,6));
 }

 Setting the struct writes s.a before evaluating it while the reverse
 is true of the array assignment. Using DMD 2.0.60. GDC does what I'd
 expect and gives both as 4,1,6.


 I think this is notorious "i = ++i + ++i".


 There is only one mutating sub-expression.


But that mutation is happening to an object that is also being read 
inside the same expression.

This is one of the definitions of undefined behavior, not a compiler bug.


 Statement s = S(4, s.a, 6) writes to s object and simultaneously reads
 it.
 http://dlang.org/expression.html states that assign expression is
 evaluated in implementation defined-manner and it is an error to depend
 on things like this.


 No evaluation order of the assignment expression can possibly lead to
 this result. This seems to be a DMD bug.


The compiler seems to be applying an optimization, which it is entitled 
to as long as the language definition is not violated.

If we are using the same object both to read and write in the same 
expression, then we should expect the consequences.

Disclaimer: I assume that D's rules are the same as C and C++ here.

Ali
Sep 29 2012
parent reply Timon Gehr <timon.gehr gmx.ch> writes:
On 09/30/2012 12:51 AM, Ali Çehreli wrote:

 On 09/29/2012 11:16 AM, Timon Gehr wrote:
  > On 09/29/2012 06:26 PM, Maxim Fomin wrote:

  >>> S s = S(1,2,3);
  >>> s = S(4, s.a, 6);
  >>>
  >>> assert(a == [4,1,6]);
  >>> assert(s == S(4,4,6));
  >>> }
  >>>
  >>> Setting the struct writes s.a before evaluating it while the reverse
  >>> is true of the array assignment. Using DMD 2.0.60. GDC does what I'd
  >>> expect and gives both as 4,1,6.
  >>
  >> I think this is notorious "i = ++i + ++i".
  >
  > There is only one mutating sub-expression.

 But that mutation is happening to an object that is also being read
 inside the same expression.

 This is one of the definitions of undefined behavior, not a compiler bug.

  >> Statement s = S(4, s.a, 6) writes to s object and simultaneously reads
  >> it.
  >> http://dlang.org/expression.html states that assign expression is
  >> evaluated in implementation defined-manner and it is an error to depend
  >> on things like this.
  >
  > No evaluation order of the assignment expression can possibly lead to
  > this result. This seems to be a DMD bug.

 The compiler seems to be applying an optimization, which it is entitled
 to as long as the language definition is not violated.


Technically there is no language definition to violate.


 If we are using the same object both to read and write in the same
 expression, then we should expect the consequences.


No. Why?


 Disclaimer: I assume that D's rules are the same as C and C++ here.


C and C++ do not have struct literals and if I am not mistaken,
constructor invocation is a sequence point.

Besides, this does not make any sense, what is the relevant part of the 
standard?

int c = 0;
c = c+1; // c is both read and written to in the same expression
Sep 29 2012
next sibling parent reply =?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com> writes:
On 09/29/2012 04:02 PM, Timon Gehr wrote:

 On 09/30/2012 12:51 AM, Ali Çehreli wrote:

 On 09/29/2012 11:16 AM, Timon Gehr wrote:

 On 09/29/2012 06:26 PM, Maxim Fomin wrote:



 S s = S(1,2,3);
 s = S(4, s.a, 6);

 assert(a == [4,1,6]);
 assert(s == S(4,4,6));
 }

 Setting the struct writes s.a before evaluating it while the reverse
 is true of the array assignment. Using DMD 2.0.60. GDC does what I'd
 expect and gives both as 4,1,6.


 I think this is notorious "i = ++i + ++i".


 There is only one mutating sub-expression.


 But that mutation is happening to an object that is also being read
 inside the same expression.

 This is one of the definitions of undefined behavior, not a compiler 




bug.

 Statement s = S(4, s.a, 6) writes to s object and simultaneously 








reads

 it.
 http://dlang.org/expression.html states that assign expression is
 evaluated in implementation defined-manner and it is an error to




 depend

 on things like this.


 No evaluation order of the assignment expression can possibly lead to
 this result. This seems to be a DMD bug.


 The compiler seems to be applying an optimization, which it is entitled
 to as long as the language definition is not violated.


 Technically there is no language definition to violate.


 If we are using the same object both to read and write in the same
 expression, then we should expect the consequences.


 No. Why?


I am confused. Of course single mutation and many reads should be fine.


 Disclaimer: I assume that D's rules are the same as C and C++ here.


 C and C++ do not have struct literals and if I am not mistaken,
 constructor invocation is a sequence point.


Yes. And in this case it is the compiler-generated constructor. The OP's 
problem goes away if there is a user-provided constructor:

     struct S {
         int a, b, c;

         this(int a, int b, int c)
         {
             this.a = a;
             this.b = b;
             this.c = c;
         }
     }

Now it is as expected:

     assert(s == S(4,1,6));


 Besides, this does not make any sense, what is the relevant part of the
 standard?

 int c = 0;
 c = c+1; // c is both read and written to in the same expression


Silly me! :p

Ali
Sep 29 2012
parent reply "ixid" <nuaccount gmail.com> writes:
On Sunday, 30 September 2012 at 00:24:34 UTC, Ali Çehreli wrote:

 On 09/29/2012 04:02 PM, Timon Gehr wrote:

 On 09/30/2012 12:51 AM, Ali Çehreli wrote:

 On 09/29/2012 11:16 AM, Timon Gehr wrote:

 On 09/29/2012 06:26 PM, Maxim Fomin wrote:



 S s = S(1,2,3);
 s = S(4, s.a, 6);

 assert(a == [4,1,6]);
 assert(s == S(4,4,6));
 }

 Setting the struct writes s.a before evaluating it while










 the reverse

 is true of the array assignment. Using DMD 2.0.60. GDC










 does what I'd

 expect and gives both as 4,1,6.


 I think this is notorious "i = ++i + ++i".


 There is only one mutating sub-expression.


 But that mutation is happening to an object that is also




 being read

 inside the same expression.

 This is one of the definitions of undefined behavior, not a




 compiler bug.

 Statement s = S(4, s.a, 6) writes to s object and








 simultaneously reads

 it.
 http://dlang.org/expression.html states that assign








 expression is

 evaluated in implementation defined-manner and it is an








 error to

 depend

 on things like this.


 No evaluation order of the assignment expression can






 possibly lead to

 this result. This seems to be a DMD bug.


 The compiler seems to be applying an optimization, which it




 is entitled

 to as long as the language definition is not violated.


 Technically there is no language definition to violate.


 If we are using the same object both to read and write in




 the same

 expression, then we should expect the consequences.


 No. Why?


 I am confused. Of course single mutation and many reads should 
 be fine.


 Disclaimer: I assume that D's rules are the same as C and




 C++ here.


 C and C++ do not have struct literals and if I am not


 mistaken,

 constructor invocation is a sequence point.


 Yes. And in this case it is the compiler-generated constructor. 
 The OP's problem goes away if there is a user-provided 
 constructor:

     struct S {
         int a, b, c;

         this(int a, int b, int c)
         {
             this.a = a;
             this.b = b;
             this.c = c;
         }
     }

 Now it is as expected:

     assert(s == S(4,1,6));


 Besides, this does not make any sense, what is the relevant


 part of the

 standard?

 int c = 0;
 c = c+1; // c is both read and written to in the same


 expression

 Silly me! :p

 Ali


This would still seem to be a very poor behaviour worth fixing. 
What is the compiler generating instead of the constructor 
example you gave?



 Sep 29 2012



  parent  =?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com>  writes:


On 09/29/2012 08:13 PM, ixid wrote:

 On Sunday, 30 September 2012 at 00:24:34 UTC, Ali Çehreli wrote:

 On 09/29/2012 04:02 PM, Timon Gehr wrote:

 On 09/30/2012 12:51 AM, Ali Çehreli wrote:

 On 09/29/2012 11:16 AM, Timon Gehr wrote:

 On 09/29/2012 06:26 PM, Maxim Fomin wrote:



 S s = S(1,2,3);
 s = S(4, s.a, 6);

 assert(a == [4,1,6]);
 assert(s == S(4,4,6));
 }

 Setting the struct writes s.a before evaluating it while










 the reverse

 is true of the array assignment. Using DMD 2.0.60. GDC










 does what I'd

 expect and gives both as 4,1,6.


 I think this is notorious "i = ++i + ++i".


 There is only one mutating sub-expression.


 But that mutation is happening to an object that is also




 being read

 inside the same expression.

 This is one of the definitions of undefined behavior, not a




 compiler bug.

 Statement s = S(4, s.a, 6) writes to s object and








 simultaneously reads

 it.
 http://dlang.org/expression.html states that assign








 expression is

 evaluated in implementation defined-manner and it is an








 error to

 depend

 on things like this.


 No evaluation order of the assignment expression can






 possibly lead to

 this result. This seems to be a DMD bug.


 The compiler seems to be applying an optimization, which it




 is entitled

 to as long as the language definition is not violated.


 Technically there is no language definition to violate.


 If we are using the same object both to read and write in




 the same

 expression, then we should expect the consequences.


 No. Why?


 I am confused. Of course single mutation and many reads should be fine.


 Disclaimer: I assume that D's rules are the same as C and




 C++ here.


 C and C++ do not have struct literals and if I am not


 mistaken,

 constructor invocation is a sequence point.


 Yes. And in this case it is the compiler-generated constructor. The
 OP's problem goes away if there is a user-provided constructor:

 struct S {
 int a, b, c;

 this(int a, int b, int c)
 {
 this.a = a;
 this.b = b;
 this.c = c;
 }
 }

 Now it is as expected:

 assert(s == S(4,1,6));


 Besides, this does not make any sense, what is the relevant


 part of the

 standard?

 int c = 0;
 c = c+1; // c is both read and written to in the same


 expression

 Silly me! :p

 Ali


 This would still seem to be a very poor behaviour worth fixing.


I agree. Would you please create a bug report:

   http://d.puremagic.com/issues/


 What is
 the compiler generating instead of the constructor example you gave?


I haven't read the generated assembly output but I am pretty sure that 
the compiler is generating the three expressions of the user-defined 
constructor "out in the open":

   s.a = 4;
   s.b = s.a;   // oops
   s.c = 6;

Ali



 Sep 29 2012



prev sibling  parent  "Maxim Fomin" <maxim maxim-fomin.ru>  writes:


On Saturday, 29 September 2012 at 23:02:08 UTC, Timon Gehr wrote:

 On 09/30/2012 12:51 AM, Ali Çehreli wrote:

 Disclaimer: I assume that D's rules are the same as C and C++ 
 here.


 C and C++ do not have struct literals and if I am not mistaken,
 constructor invocation is a sequence point.


S(4, s.a, 6) is a struct literal here, not a constructor call 
(because structure S doesn't define any constructors). C has 
compound literals which is close to D struct literals.


 Besides, this does not make any sense, what is the relevant 
 part of the standard?


The standard states for assign expression (in $6.15.6) that "the 
side effect of updating the stored value is sequenced after the 
value computations of the left and the rights operands. The 
evaluations of the operands are unsequenced". This means that a 
compiler can evaluate either first and the second operand, or the 
second and the first. In any case it can store value only after 
evaluations of both operands which means that value 4 cannot be 
assigned to s.a when S(4, s.a, 6) is evaluated. Actually, it is a 
bug.



 Sep 29 2012