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

• ixid (15/15) Sep 29 2012 This behaviour seems inconsistent and unintuitive:
• Maxim Fomin (7/22) Sep 29 2012 I think this is notorious "i = ++i + ++i".
• Timon Gehr (4/30) Sep 29 2012 No evaluation order of the assignment expression can possibly lead to
"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
"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
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
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
"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
=?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
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
=?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

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
"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

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

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

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
=?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

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

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

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
"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