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January 22, 2010
op=
according to the spec,

a op= b;

is semantically equivalent to

a = a op b;

but this doesn't seem to be strictly true. for example:

char c = 'a';
real r = 3.14;

c = c + r; // error

c += r;  // accepted; seems to be doing c += floor(r);

is this behavior intentional?
January 22, 2010
Re: op=
Ellery Newcomer wrote:
> according to the spec,
> 
> a op= b;
> 
> is semantically equivalent to
> 
> a = a op b;
> 
> but this doesn't seem to be strictly true.


It's untrue in several ways.  For example:

SomeReferenceType a, b;

a = a + b; // Creates new object.
b += c; // Modifies an object in place.

and of course:

a = a = c; // Dual assignment.
a == b; // Equality comparison.


-- 
Rainer Deyke - rainerd@eldwood.com
January 22, 2010
Re: op=
On Thu, 21 Jan 2010 22:44:24 -0500, Ellery Newcomer  
<ellery-newcomer@utulsa.edu> wrote:

> according to the spec,
>
> a op= b;
>
> is semantically equivalent to
>
> a = a op b;
>
> but this doesn't seem to be strictly true. for example:
>
> char c = 'a';
> real r = 3.14;
>
> c = c + r; // error
>
> c += r;  // accepted; seems to be doing c += floor(r);
>
> is this behavior intentional?

There are two issues, I think recently discussed, a op= b is actually  
equivalent to:

a = a op cast(typeof(a))b;

when dealing with primitives.  I'm not sure if this happens with user  
defined types.

And the second issue (if you want to get technical) is that a op= b is  
considered a different operation than a = a op b because one can manually  
optimize a op= b more than a = a op b.  So technically the seemingly  
analogous calls can do something different.  A good example is array  
appending, a ~= b is not equivalent to a = a ~ b, because the former may  
allocate in place and the latter always reallocates.

-Steve
January 23, 2010
Re: op=
On 01/22/2010 12:23 PM, Steven Schveighoffer wrote:
> On Thu, 21 Jan 2010 22:44:24 -0500, Ellery Newcomer
> <ellery-newcomer@utulsa.edu> wrote:
>
>> according to the spec,
>>
>> a op= b;
>>
>> is semantically equivalent to
>>
>> a = a op b;
>>
>> but this doesn't seem to be strictly true. for example:
>>
>> char c = 'a';
>> real r = 3.14;
>>
>> c = c + r; // error
>>
>> c += r; // accepted; seems to be doing c += floor(r);
>>
>> is this behavior intentional?
>
> There are two issues, I think recently discussed, a op= b is actually
> equivalent to:
>
> a = a op cast(typeof(a))b;
>
> when dealing with primitives. I'm not sure if this happens with user
> defined types.
>
> And the second issue (if you want to get technical) is that a op= b is
> considered a different operation than a = a op b because one can
> manually optimize a op= b more than a = a op b. So technically the
> seemingly analogous calls can do something different. A good example is
> array appending, a ~= b is not equivalent to a = a ~ b, because the
> former may allocate in place and the latter always reallocates.
>
> -Steve

Alright, that makes sense enough, thanks.

Spec needs to be changed, though

http://d.puremagic.com/issues/show_bug.cgi?id=3735
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