• Ellery Newcomer (10/10) Jan 21 2010 according to the spec,
• Rainer Deyke (10/19) Jan 21 2010 It's untrue in several ways. For example:
• Steven Schveighoffer (14/24) Jan 22 2010 There are two issues, I think recently discussed, a op= b is actually
• Ellery Newcomer (4/36) Jan 23 2010 Alright, that makes sense enough, thanks.

Ellery Newcomer <ellery-newcomer utulsa.edu> writes:

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?

Rainer Deyke <rainerd eldwood.com> writes:

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

"Steven Schveighoffer" <schveiguy yahoo.com> writes:

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

Ellery Newcomer <ellery-newcomer utulsa.edu> writes:

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