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Date: Thu, 2 Feb 2012 05:33:17 -0800 (PST)
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Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:19790

=E5=9C=A8 2012=E5=B9=B41=E6=9C=8814=E6=97=A5=E6=98=9F=E6=9C=9F=E5=85=ADUTC+=
8=E4=B8=8A=E5=8D=886=E6=97=B648=E5=88=8629=E7=A7=92=EF=BC=8CEvan Driscoll=
=E5=86=99=E9=81=93=EF=BC=9A
> On 01/13/2012 03:20 PM, Neil Cerutti wrote:
> > They perform the same action, but their semantics are different.
> > operator+ will always return a new object, thanks to its
> > signature, and operator+=3D shall never do so. That's the main
> > difference I was getting at.
>=20
> I was talking about the combination of + and =3D, since the discussion is=
=20
> about 'a =3D a + b' vs 'a +=3D b', not 'a + b' vs 'a +=3D b' (where the=
=20
> differences are obvious).
>=20
> And I stand by my statement. In 'a =3D a + b', operator+ obviously return=
s=20
> a new object, but operator=3D should then go and assign the result to and=
=20
> return a reference to 'a', just like how 'a +=3D b' will return a=20
> reference to 'a'.
>

The operation a+b means add(a,b) and returns a result instance, furthermore=
 a and b can't be modified.

The expression a =3D a+b are two operations not one. But in C or C++ the  p=
roblem is mixing operations and expressions in a free style allowed.
=20
The operation a+=3Db means a modified by b and b can't be changed.
Note that no new instance is necessary in a+=3Db.
=20
=20

> If you're working in C++ and overload your operators so that 'a +=3D b'=
=20
> and 'a =3D a + b' have different observable behaviors (besides perhaps=20
> time), then either your implementation is buggy or your design is very=20
> bad-mannered.
>=20
> Evan

Do you mean the result instances after 'a+=3Db' and 'a=3Da+b' or=20
the actions of behaviors of instances involved  in performing 'a+=3Db' and =
'a=3Da+b'?