Re: undefined compound chaining behavior

[Stefan Seefeld]

On 2014-06-12 21:24, Hossein Sajjadi wrote:
> Again we analyse the expression
> a=1
> a+=(a+=4)
> It is shown that the operator in both langauges is the same.

Even if you can show that operator+= is the same in both, that doesn't
mean that the ternary expression "a+=a+=4" maps from Octave straight to
C++. In fact, as others have said, such compound expressions are broken
up, so the C++ compiler never sees this as a single expression, but
rather as two binary ones (with a temporary, just as Jordi pointed out.)

Thus, these expressions are in fact not equivalent. (You could think of
it as Octave injecting synthetic sequence points, if that helps.)

        Stefan

-- 

      ...ich hab' noch einen Koffer in Berlin...