Re: Order of evaluation of a += a++;

[Andrei Alexandrescu via Digitalmars-d]

On 3/31/15 12:18 AM, Orvid King wrote:

I might be a bit confused here, but from what you're saying:
b += a++;
and
b += ++a;

Are doing the same thing in DMD?

No. Lowering of e1 += e2++ is:

e1 += e2++

-\>

(ref a, b) { a = (cast(typeof(a)) (a + b); }(
  e1,
  (ref a) { auto x = a; ++a; return x; }(e2)
)

whereas lowering of e1 += ++e2 is:

e1 += ++e2

-\>

(ref a, b) { a = (cast(typeof(a)) (a + b); }(
  e1,
  (ref a) { ++a; return a; }(e2)
)


Andrei