commutative Keyword
davidl
davidl at 126.com
Wed Jun 27 20:52:18 PDT 2007
This bug inspired me a little bit:
http://d.puremagic.com/issues/show_bug.cgi?id=124
Actually the operator overloading is somewhat lack of important semantic.
Commutative operators work differently from the non-commutative operators.
consider OPER is a non-commutative operator.
operand2 = operand1 OPER operand2
there's a chance of optimizing as in the
http://www.digitalmars.com/pnews/read.php?server=news.digitalmars.com&group=digitalmars.D&artnum=35949
The main idea is provide OPER_r overloading.
so operand2 = operand2.OPER_r(operand1)
Let's review OPER_r and OPER , actually they are quite close.
consider the plain function calling(the class object get pushed):
typeof(this) OPER(this, operand1)
and OPER_r can actually be
typeof(this) OPER(operand1, this)
so the binary form of OPER_r could actually be deduced from OPER
impelmenation
We don't need to implement OPER_r explicitly now. And with commutative
keyword
we have something like:
commutative opAdd(int operand) {}
opSub(int operand){}
when compiler meet the expression
X = A - ((B*C)+D) * E;
it evaluates it with the first priority in this expression.
X = A - ((B*C)+D) * E; would get A evaluated first, then
T2 = ((B*C)+D)*E // since - is non commutative, so we need T2 to store the
// *EXPRESSION* result
// so compiler optimizes it to :
// X = T2.opSubAssign_r (A)
to evaluate T2
T3 = ((B*C)+D) evaluated , and then
E
to evaluate T3
(B*C) evaluated , and then
D
then
B
then
C
The optimization as we can see is not complicated.
Compiler could optimize it from top to buttom.
OPTIMIZATION RULE:
Once find an operator is non commutative, the compiler optimizes it to the
operator_r calling form.
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