標題：
Proposal: Operator overloading without temporaries作者：
Don Clugston <dac@nospam.com.au>
正文：
Background: Operator overloading, in the form it exists in C++ and
currently in D, inherently results in sub-optimal code, because it
always results in unnecessary temporary objects being created.
For example,
X = A - ((B*C) + D)* E;
becomes:
T1 = B * C;
T2 = T1 + D;
T3 = T2 * E;
T4 = A - T3;
X = T4;
Four objects were created, whereas only one was strictly required.
In C++, there are libraries like Blitz++ which use complicated
expression templates in order to avoid these creating these temporaries,
and provide performance comparable with FORTRAN. I think D can do much
better...
Note that temporaries are avoided when using the opXXXAssign() operators
like +=.
===========
?? Proposal
===========
(1) Allow the compiler to assume that b = b + c? can be replaced with b
+= c. (In C++, operator + and operator += are just symbols, the compiler
doesn't know that there is any relationship between them).
In the example above, this would allow the compiler to generate:
T1 = B * C;
T1 += D;
T1 *= E;
and we have eliminated two of the three temporaries.
(2). Fill in the gaps in the operator overloading table by introducing
opAddAssign_r, opSubAssign_r, etc.
Just as A.opSubAssign(B)
is the operation? A -= B? or equivalently? A = A - B, similarly
A.opSubAssign_r(B)
would mean
A = B - A.
and would only occur when temporaries are generated in expressions. Like
-=, it's an operation which can frequently be performed very
efficiently, but at present the language has no way of expressing it.
Our original example then becomes:
T1 = B.opMul(C);
T1.opAddAssign(D);
T1.opMulAssign(E);
T1.opSubAssign_r(A);
X = T1;
... and all the useless temporaries are gone!
More formally, when the expression tree for an expression is generated:
With a binary operator XXX, operating on left & right nodes:
if (the left node is *not* an original leaf node) {
??? // the left node is a temporary, does not need to be preserved.
??? // we don't care if the right node is a temporary or not
??? look for opXXXAssign().
} else if (the the right node is not an original leaf node) {
??? // the right node is a temporary
??? look for opXXXAssign_r()
} else {
?? // both left and right nodes are leaf nodes, we have to
?? // create a temporary
??? look for opXXX(), just as it does now.
}
These rules also cope with the situation where temporaries are required:
eg
X = (A*B) + (C*D);
becomes
T1 = A*B;
T2 = C*D;
T1 += T2;
X = T1;
If this were implemented, it would permanently eradicate (for D) the
most significant advantage which Fortran has managed to retain over
object-oriented languages. And I really don't think it would be
difficult to implement, or have negative side-effects.
There are a couple of decisions to be made:
(I) should the compiler use opAdd() and generate a temporary, if
opAddAssign_r() doesn't exist, to preserve existing behaviour? I think
the answer to this is YES.
(II) should the compiler use opAdd() and generate a temporary, if
oppAddAssign() doesn't exist, to preserve existing behaviour? Again, I'm
inclined to answer YES.
(III) If the code includes +=, and there is an opAdd() but no
opAddAssign(), should the compiler accept this, and just generate an
opAdd() followed by an assignment?? This would mean that opAdd() would
generate the += operation as well as +, while opAddAssign() would be a
performance enhancement. (It would still be possible to have
opAddAssign() without opAdd(), to have += but not +, but it would not be
possible to have + without +=). This would mean that += would be
*purely* syntactic sugar.
Decision III would be a little more difficult to implement and is of
less obvious merit, I only mention it as a possibility.
Comments?