[Issue 3378] [tdpl] ++x should be an lvalue

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http://d.puremagic.com/issues/show_bug.cgi?id=3378



--- Comment #6 from Andrei Alexandrescu <andrei at metalanguage.com> 2010-03-04 04:17:15 PST ---
(In reply to comment #5)
> C99 says this about ++x:
> 
> -------6.5.3.1-------------
> The operand of the prefix increment or decrement operator shall have qualified
> or unqualified real or pointer type and shall be a modifiable lvalue.
> 
> Semantics
> 
> The value of the operand of the prefix ++ operator is incremented. The result
> is the new value of the operand after incrementation. The expression ++E is
> equivalent to (E+=1). See the discussions of additive operators and compound
> assignment for information on constraints, types, side effects, and conversions
> and the effects of operations on pointers. The prefix -- operator is analogous
> to the prefix ++ operator, except that the value of the operand is decremented.
> --------6.5.16-------------------
> An assignment expression has the value of the left operand after the
> assignment, but is not an lvalue.
> ---------------------------------
> 
> It is equivalent to x+=1, and therefore not an lvalue.
> 
> The C++98 spec also says that ++x is equivalent to x+=1, but says that the
> result of x+=1 is an lvalue.

(Still scantily connected.) Wait, I'm confused. C and C++ do not define ++x as
x+=1, right? We shouldn't either, for reasons that we've discussed at length
before (i.e. there are UDTs for which increment makes sense but addition does
not.)

So:

(a) x+=1 is not a part of the discussion about ++x.

(b) Keeping ++x an rvalue is a gratuitous incompatibility with C and C++

(c) Keeping ++x an rvalue requires extensive rework of TDPL (the bump example
is taken from there)

There isn't much reason to debate. It's a trivial change that has only benefits
(albeit minor), I suggest we simply make it and move on.

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