🚀 go-pugleaf

In article <1992Nov03.195240.14360@microsoft.com> jimad@microsoft.com (Jim Adcock) writes:
>In article <1992Oct31.025229.6520@cs.brown.edu> sdm@cs.brown.edu (Scott Meyers) writes:
>|Is the return type of operator= an lvalue?  
>
>No.

        Jim is mistaken; the built-in types return an lvalue in C++ (except for
bit-field assignment, which I don't know about offhand).

>|In particular, if a, b, and c are of built-in type, is this legal?
>|
>|    (a = b) = c;
>
>No, it is an error.  If (a = b) were an lvalue the same
>as "a" then "a" would be being accessed twice for modification in one
>expression, which in turn is an unconstrained error.  

        It is not an error.  ((a=b)=c) is equivalent to (a=c), except for side
effects, of course.

>My answers are based on the ANSI-C spec

        In Standard C, = returns an rvalue.  In C++, = returns an lvalue.

        Stroustrup talks about this in the "Lvalues" section of "The Evolution
of C++: 1985 to 1989".  I quote:

                  "Note that the default definition of assignment of
          an X as a call of

              X& operator=(const X&)

	  makes assignment of Xs produce an lvalue.  For uniformity,
	  this rule has been extended to assignments of built-in
	  types.  By implication, +=, -=, *=, etc.  now also produce
	  lvalues.  So -- again by implication -- does prefix ++ and
	  -- (but not the postfix versions of these operators).

                  "In addition, the comma and ?: can also produce
          lvalues...."

>The "right" thing to do is to return a value, not a reference.
>...Don't return a non-const reference.

        Return a non-const reference.  C++ does it with its built-in types.


-- 
John Davison
davisonj@ecn.purdue.edu