[Bug c++/45437] Loses reference during update

["igodard at pacbell dot net" <gcc-bugzilla at gcc dot gnu dot org>]

------- Comment #11 from igodard at pacbell dot net  2010-08-29 18:24 -------
Note to Nelson, for the record here:
There is a disagreement about C++ sequence semantics happening in
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45437

The gist is the following code:
    bool b = false;
    bool f() { b = true; return false; }
    b |= f();
The question is whether this is well defined (with b having the value true
after the assignment) or undefined. Your proposal states:

5.17p1 (assignment and compound assignment operators):

    The assignment operator (=) and the compound assignment operators all group
right-to-left. All require a modifiable lvalue as their left operand and return
an lvalue with the type and value of the left operand after the assignment has
taken place an lvalue referring to the left operand. The result in all cases is
a bit-field if the left operand is a bit-field. In all cases, the assignment is
sequenced after the value computation of the right and left operands, and
before the value computation of the assignment expression. With respect to an
indeterminately-sequenced function call, the operation of a compound assignment
is a single evaluation. [ Note: Therefore, a function call shall not intervene
between the lvalue-to-rvalue conversion and the side effect associated with any
single compound assignment operator. ?end note ]

All agree that this governs; the question is what "the value computation of the
right and left operands" of the |= assignment means.

In one interpretation, it means the rvalue evaluation of b and f(); in this
interpretation the ordering of the operand evaluations is undefined and hence
the value of b at and after the assignment is also undefined. It is unclear how
the assignment receives an lvalue to assign to in this interpretation.

In the second interpretation, it means the lvalue evaluation of b and the
rvalue evaluation of f(); in this interpretation the lvalue-to-rvalue
conversion of b takes place within the assignment, so consequently f() must be
evaluated before the conversion takes place. Hence, the assignment receives the
(lvalue) b with a well-defined rvalue of true, and the value after the
assignment is also well-defined and true.

Whichever of these interpretations you had intended, the extent of the
disagreement suggests that the proposed wording, while fine for simple
assignment, might be clarified for the compound assignment operators.

Respectfully,

Ivan Godard


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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45437