Alexandre Oliva <[EMAIL PROTECTED]> writes:
| On Jan 26, 2006, Gabriel Dos Reis <[EMAIL PROTECTED]> wrote:
|
| > I don't see anything in the ISO C standard that implies that.
|
| > This
|
| > x.v1 = 384;
| > x.v2 = 94.08;
| > int v = x.v2;
| > x.v1 = v;
|
| > is valid fragment.
|
| But can you see anything in it that makes it undefined?
Excessive snippage is a perilious excercise. In this specific case,
your leaving out the context misled you. I was specifically replying
to Michael Veksler's assertion:
# So, is union is a very useful feature in ISO C, without
# gcc's extension? It seems that the only legal use of union
# is to use the same type through the whole life of the object.
#
# Here is the rationale:
#
# Quoting Richard Guenther <[EMAIL PROTECTED]>:
# > On 1/25/06, Alexandre Oliva <[EMAIL PROTECTED]> wrote:
# > > On Jan 22, 2006, Richard Guenther <[EMAIL PROTECTED]> wrote:
# > >
# [...]
# > >
# > > > int ii; double dd; void foo (int *ip, double *dp) {
# > > > *ip = 15; ii = *ip; *dp = 1.5; dd = *dp; }
# > > > void test (void) { union { int i; double d; } u;
# > > > foo (&u.i, &u.d); }
# > >
# > > So it is perfectly valid, but if GCC reorders the read from *ip past
# > > the store to *dp, it turns the valid program into one that misbehaves.
# >
# > *ip = 15; ii = *ip; *dp = 1.5; dd = *dp;
# > Here ^^^
# > you are accessing memory of type integer as type double. And gcc will
# > happily reorder the read from *ip with the store to *dp based on TBAA
# > unless it inlines the function and applies the "special" gcc rules about
# > unions.
# > So this example is invalid, too.
#
# So in theory, if there is a union of two non-char types:
# union { T1 v1, T2 v2} x;
# it is illegal to access both x.v1 and x.v2 for the same variable x
# anywhere in the whole program.
which I happen not to find any support for in the C standard.
| Failing that, regular assignment and access rules apply, and so it is
| valid.
I could not agree more; and in fact, that was my point.
-- Gaby
