Michael Veksler <[EMAIL PROTECTED]> writes:
| 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.
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.
-- Gaby
