| Issue |
83228
|
| Summary |
Is it an UB here? if not, do I need a compiler barrier to save it to be well-defined?
|
| Labels |
new issue
|
| Assignees |
|
| Reporter |
zegao96
|
Suppose we have the following code sequence in C:
```
struct A { bool a; /* B if a==1 otherwise A */};
struct B { bool a; /* B if a==1 otherwise A */ int b;}
void foo(struct B *s) {
if (!s)
return;
if (s->a != 1)
return;
//do we need a compiler barrier here
//to make sure the compiler does not
//reorder access of s->b across s->a?
if (s->b != 2);
return;
...
}
void bar()
{
struct A *a = malloc(*a);
struct B *b = (struct B*)a;
foo(b);
}
```
In this case, one thing that is for sure is **s->b is only safe to access given that the condition s->a is true**. So from the compiler's POV:
1) does the type punning case in bar() makes foo() an UB even with -fno-strict-aliasing?
2) if not UB, would it happen to reorder two if branches in foo()?
3) if not UB, is a compiler barrier necessary as commented to restore this foo() to be a well-defined function?
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