On Tuesday, 4 June 2024 at 16:58:50 UTC, Basile B. wrote:
question in the header, code in the body, execute on a X86 or
X86_64 CPU
```d
module test;
void setIt(ref bool b) @safe
{
b = false;
}
void main(string[] args)
{
ushort a = 0b1111111111111111;
bool* b = cast(bool*)&a;
setIt(*b);
assert(a == 0b1111111100000000); // what actually happens
assert(a == 0b1111111111111110); // what would be safe
}
```
I understand that the notion of `bool` doesn't exist on X86,
hence what will be used is rather an instruction that write on
the lower 8 bits, but with a 7 bits corruption.
Do I corrupt memory here or not ?
I don't think so. You passed an address to a bool, which uses 8
bits of space, even though the compiler treats it as a 1-bit
integer.
In order for your code to do what you expect, all bool writes
would have to be read/modify/write operations. I don't think
anyone would prefer this.
Is that a safety violation ?
No, you are not writing to memory you don't have access to. An
address is pointing at a byte level, not a bit level.
-Steve
