On Saturday, 1 June 2013 at 21:02:44 UTC, Jonathan M Davis wrote:
@safe is for memory safety, meaning that @safe code cannot
corrupt memory. You
can get segfaults due to null pointers and the like, but you
can't have code
which writes passed the end of a buffer, or which uses a freed
memory, or does
anything else which involves writing or reading from memory
which variables
aren't supposed to have access to.
Not true.
void foo(int* p) @safe
{
*p = 0;
}
void main()
{
int[3] buf1 = [1, 2, 3];
int[1] buf2;
int* p = buf2.ptr;
--p;
foo(p);
import std.stdio;
writeln(buf1);
}
For me, this prints [1, 2, 0]. You could easily come up with an
example which writes to freed memory.
You can argue that foo didn't "cause" this problem (the undefined
behaviour from the pointer arithmetic in main did), but that's
irrelevant: what guarantees do I have when I call a @safe
function that I don't have with any non-@safe function?
Do @safe functions only provide guarantees when the inputs are
valid, or is it the case the @safe functions are guaranteed to
not *introduce* any new undefined behaviour?
