What exactly does "@safe" mean?

[Peter Alexander]

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- at 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?