Re: Exception Specification Compromise
On Feb 13, 3:39 am, "Nevin :-] Liber" <ne...@eviloverlord.com> wrote:
In article
<6bb964cd-04bb-49b0-b344-a17a2e943...@b7g2000yqd.googlegroups.com>,
ThosRTanner <ttann...@bloomberg.net> wrote:
On Feb 12,> It's easy to do static checking of throw specifications at compile
time because the compiler knows the throw spec of all the called
functions, because it has to see the prototype before it can compile
the call.
Again, I just don't see how you can analyze this without whole program
analysis, which isn't trivial.
GIven the following:
struct Foo
{
virtual void Bar() const = 0;
};
bool DoIThrow(Foo const& f)
{
try { f.Bar(); return false; }
catch (...) { return true; }
}
So, what does DoIThrow() return: true or false?
Personally, I don't know how to do the analysis w/o looking at *every*
derived class of DoIThrow. What is your "easy" way to figure this out?
And this is still a fairly simple case...
It's not really necessary to know. From the throw specs, it can throw
anything. As far as the compilation goes, that's all that you need to
know.
So this:
void wibble() throw(std::exception)
{
Foo f;
DoIThrow(f);
}
can terminate your program. What further analysis is necessary?
I don't care what it throws. All I care is that it can throw something
that wibble() says it can't throw.
It's exactly the same as const:
void fiddle(char *f);
char const *s = "abd";
fiddle(s);
won't compile. Not that you know whether or not fiddle changes s. The
function prototype says it can. And so it won't compile, and a good
thing too.
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