Re: std::terminate() and std::unexpected()
marlow.andrew@googlemail.com wrote:
wasti.r...@gmx.net wrote:
Consider this snippet:
------------------------
void fn1() { std::cerr << "fn1\n"; std::terminate(); }
void fn2() { std::cerr << "fn2\n"; std::terminate(); }
void foo() throw ()
{
try {
std::set_unexpected(&fn1);
throw 0;
} catch(...) {
std::set_unexpected(&fn2);
throw;
}
}
------------------------
Should this program print fn1 or fn2?
fn1. It seems like an open-and-shut case to me, unless I missing
something. when flow of control reaches 'throw 0' it has to call the
fn1 unexpected handler because that is the handler set up via the
previous line. fn2 wont be setup until it enters the catch(...) and it
wont do that because it calls fn1 first.
To, the case is not so open-and-shut, because the initial exception is
being caught within the function (and then re-thrown).
If you think that does not matter, do you also think that
std::unexpected must be called for this function:
void foo() throw ()
{
try {
std::set_unexpected(&fn1);
throw 0;
} catch(...) {
std::set_unexpected(&fn2);
if (std::rand() == 1)
{
throw;
}
}
}
GCC currently does not reload the handlers on rethrow, i.e. the above
program prints fn1, but I think this is a bug.
From my reading of the standard, I can't find any conclusive evidence
that favours one option over the other. I regard the behaviour to be
unspecified.
Bart v Ingen Schenau
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