D
Dave
Is it possible for the nothrow form of the new operator to ever throw?
D
Dave
Dave said:
Oh Lord, what on Earth is wrong with me...
Is it possible for *PLACEMENT* new to ever throw???
Somebody shoot me!!!
R
Ron Natalie
Dave said:
No...it has a throw() exception specification on it. If something inside it were to
throw, terminate() would be called.
R
Ron Natalie
Dave said:
Sure, the constructor for the object can throw. But the placment allocation
function can not throw. It's defined with a throw() exception specification.
Terminate() would be called even if the the allocation function was replaced
with something that would throw.
Bang.
H
Howard Hinnant
Dave said:
<chuckle> Actually your original question has interesting answers. Just
goes to show: there are no dumb questions!
Yes, the nothrow form of the new operator can throw. Well, that's
probably a bit too inflamatory. A new-nothrow expression can throw.
Consider:
struct X
{
X() {throw "up";}
};
void foo()
{
X* xp = new (std::nothrow) X;
}
The new expression in foo() will throw, even though it is using the
nothrow expression. Same deal for placement new.
-Howard