Consider these two possible definitions for a class:
Exhibit A:
struct A
{
A() = delete;
};
Exhibit A′:
struct A
{
A() noexcept = delete;
}
Is there any point in declaring a deleted function as noexcept
?
Consider these two possible definitions for a class:
Exhibit A:
struct A
{
A() = delete;
};
Exhibit A′:
struct A
{
A() noexcept = delete;
}
Is there any point in declaring a deleted function as noexcept
?
(Posted this initially as a comment, but encouraged to post as an answer.)
Simply, no. A function that is deleted cannot be called (or, in the case of a constructor, used to initialise an object) let alone throw an exception.
Edit:
hvd mentioned in comments below that noexcept(f())
does not call f()
. If the constructor of class A
is delete
d, then noexcept(A())
will fail to compile, regardless of whether the constructor is declared noexcept
. This is (essentially) a consequence of the requirement that noexcept(expression)
be given a valid expression - and an expression A()
for a class A
requires a valid constructor.
Revolver_Ocelot also correctly points out that it is not possible to overload on noexcept
(i.e. it is not possible to have two functions with the same signature, except that one is noexcept
and one isn't). So, within a definition of class A
, both A() = delete
and A() noexcept = delete
both have the same effect i.e. class A
not having a non-argument constructor.