问题
How can the operator()
of a lambda be declared as noreturn
?
Ideone accepts the following code:
#include <cstdlib>
int main() {
[]() [[noreturn]] { std::exit(1); }();
return 0;
}
Clang 3.5 rejects it with:
error: 'noreturn' attribute cannot be applied to types
You can try it in godbolt: http://goo.gl/vsuCsF
Which one is right?
Update: the relevant standard sections appear to be 5.1.2.5, 7.6.3, 7.6.4 but after reading does it still isn't 100% clear to me (i) what is the right behavior, (ii) how to mark the operator() of a lambda as noreturn
.
回答1:
Clang is correct. An attribute can appertain to a function being declared, or to its type; the two are different. [[noreturn]]
must appertain to the function itself. The difference can be seen in
// [[noreturn]] appertains to the entity that's being declared
void f [[noreturn]] (); // §8.3 [dcl.meaning]/p1:
// The optional attribute-specifier-seq following a
// declarator-id appertains to the entity that is declared."
[[noreturn]] void h (); // §7 [dcl.dcl]/p2:
// "The attribute-specifier-seq in a simple-declaration
// appertains to each of the entities declared by
// the declarators of the init-declarator-list."
// ill-formed - [[noreturn]] appertains to the type (§8.3.5 [dcl.fct]/p1:
// "The optional attribute-specifier-seq appertains to the function type.")
void g () [[noreturn]] {}
Indeed if you compile this in g++ it tells you that
warning: attribute ignored [-Wattributes]
void g () [[noreturn]] {}
^
note: an attribute that appertains to a type-specifier is ignored
Note that it doesn't emit a warning that g()
actually does return.
Since an "attribute-specifier-seq in the lambda-declarator appertains to the type of the corresponding function call operator or operator template" (§5.1.2 [expr.prim.lambda]/p5) rather than to that operator/operator template itself, you can't use [[noreturn]]
there. More generally, the language provides no way for you to apply an attribute to the operator ()
of a lambda itself.
回答2:
So a lambda delcarator has the following grammar the draft C++ standard section 5.1.2
Lambda expressions:
( parameter-declaration-clause ) mutableopt exception-specificationopt attribute-specifier-seqopt trailing-return-typeopt
and the noreturn attribute is indeed a valid attribute-specifier-seq so from a grammar perspective I don't see a restriction from section 7.6.3
Noreturn attribute it says (emphasis mine going forward):
[...]The attribute may be applied to the declarator-id in a function declaration.[...]
which does not seem to forbid your use but it does suggest that it is not allowed. If we look at section 7.6.4
Carries dependency attribute it says:
[...]The attribute may be applied to the declarator-id of a parameter-declaration in a function declaration or lambda[...]
the fact that it explicitly includes the lamda case strongly indicates that section 7.6.3
is meant to exclude lambdas and therefore clang
would be correct. As a side note Visual Studio also rejects this code.
回答3:
[C++11: 7.6.3/1]:
The attribute-tokennoreturn
specifies that a function does not return. It shall appear at most once in each attribute-list and no attribute-argument-clause shall be present. The attribute may be applied to the declarator-id in a function declaration. The first declaration of a function shall specify thenoreturn
attribute if any declaration of that function specifies thenoreturn
attribute. If a function is declared with thenoreturn
attribute in one translation unit and the same function is declared without thenoreturn
attribute in another translation unit, the program is ill-formed; no diagnostic required.
I concede that this wording, as is, doesn't prohibit the attribute from appearing elsewhere, but in concert with seeing no evidence anywhere in the standard for it, I don't think this is intended to work with lambda declarations.
Therefore, Clang would be correct.
It may or may not be telling that there was a patch proposal to Clang to allow GCC-style noreturn attributes on lambdas, but not the standard form.
Unfortunately, this feature is not included in GCC's list of extensions, so I can't really see exactly what's going on here.
来源:https://stackoverflow.com/questions/26888805/how-to-declare-a-lambdas-operator-as-noreturn