#include <exception>
constexpr bool foo(bool x)
{
return x ? true : throw std::exception();
}
int main()
{
// 1) must never be compiled
// static_assert(foo(false), "");
// 2) must always be compiled?
const bool x = foo(false);
// 3) must never compile?
constexpr bool y = foo(false);
return 0;
}
I'm sure that (1) must lead to a compile error. I'm quite sure that (2) must not be rejected at compile time, though it will fail at runtime.
The interesting case is the constexpr variable (3). In this simple example, gcc and clang actually evaluate the expression, and will therefore reject the program. (Error message: y is not a constant expression).
Is every C++11 compiler forced to reject the program? What if foo(false) was replaced by a more complex expression?
I was suprised to find out that constexpr were not turing-complete, though it will be after a change in the specification:
Is constexpr-based computation Turing complete?
Maybe this is related to my question. As far as I understand, the compiler is allowed to postpone the actual evaluation of the constexpr (3) in this example until runtime. But if constexpr are turing-complete, I find it hard to believe that the compiler can decide for all constexpr whether an exception will be thrown (which means that the constexpr is invalid).
By my reading, yes, every compiler must complain about statement (3).
N3242 7.1.5 paragraph 9:
A constexpr specifier used in an object declaration declares the object as const. Such an object shall have literal type and shall be initialized. If it initialized by a constructor call, that call shall be a constant expression (5.19). Otherwise, every full-expression that appears in its initializer shall be a constant expression. Each implicit conversion used in converting the initializer expressions and each constructor call used for the initialization shall be one of those allowed in a constant expression (5.19).
I think of a constexpr object as "evaluated at compile time", and a constexpr function or constexpr constructor as "might be evaluated at compile time". A compiler must determine the semantic validity of statements like (3) at compile time. You could argue that the "evaluation" can still be done at run time, but checking for validity does most of that work anyway. Plus, the code could then continue to instantiate a template like Check<y>, which pretty much guarantees the compiler needs to figure out the value of y at compile-time.
This does mean you could write a diabolical program to make the compiler take a really long or infinite time. But I suspect that was already possible with operator-> tricks.
I'm sure that (1) must lead to a compile error. I'm quite sure that (2) must not be rejected at compile time, though it will fail at runtime.
Correct. The throw part of the conditional operator is not a constant expression, and in (1), it's not unevaluated. For (2), foo is not forced to be evaluated at compile-time.
For (3), how would the compiler be allowed to post-pone evaluation? The constexpr decl-specifier forces foo to be evaluated at compile-time. It's basically the same as (1), initialization of y is a context where a constant expression is required.
§7.1.6 [dcl.constexpr] p9
A constexpr specifier used in an object declaration declares the object as const. Such an object shall have literal type and shall be initialized. If it is initialized by a constructor call, that call shall be a constant expression (5.19). Otherwise, or if a constexpr specifier is used in a reference declaration, every full-expression that appears in its initializer shall be a constant expression. [...]
Related
Recently I was surprised that the following code compiles in clang, gcc and msvc too (at least with their current versions).
struct A {
static const int value = 42;
};
constexpr int f(A a) { return a.value; }
void g() {
A a; // Intentionally non-constexpr.
constexpr int kInt = f(a);
}
My understanding was that the call to f is not constexpr because the argument i isn't, but it seems I am wrong. Is this a proper standard-supported code or some kind of compiler extension?
As mentioned in the comments, the rules for constant expressions do not generally require that every variable mentioned in the expression and whose lifetime began outside the expression evaluation is constexpr.
There is a (long) list of requirements that when not satisfied prevent an expression from being a constant expression. As long as none of them is violated, the expression is a constant expression.
The requirement that a used variable/object be constexpr is formally known as the object being usable in constant expressions (although the exact definition contains more detailed requirements and exceptions, see also linked cppreference page).
Looking at the list you can see that this property is required only in certain situations, namely only for variables/objects whose lifetime began outside the expression and if either a virtual function call is performed on it, a lvalue-to-rvalue conversion is performed on it or it is a reference variable named in the expression.
Neither of these cases apply here. There are no virtual functions involved and a is not a reference variable. Typically the lvalue-to-rvalue conversion causes the requirement to become important. An lvalue-to-rvalue conversions happens whenever you try to use the value stored in the object or one of its subobjects. However A is an empty class without any state and therefore there is no value to read. When passing a to the function, the implicit copy constructor is called to construct the parameter of f, but because the class is empty, it doesn't actually do anything. It doesn't access any state of a.
Note that, as mentioned above, the rules are stricter if you use references, e.g.
A a;
A& ar = a;
constexpr int kInt = f(ar);
will fail, because ar names a reference variable which is not usable in constant expressions. This will hopefully be fixed soon to be more consistent. (see https://github.com/cplusplus/papers/issues/973)
struct A {
consteval A() {};
};
constexpr bool g() {
auto a = new A;
delete a;
return true;
}
int main() {
static_assert(g());
}
https://godbolt.org/z/jsq35WxKs
GCC and MSVC reject the program, ICC and Clang accept it:
///MSVC:
<source>(6): error C7595: 'A::A': call to immediate function is not a constant expression
Compiler returned: 2
//GCC:
<source>: In function 'constexpr bool g()':
<source>:6:18: error: the value of '<anonymous>' is not usable in a constant expression
6 | auto a = new A;
| ^
<source>:6:18: note: '<anonymous>' was not declared 'constexpr'
<source>:7:12: error: type '<type error>' argument given to 'delete', expected pointer
7 | delete a;
| ^
Compiler returned: 1
Although, replacing new A by new A() results in GCC accepting the program as well (but not for new A{} either).
Making at least one of the following changes results in all four compilers accepting the program:
Replace consteval with constexpr
Replace constexpr with consteval
Replace
auto a = new A;
delete a;
with
auto alloc = std::allocator<A>{};
auto a = alloc.allocate(1);
std::construct_at(a);
std::destroy_at(a);
alloc.deallocate(a, 1);
with A a;, with auto&& a = A{}; or with A{};
Only exceptions:
Clang trunk with libstdc++ seems to fail compilation with the std::allocator version seemingly due to an unrelated bug. With Clang 13 or libc++ it is accepted as well.
In file included from <source>:1:
In file included from [...]/memory:78:
[...]/shared_ptr_atomic.h:459:14: error: missing 'typename' prior to dependent type name '_Atomic_count::pointer'
static _Atomic_count::pointer
MSVC rejects the std::allocator version as long as there is consteval on the constructor:
error C7595: 'A::A': call to immediate function is not a constant expression
<source>(10): note: see reference to function template instantiation '_Ty *std::construct_at<_Ty,,void>(_Ty *const ) noexcept(false)' being compiled
with
[
_Ty=A
]
Replacing static_assert(g()); with g() or removing the call completely does not seem to have any impact on these results.
Which compilers are correct and if the original is ill-formed, why is only that particular combination of qualifiers and construction method disallowed?
Motivated by the comments under this answer.
The relevant wording is [expr.const]/13:
An expression or conversion is an immediate invocation if it is a potentially-evaluated explicit or implicit invocation of an immediate function and is not in an immediate function context. An immediate invocation shall be a constant expression.
Note the words 'or conversion' and 'implicit invocation' - this seems to imply that the rule is intended to apply on a per-function-call basis.1 The evaluation of a single atomic expression can consist of multiple such calls, as in the case of e.g. the new-expression which may call an allocation function, a constructor, and a deallocation function. If the selected constructor is consteval, the part of the evaluation of the new-expression that initializes the object (i.e. the constructor call), and only that part, is an immediate invocation. Under this interpretation, using new with a consteval constructor should not be ill-formed regardless of context - even outside of a constant expression - as long as the initialization of the object is itself constant, of course.
There is an issue with this reading, however: the last sentence clearly says that an immediate invocation must be an expression. A 'sub-atomic call' as described above isn't one, it does not have a value category, and could not possibly satisfy the definition of a constant expression ([expr.const]/11):
A constant expression is either a glvalue core constant expression that refers to an entity that is a permitted result of a constant expression (as defined below), or a prvalue core constant expression whose value satisfies the following constraints [...]
A literal interpretation of this wording would preclude any use of a consteval constructor outside of an immediate function context, since a call to it can never appear as a standalone expression. This is clearly not the intended meaning - among other things, it would render parts of the standard library unusable.
A more optimistic (but also less faithful to the words as written) version of this reading is that the atomic expression containing the call (formally: the expression which the call is an immediate subexpression of 2) must be a constant expression. This still doesn't allow your new A construct because it is not a constant expression by itself, and also leaves some uncertainty in cases like initialization of function parameters or variables in general.
I'm inclined to believe that the first reading is the intended one, and that new A should be fine, but clearly there's implementation divergence.
As for the contradictory 'shall be a constant expression' requirement, this isn't the only place in the standard where it appears like this. Earlier in the same section, [expr.const]/2.2:
A variable or temporary object o is constant-initialized if [...]
the full-expression of its initialization is a constant expression when interpreted as a constant-expression [...]
Clearly, the following is supposed to be valid:
constinit A a;
But there's no constant expression in sight.
So, to answer your question:
Whether the call to g is being evaluated as part of a manifestly constant-evaluated expression does not matter3 regardless of which interpretation of [expr.const]/13 you go with. new A is either well-formed even during normal evaluation or ill-formed anywhere outside of an immediate function context.
By the looks of it, Clang and ICC implement the former set of rules while GCC and MSVC adhere to the latter. With the exception of GCC accepting new A() as an outlier (which is clearly a bug), neither are wrong, the wording is just defective.
[1] CWG2410 fixes the wording to properly include things like constructor calls (which are neither expressions nor conversions).
[2] Yes, a non-expression can be a subexpression.
[3] Such a requirement would be impossible to enforce.
In C++17, this code is illegal:
constexpr int foo(int i) {
return std::integral_constant<int, i>::value;
}
That's because even if foo can be evaluated at compile-time, the compiler still needs to produce the instructions to execute it at runtime, thus making the template instantiation impossible.
In C++20 we will have consteval functions, which are required to be evaluated at compile-time, so the runtime constraint should be removed. Does it mean this code will be legal?
consteval int foo(int i) {
return std::integral_constant<int, i>::value;
}
No.
Whatever changes the paper will entail, which is little at this point, it cannot change the fact that a non-template function definition is only typed once. Moreover, if your proposed code would be legal, we could presumably find a way to declare a variable of type std::integral_constant<int, i>, which feels very prohibitive in terms of the ODR.
The paper also indicates that parameters are not intended to be treated as core constant expressions in one of its examples;
consteval int sqrsqr(int n) {
return sqr(sqr(n)); // Not a constant-expression at this point,
} // but that's okay.
In short, function parameters will never be constant expressions, due to possible typing discrepancy.
Does it mean this code will be legal?
consteval int foo(int i) {
return std::integral_constant<int, i>::value;
}
No. This is still ill-formed. While consteval requires the call itself to be a constant expression, so you know that the argument that produces i must be a constant expression, foo itself is still not a template. Template?
A slight variation in your example might make this more obvious:
consteval auto foo(int i) {
return std::integral_constant<int, i>();
}
Were this to be valid, foo(1) and foo(2) would... return different types. This is an entirely different language feature (constexpr function parameters) - because in order for this to work, such functions would really need to behave like templates.
It may seem a little unintuitive. After all, if the argument that produced i was a constant expression, surely i should be usable as one as well? But it's still not - there are no additional exceptions in [expr.const] that permit parameters for immediate functions. An immediate function is still just a function, and its parameters still aren't constant expressions -- in the same way that a normal constexpr function's parameters aren't constant expressions.
Of course with int, we can just rewrite the function to lift the function parameter into a template parameter:
template <int i>
consteval int foo() {
return std::integral_constant<int, i>::value;
}
And C++20 gives us class types as non-type template parameters, so we can actually do this for many more types than we could before. But there are still plenty of types that we could use as a parameter to an immediate function that we cannot use as a template parameter - so this won't always work (e.g. std::optional or, more excitingly in C++20, std::string).
It would seem that this will not be legal in C++20. A good explanation for why this would be problematic to support has already been given in the answers by #Barry and #Columbo (it doesn't really work with the type system). I'll just add what I believe to be the relevant quotes from the standard here that actually make this illegal.
Based on [temp.arg.nontype]/2
A template-argument for a non-type template-parameter shall be a converted constant expression […]
A converted constant expression is a constant expression that is implicitly converted to a particular type [expr.const]/7 (here, the type of the template parameter). So your question boils down to the question of whether a variable inside a consteval function is a constant expression. Based on [expr.const]/8
A constant expression is either a glvalue core constant expression that refers to an entity that is a permitted result of a constant expression (as defined below), or a prvalue core constant expression whose value satisfies the following constraints: […]
The expression i is a glvalue id-expression that is a core constant expression (because its evaluation does not do any of the things listed in [expr.const]/4). However, the entity this core constant expression refers to is not a permitted result of a constant expression [expr.const]/8:
An entity is a permitted result of a constant expression if it is an object with static storage duration that either is not a temporary object or is a temporary object whose value satisfies the above constraints, or if it is a non-immediate function.
The object in question is neither of static storage duration nor is it a temporary object…
According to this answer apparently there is no good reason why structured bindings are not allowed to be constexpr, yet the standard still forbids it. In this case, however, shouldn't the use of the structured bindings inside the constexpr function also be prohibited? Consider a simple snippet:
#include <utility>
constexpr int foo(std::pair<int, int> p) {
auto [a, b] = p;
return a;
}
int main() {
constexpr int a = foo({1, 2});
static_assert(a == 1);
}
Both gcc and clang does not cause trouble compiling the code. Is the code ill-formed either way or is this one actually allowed?
In the case of function declaration, the constexpr specifier is an assertion made to the compiler that the function being declared may be evaluated in a constant expression, i.e. an expression that can be evaluated at compile-time. Nevertheless the object initialization inside a declaration does not need to have constexpr inside its declaration specifier to be a constant expression.
Shorter: constexpr function may imply constant expression but constant expression initialization does not need that the associated declaration has a constexpr specifier.
You can check this in the C++ standard [dcl.constexpr]:
A call to a constexpr function produces the same result as a call to an equivalent non-constexpr function in
all respects except that
— a call to a constexpr function can appear in a constant expression[...]
This is the evaluation of an expression that determines if an expression is a
constant expression [expr.const]:
An expression e is a core constant expression unless the evaluation of e [...] would evaluate one of the following expression[...]
A declaration is not an expression, so an initialization of an object being declared is a constant expression irrespective of the presence or not of a constexpr specifier in the declaration.
Finally, in [dcl.constexpr], it is specified that a constexpr function must be such that there exist parameters for which its body can be evaluated as a constant expression:
For a constexpr function or constexpr constructor that is neither defaulted nor a template, if no argument
values exist such that an invocation of the function or constructor could be an evaluated subexpression of
a core constant expression (8.20), or, for a constructor, a constant initializer for some object (6.6.2), the
program is ill-formed, no diagnostic required.
When you declare constexpr int a the compiler expects a to be inialized by a constant expression and the expression foo({1,2}) is a constant expression, so your code is well formed.
PS: Nevertheless, declaration specifiers (static, thread_local=>static) in the the declaration of function local variable implies that the function cannot be declared constexpr.
There are several requirements that a constexpr function must meet. There are some requirements for the body of a constexpr function, and the shown code does not appear to violate any of them. The key point is that there is no requirement that every statement in a function must be a constexpr. The only interesting requirement in question here, is this one:
there exists at least one set of argument values such that an
invocation of the function could be an evaluated subexpression of a
core constant expression (for constructors, use in a constant
initializer is sufficient) (since C++14). No diagnostic is required
for a violation of this bullet.
Note the last sentence. The compiler may, but is not required to, throw a red flag.
The key requirement is merely that there is some assortment of parameter values to the function that results in a constant result from the function (and the function body meets the listed requirements). For example, the function might use a structured binding conditionally; but for some set of parameter values do something else, producing a constant result. This would tick this checkbox for a constexpr function.
But, despite the sophistication of modern C++ compilers, they may not necessarily be capable of reaching this determination in every possible instance, so, in practice, it would be hard to enforce such a requirement, hence the compilers are permitted to just take this for granted.
On this site, it is specified that:
"A constexpr function must satisfy the following requirements:
[...]
there exists at least one set of argument values such that an invocation of the function could be an evaluated subexpression of a core constant expression (for constructors, use in a constant initializer is sufficient) (since C++14). No diagnostic is required for a violation of this bullet."
What is the meaning of the bolded statement?
Looking at the linked defect report
struct X {
std::unique_ptr<int> p;
constexpr X() { }
};
Before C++14, this would be ill-formed due to [dcl.constexpr]
For a constexpr constructor, if no argument values exist such that after function invocation substitution, every constructor call and full-expression in the mem-initializers would be a constant expression (including conversions), the program is ill-formed; no diagnostic required.
Which mandates that there exists some argument (in this case, only the empty set) that can create a constant expression for the invocation of X::X, as in
constexpr X x; // must be valid before C++14
Since std::unique_ptr isn't a literal type, it has a non-trivial destructor, this is impossible. Yet the defect report proposed that constexpr constructors should still be well-formed in such cases due to this kind of use case
X x; // not constexpr, but initialization should be constant
Hence the rewording
For a constexpr function or constexpr constructor that is neither defaulted nor a template, if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression, or, for a constructor, a constant initializer for some object , the program is ill-formed, no diagnostic required.
Translated, it means: a constexpr constructor is well-formed as long as it is a constexpr function, and its member initializations are also constexpr functions, even if the type itself can never be constexpr.