When C++14 lifted restrictions on constexpr it seemed to include the following (copied from Wikipedia):
Expressions may change the value of an object if the lifetime of that
object began within the constant expression function. This includes
calls to any non-const constexpr-declared non-static member functions.
That seems to imply that you could create an object using new and as long as you delete it within the expression, then it would be allowed.
Language lawyer answer. All references to N3797.
7.1.5/5 states:
For a non-template, non-defaulted constexpr function or a non-template, non-defaulted, non-inheriting constexpr constructor, if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression (5.19), the program is ill-formed; no diagnostic required.
Jumping over to 5.19, we see:
A conditional-expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine (1.9), would evaluate one of the following expressions:
... [lots of bullets]...
a new-expression (5.3.4);
a delete-expression (5.3.5);
... [lots more bullets]...
So no: a program containing a constexpr function with an unconditional invocation of new or delete in it is ill-formed, no diagnostic required. (I'd be surprised, however, if any half-decent compiler failed to diagnose such invocations of new or delete in a constexpr function, required or not.)
I don't think so. You are still limited to calling other constexpr functions. new is actually a function call to operator new() which is not a constexpr function. The same goes for delete.
Related
What std::is_constructible shall return for an aggregate type that does not allow creation of objects due to invalid initializer of a member field?
Consider for example
#include <type_traits>
template <class T>
struct A {
T x{};
};
static_assert( !std::is_constructible_v<A<int&>> );
A<int&> obj; is not well-formed since it cannot initialize int& from {}. So I would expect that the example program above compiles fine, as it does in GCC. But MSVC accepts the opposite statement static_assert( std::is_constructible_v<A<int&>> ); presumable since the default constructor of A was not formally deleted. And Clang behaves in the third way stopping the compilation with the error:
error: non-const lvalue reference to type 'int' cannot bind to an initializer list temporary
T x{};
^~
: note: in instantiation of default member initializer 'A<int &>::x' requested here
struct A {
^
Online demo: https://gcc.godbolt.org/z/nnxcGn7WG
Which one of the behaviors is correct according to the standard?
std::is_constructible_v<A<int&>> checks whether a variable initialization with () initializer is possible. That's never aggregate initialization (not even in C++20), but always value initialization, which will use the default constructor.
Because your class doesn't declare any constructor, it still has an implicit default constructor declared. That one will be called during value initialization.
The implicit default constructor is also not defined as deleted, because none of the points in [class.default.ctor]/2 apply. There is one point for reference members without any default member initializer though.
So this constructor will be chosen in overload resolution and therefore std::is_constructible_v<A<int&>> is true. That the instantiation of the default constructor might be ill-formed is irrelevant. Only declarations are checked (the immediate context). So GCC's behavior is not correct.
The only remaining question now is whether the implicit default constructor (or rather the default member initializer?) is supposed to be instantiated from the std::is_constructible test itself, causing the program to be ill-formed.
As far as I can tell the standard doesn't specify that clearly. The standard says that the noexcept-specifier of a function is implicitly instantiated when it is needed. (see [temp.inst]/15)
It also says that it is needed if it is used in an unevaluated operand in a way that would be an ODR use if it was potentially-evaluated. (see [except.spec]/13.2
Arguably the type trait has to make such a use.
Then [except.spec]/7.3 specifies that it needs to be checked whether the default member initializers are potentially-throwing in order to check whether the implicit default constructor has a potentially-throwing exception specification.
Clang seems to then follow the idea that this requires instantiation of the default member initializer and therefore causes the compilation error because that instantiation is invalid.
The problems I see with that is:
I don't see anything in the [temp.inst] about when default member initializers are instantiated or what that would mean exactly.
[temp.inst]/15 speaks of the noexcept-specifier grammar construct and since the default constructor is implicit, that doesn't really work out.
The standard states:
Only the validity of the immediate context of the variable initialization is considered.
and
The evaluation of the initialization can result in side effects such as the instantiation of class template specializations and function template specializations, the generation of implicitly-defined functions, and so on.
Such side effects are not in the “immediate context” and can result in the program being ill-formed.
It is the generation of an the definition for an implicity-defined function (the constructor) that fails in your example. So this is not the "immediate context". That would allow:
The msvc interpretation: It doesn't consider the validity of the non-immediate context
The clang interpretation: The side effects result in your program being ill-formed
If anything it seems to me that gcc's result might be incorrect. But I'm not sure if the standard wants to forbid the consideration of the "non-immediate context" or if it simply doesn't require it.
cppreference said the following about the body of a constexpr function:
the function body must not contain:
a definition of a variable of non-literal type
a definition of a variable of static or thread storage duration.
All I understood about a constexpr function is that the statements in its body should be evaluated at compile-time so that the call expression can be evaluated at compile-time. Am I true?
Since C++14, the standard allows the body to contain variable definitions of literal types. So what about those definitions, are they evaluated at compile-time or runtime? since non-constexpr variables are allowed also.
What I think I misunderstood is that the compiler shall be able to evaluate at compile-time every statement in a constexpr function body. Does this true since C++14?
To make my confusion clear, I have this simple example:
// assuming std::is_literal_type<T> is true.
constexpr T f() { T t{}; return t; };
I can't understand how the compile behaves with the above snippet. Variable t is an automatic non-const variable and gets defined at run-time not compile-time; so it basically cannot appear in a constexpr declaration. Does this means, the compiler will never evaluate f() at compile-time because it has a statement, that's T t;, which will be evaluated at runtime only.
My confusion is increased when I have tried:
constexpr T result = f();
and it compiles successfully!. Does this mean f() are evaluated at compile-time? if yes, what about the runtime definition of t?
My second question is about why static-duration variables are not allowed in the body of a constexpr function.
the statements in its body should be evaluated at compile-time
...along at least one possible code path.
Or, as the cpprefenrce page you're quoting puts it,
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
The function can be doing disk I/O if it wants to, as long as there is an if branch that skips it.
Variable t is ... defined at run-time not compile-time
It's defined in both. There is a run-time definition of the function (actual machine code compiled and written to the object file) and also there is a compile-time version of the function, a bunch of AST nodes or some sort of internal compiler representation of them. In a constexpr context, compile-time version will be evaluated and replaced by its result. In runtime context, a function call instruction will be compiled.
If a non-literal class type has no constexpr constructor (it is not constexpr constructible), does a non-static constexpr member function make any sense? I mean if you cannot construct the object at compile time, how would you able to use its member functions?
Anyway, the major compilers don't complain about it, which makes me think it is allowed by the standard.
Nevertheless, you are able to use such constexpr member functions in runtime without any problem. The only question now what is the effect of constexpr in this case, if any. My best guess is that the return value of the constexpr member is being evaluated at compile-time (if possible), so on a run-time call it have to do a simple copy.
Is my guess correct, or is the constexpr specifier absolutely meaningless in this case (i.e. the member function is being evaluated at runtime)?
The premise of your question seems to be that only constexpr functions can be evaluated at compile-time.
This premise is incorrect. The compiler can precompute anything it can figure out a way to do, as long as the exact side result and side-effects are produced (as-if rule).
What constexpr provides is a guarantee that certain expressions will be evaluated at compile-time by every compiler (it's not a "quality of implementation" issue), which makes it possible to use them in contexts where a compile-time value is needed, such as non-type template arguments, operands of case clauses in switch statements, etc.
The specific details around constexpr functions include that there has to be at least one set of arguments (the target instance is an implied argument) such that the constexpr evaluation rules are met. If that isn't true, your program is ill-formed and its runtime behavior is not specified at all, so don't go adding constexpr where it doesn't logically belong.
However, compilers aren't required to diagnose violations of this rule. That means that "major compilers don't complain about it" should not be in any way interpreted as assurance that the code is correct.
Standard's wording, section 7.1.5 (draft n4582)
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.
This is probably a bit of an unusual question, in that it asks for a fuller explanation of a short answer given to another question and of some aspects of the C++11 Standard related to it.
For ease of reference, I shall sum up the referenced question here. The OP defines a class:
struct Account
{
static constexpr int period = 30;
void foo(const int &) { }
void bar() { foo(period); } //no error?
};
and is wondering why he gets no error about his usage of an in-class initialized static data member (a book mentioned this to be illegal). Johannes Schaub's answer states, that:
This violates the One Definition Rule;
No diagnostics is required.
As much as I rely the source and validity of this answer, I honestly dislike it because I personally find it too cryptic, so I tried to work out a more meaningful answer myself, with only partial success. Relevant seems to be § 9.4.2/4:
"There shall be exactly one definition of a static data member that is odr-used (3.2) in a program; no diagnostic is required" [Emphases are mine]
Which gets me a bit closer to the point. And this is how § 3.2/2 defines an odr-used variable:
"A variable whose name appears as a potentially-evaluated expression is odr-used unless it is an object that satisfies the requirements for appearing in a constant expression (5.19) and the lvalue-to-rvalue conversion (4.1) is immediately applied" [Emphases are mine]
In the OP's question, variable period clearly satisfies the requirements for appearing in a constant expression, being a constexpr variable. So the reason must be certainly found in the second condition: "and the lvalue-to-rvalue conversion (4.1) is immediately applied".
This is where I have troubles interpreting the Standard. What does this second condition actually mean? What are the situations it covers? Does it mean that a static constexpr variable is not odr-used (and therefore can be in-class initialized) if it is returned from a function?
More generally: What are you allowed to do with a static constexpr variable so that you can in-class initialize it?
Does it mean that a static constexpr variable is not odr-used (and
therefore can be in-class initialized) if it is returned from a
function?
Yes.
Essentially, as long as you treat it as a value, rather than an object, then it is not odr-used. Consider that if you pasted in the value, the code would function identically- this is when it is treated as an rvalue. But there are some scenarios where it would not.
There are only a few scenarios where lvalue-to-rvalue conversion is not performed on primitives, and that's reference binding, &obj, and probably a couple others, but it's very few. Remember that, if the compiler gives you a const int& referring to period, then you must be able to take it's address, and furthermore, this address must be the same for each TU. That means, in C++'s horrendous TU system, that there must be one explicit definition.
If it is not odr-used, the compiler can make a copy in each TU, or substitute the value, or whatever it wants, and you can't observe the difference.
You missed a part of the premise. The class definition give above is completely valid, if you also define Account::period somewhere (but without providing an initializer). See the last sentance of 9.4.2/3:
The member shall still be defined in a namespace scope if it is odr-used
(3.2) in the program and the namespace scope definition shall not
contain an initializer.
This entire discussion only applies to static constexpr data members, not to namespace-scope static variables. When static data members are constexpr (rather than simply const) you have to initialize them in the class definition. So your question should really be: What are you allowed to do with a static constexpr data member so that you don't have to provide a definition for it at namespace scope?
From the preceding, the answer is that the member must not be odr-used. And you already found relevant parts of the definition of odr-used. So you can use the member in a context where it is not potentially-evaluated - as an unevaluated operand (for example of sizeofor decltype) or a subexpression thereof. And you can use it in an expression where the lvalue-to-rvalue conversion is immediately applied.
So now we are down to What uses of a variable cause an immediate lvalue to rvalue conversion?
Part of that answer is in §5/8:
Whenever a glvalue expression appears as an operand of an operator
that expects a prvalue for that operand, the lvalue-to-rvalue (4.1),
array-to-pointer (4.2), or function-to-pointer (4.3) standard
conversions are applied to convert the expression to a prvalue.
For arithmetic types that essentially applies to all operators that apply standard arithmetic conversions. So you can use the member in various arithmetic and logical operations without needing a definition.
I can't enumerate all things, you can or can't do here, because the requirements that something be a [g]lvalue or [p]rvalue are spread across the standard. The rule of thumb is: if only the value of the variable is used, a lvalue to rvalue conversion is applied; if the variable is used as an object, it is used as lvalue.
You can't use it in contexts where an lvalue is explicitly required, for example as argument to the address-of operator or mutating operators). Direct binding of lvalue references (without conversion) is such a context.
Some more examples (without detailed standardese analysis):
You can pass it to functions, unless the function parameter is a reference to which the variable can be directly bound.
For returning it from a function: if implicit conversion to the return type involves a user-defined conversion function, the rules for passing to a function apply. Otherwise you can return it, unless the function returns an lvalue (a reference) referring directly to the variable.
The key rule for odr-used variables, the "One Definition Rule" is in 3.2/3:
Every program shall contain exactly one definition of every non-inline
function or variable that is odr-used in that program; no diagnostic
required.
The "no diagnostic required" part means that programs violating this rule cause undefined behavior, which may range from failing to compile, compiling and failing in surprising ways to compiling and acting as if everything was OK. And your compiler need not warn you about the problem.
The reason, as others have already indicated, is that many of these violations would only be detected by a linker. But optimizations may have removed references to objects, so that no cause for linkage failure remains or else linking may sometimes occur only at runtime or be defined to pick an arbitrary instance from multiple definitions of a name.
Just because a function (or constructor)...
is declared constexpr and
the function definition meets the constexpr requirements
...doesn't mean that the compiler will evaluate the constexpr function during translation. I've been looking through the C++11 FDIS (N3242, available at http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2011/) to try and determine two things:
When is the compiler obligated to evaluate a constexpr function during translation?
When is the compiler allowed to evaluate a constexpr function during translation?
Section 5.19 Paragraph 1 says that constant expressions can be evaluated during translation. As far as I can comprehend, the remainder of Section 5.19 sets forth the rules for what is valid in the definition of a constexpr function.
I understand that I can force constexpr evaluation during translation by declaring the result of the constexpr function as constexpr. Like this:
// Declaration
constexpr double eulers_num() { return 2.718281828459045235360287471; }
// Forced evaluation during translation
constexpr double twoEulers = eulers_num() * 2.0;
static_assert(twoEulers > 5.0, "Yipes!");
So far I've been unable to find the paragraphs in the FDIS that:
Force twoEulers to be evaluated during translation or
Specify other situations when the compiler may or must evaluate a constexpr function during translation.
What I'm particularly interested in discovering is whether constexpr evaluation during translation is triggered by:
When all parameters passed to the constexpr function are literals, or
The implied object argument during overload resolution (Section 13.3.1 Paragraph 3) is either constexpr or requires a literal (such as for array dimensions), or
Something else entirely.
Please, if possible, in your responses cite sections of the FDIS that I can look up or key phrases I can search in the FDIS. The English in the standard is somewhat obtuse, so I may have been reading the relevant paragraphs and have entirely missed their meaning or intent.
It is "allowed" to evaluate the constexpr call at compile time whenever it is actually possible to do so. Remember that the specification operates under the "as if" rule. So if you can't tell the difference, the compiler can do whatever it wants.
The compiler is required to evaluate constexpr calls at compile time when it actually needs the answer at compile time. For example:
constexpr int foo() {return 5;}
std::array<float, foo()> arr;
The compiler needs to know the array size at compile time. Therefore, it must evaluate the constant expression at compile time. If the constexpr function cannot be executed at compile time, you get a compile-time error.
Nicol Bolas is 100% correct, but there is one other interesting aspect: whether the expression is evaluated at translation-time and whether it is evaluated at run-time are completely independent questions. Since the constant expression cannot have side-effects, it can be evaluated an arbitrary number of times, and nothing stops it from being evaluated at both translation-time and run-time.
Suppose the constant expression were a large array (not a std::array, just an array), which is entirely legal, and the program does not indicate that it has static storage. Suppose also that only element 7 of the array is used in a context in which compile-time computation is necessary. It is quite reasonable for the compiler to compute the entire array, use element 7, discard it, and insert code to compute it at run-time in the scope in which it is used, rather than bloating the binary with the whole computed array. I believe this is not a theoretical issue; I've observed it with various compilers in various contexts. constexpr does not imply static.
Of course, if the compiler can determine that the array is not used at runtime, it might not even insert code to compute it, but that's yet another issue.
If you do use such an object at run-time, and you want to indicate to the compiler that it would be worth keeping it around for the duration of the program, you should declare it as static.
By combing the FDIS I have found three places that specify where a constexpr expression must be evaluated during translation.
Section 3.6.2 Initialization of non-local variables, paragraph 2 says if an object with static or thread local storage duration is initialized with a constexpr constructor then the constructor is evaluated during translation:
Constant initialization is performed:
if an object with static or thread storage duration is initialized by a constructor call, if the constructor is a constexpr constructor, if all constructor arguments are constant expressions (including conversions), and if, after function invocation substitution (7.1.5), every constructor call and full-expression in the mem-initializers is a constant expression;
Section 7.1.5 The constexpr specifier, paragraph 9 says if an object declaration includes the constexpr specifier, that object is evaluated during translation (i.e., is a literal):
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, 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’ve heard people argue that this paragraph leaves room for an implementation to postpone the initialization until runtime unless the effect can be detected during translation due to, say, a static_assert. That is probably not an accurate view because whether a value is initialized during translation is, under some circumstances, observable. This view is reinforced by Section 5.19 Constant expressions paragraph 4:
[ Note: Although in some contexts constant expressions must be evaluated during program translation, others may be evaluated during program execution. Since this International Standard imposes no restrictions on the accuracy of floating-point operations, it is unspecified whether the evaluation of a floating-point expression during translation yields the same result as the evaluation of the same expression (or the same operations on the same values) during program execution... — end note ]
Section 9.4.2 Static data members, paragraph 3 says if a const static data member of literal type is initialized by a constexpr function or constructor, then that function or constructor must be evaluated during translation:
If a static data member is of const literal type, its declaration in the class definition can specify a brace-orequal-initializer in which every initializer-clause that is an assignment-expression is a constant expression. A static data member of literal type can be declared in the class definition with the constexpr specifier; if so, its declaration shall specify a brace-or-equal-initializer in which every initializer-clause that is an assignment-expression is a constant expression. [ Note: In both these cases, the member may appear in constant expressions. — end note ]
Interestingly, I did not find anything in the FDIS that required a constexpr expression to be evaluated if its result is used as an array dimension. I'm quite sure the standard committee expects that to be the case. But I also could have missed that in my search.
Outside of those circumstances the C++11 standard allows computations in constexpr functions and constructors to be performed during translation. But it does not require it. The computations could occur at runtime. Which computations the compiler performs during translation are, to a certain extent, a quality of implementation question.
In all three of the situations I located, the trigger for translation-time evaluation is based on the requirements of the target using the result of the constexpr call. Whether or not the arguments to the constexpr function are literal is never considered (although it is a pre-requisite for valid evaluation).
So, to get to the real point of this, it appears that constexpr evaluation during translation is triggered by:
The implied object argument during overload resolution (Section 13.3.1 Paragraph 3) is either constexpr or requires a literal.
I hope that's helpful to someone besides me. Thanks to everyone who contributed.
I don't think it's forced anywhere. I had a look too, it's tricky because there's not one paper on constexpr in that list; they all seem to add/remove from the previous collection of papers.
I think the general idea is when the inputs to the constexpr function are constexpr themselves, it'll all be done at compile time; and by extension non-function constexpr statements, which are literal anyway will be run at compile time if you're using a half intelligent compiler.
If a constexpr function or constructor is called with arguments which
aren't constant expressions, the call behaves as if the function were
not constexpr, and the resulting value is not a constant expression.
from wikipedia
which in seem to get the info from this pdf:
constexpr functions: A constexpr function is one which is “suf-
ficiently simple” so that it delivers a constant expression when
called with arguments that are constant values (see §2.1).