The keyword constexpr enforced pretty tight restrictions on functions on its introduction into the C++11 standard. These restrictions were loosened with C++14 and C++20 (most noteworthy):
C++14 allowed multiple return statements, static_asserts etc.
C++20 allowed try and asm
C++23 further softens these restrictions. From what I can see in cppreference, constexpr for functions seems to only have the following meaning left:
it must not be a coroutine
for constructor and destructor, the class must have no virtual base classes
For constexpr function templates and constexpr member functions of class templates, at least one specialization must satisfy the abovementioned requirements.
C++23 even removed the restriction that a constexpr function must be "evaluatable" at compile time for any type in p2448r2. From my understanding this completely removed the idea of a constexpr function to be evaluated at compile time.
Is that it? If so, how is a constexpr function even useful anymore?
What you actually seem to ask is: why not make anything constexpr by default?
Because you might want others to not use a function at compile-time, to give you a possibility to switch to a non-constexpr implementation later.
Imagine this:
You see a library function, which you'd like to use at compile-time.
Let's say, size_t RequiredBufferSize();. If it happens to be constexpr, you can allocate the buffer on the stack, or something like that.
You're not sure if it's supposed to work at compile-time, because there's no constexpr in our imaginary language.
You try it, and it does work at compile-time. You start using it this way.
Let's say the implementation was {return 42;}, which is constexpr.
A new version of the library is released, the function no longer works at compile-time (e.g. the size is loaded from a config file).
You complain to the developer, and he argues that the function was never intended to work at compile-time, and you relied on an implementation detail.
Related
C++11 allows functions declared with the constexpr specifier to be used in constant expressions such as template arguments. There are stringent requirements about what is allowed to be constexpr; essentially such a function encapsulates only one subexpression and nothing else. (Edit: this is relaxed in C++14 but the question stands.)
Why require the keyword at all? What is gained?
It does help in revealing the intent of an interface, but it doesn't validate that intent, by guaranteeing that a function is usable in constant expressions. After writing a constexpr function, a programmer must still:
Write a test case or otherwise ensure it's actually used in a constant expression.
Document what parameter values are valid in a constant expression context.
Contrary to revealing intent, decorating functions with constexpr may add a false sense of security since tangential syntactic constraints are checked while ignoring the central semantic constraint.
In short: Would there be any undesirable effect on the language if constexpr in function declarations were merely optional? Or would there be any effect at all on any valid program?
Preventing client code expecting more than you're promising
Say I'm writing a library and have a function in there that currently returns a constant:
awesome_lib.hpp:
inline int f() { return 4; }
If constexpr wasn't required, you - as the author of client code - might go away and do something like this:
client_app.cpp:
#include <awesome_lib.hpp>
#include <array>
std::array<int, f()> my_array; // needs CT template arg
int my_c_array[f()]; // needs CT array dimension
Then should I change f() to say return the value from a config file, your client code would break, but I'd have no idea that I'd risked breaking your code. Indeed, it might be only when you have some production issue and go to recompile that you find this additional issue frustrating your rebuilding.
By changing only the implementation of f(), I'd have effectively changed the usage that could be made of the interface.
Instead, C++11 onwards provide constexpr so I can denote that client code can have a reasonable expectation of the function remaining a constexpr, and use it as such. I'm aware of and endorsing such usage as part of my interface. Just as in C++03, the compiler continues to guarantee client code isn't built to depend on other non-constexpr functions to prevent the "unwanted/unknown dependency" scenario above; that's more than documentation - it's compile time enforcement.
It's noteworthy that this continues the C++ trend of offering better alternatives for traditional uses of preprocessor macros (consider #define F 4, and how the client programmer knows whether the lib programmer considers it fair game to change to say #define F config["f"]), with their well-known "evils" such as being outside the language's namespace/class scoping system.
Why isn't there a diagnostic for "obviously" never-const functions?
I think the confusion here is due to constexpr not proactively ensuring there is any set of arguments for which the result is actually compile-time const: rather, it requires the programmer to take responsibility for that (otherwise §7.1.5/5 in the Standard deems the program ill-formed but doesn't require the compiler to issue a diagnostic). Yes, that's unfortunate, but it doesn't remove the above utility of constexpr.
So, perhaps it's helpful to switch from the question "what's the point of constexpr" to consider "why can I compile a constexpr function that can never actually return a const value?".
Answer: because there'd be a need for exhaustive branch analysis that could involve any number of combinations. It could be excessively costly in compile time and/or memory - even beyond the capability of any imaginable hardware - to diagnose. Further, even when it is practical having to diagnose such cases accurately is a whole new can of worms for compiler writers (who have better uses for their time). There would also be implications for the program such as the definition of functions called from within the constexpr function needing to be visible when the validation was performed (and functions that function calls etc.).
Meanwhile, lack of constexpr continues to forbid use as a const value: the strictness is on the sans-constexpr side. That's useful as illustrated above.
Comparison with non-`const` member functions
constexpr prevents int x[f()] while lack of const prevents const X x; x.f(); - they're both ensuring client code doesn't hardcode unwanted dependency
in both cases, you wouldn't want the compiler to determine const[expr]-ness automatically:
you wouldn't want client code to call a member function on a const object when you can already anticipate that function will evolve to modify the observable value, breaking the client code
you wouldn't want a value used as a template parameter or array dimension if you already anticipated it later being determined at runtime
they differ in that the compiler enforces const use of other members within a const member function, but does not enforce a compile-time constant result with constexpr (due to practical compiler limitations)
When I pressed Richard Smith, a Clang author, he explained:
The constexpr keyword does have utility.
It affects when a function template specialization is instantiated (constexpr function template specializations may need to be instantiated if they're called in unevaluated contexts; the same is not true for non-constexpr functions since a call to one can never be part of a constant expression). If we removed the meaning of the keyword, we'd have to instantiate a bunch more specializations early, just in case the call happens to be a constant expression.
It reduces compilation time, by limiting the set of function calls that implementations are required to try evaluating during translation. (This matters for contexts where implementations are required to try constant expression evaluation, but it's not an error if such evaluation fails -- in particular, the initializers of objects of static storage duration.)
This all didn't seem convincing at first, but if you work through the details, things do unravel without constexpr. A function need not be instantiated until it is ODR-used, which essentially means used at runtime. What is special about constexpr functions is that they can violate this rule and require instantiation anyway.
Function instantiation is a recursive procedure. Instantiating a function results in instantiation of the functions and classes it uses, regardless of the arguments to any particular call.
If something went wrong while instantiating this dependency tree (potentially at significant expense), it would be difficult to swallow the error. Furthermore, class template instantiation can have runtime side-effects.
Given an argument-dependent compile-time function call in a function signature, overload resolution may incur instantiation of function definitions merely auxiliary to the ones in the overload set, including the functions that don't even get called. Such instantiations may have side effects including ill-formedness and runtime behavior.
It's a corner case to be sure, but bad things can happen if you don't require people to opt-in to constexpr functions.
We can live without constexpr, but in certain cases it makes the code easier and intuitive.
For example we have a class which declares an array with some reference length:
template<typename T, size_t SIZE>
struct MyArray
{
T a[SIZE];
};
Conventionally you might declare MyArray as:
int a1[100];
MyArray<decltype(*a1), sizeof(a1)/sizeof(decltype(a1[0]))> obj;
Now see how it goes with constexpr:
template<typename T, size_t SIZE>
constexpr
size_t getSize (const T (&a)[SIZE]) { return SIZE; }
int a1[100];
MyArray<decltype(*a1), getSize(a1)> obj;
In short, any function (e.g. getSize(a1)) can be used as template argument only if the compiler recognizes it as constexpr.
constexpr is also used to check the negative logic. It ensures that a given object is at compile time. Here is the reference link e.g.
int i = 5;
const int j = i; // ok, but `j` is not at compile time
constexprt int k = i; // error
Without the keyword, the compiler cannot diagnose mistakes. The compiler would not be able to tell you that the function is an invalid syntactically as aconstexpr. Although you said this provides a "false sense of security", I believe it is better to pick up these errors as early as possible.
The standard says about template constexpr functions/constructors in dcl.constexpr/6:
If the instantiated template specialization of a constexpr function template or member function of a class template would fail to satisfy the requirements for a constexpr function or constexpr constructor, that specialization is still a constexpr function or constexpr constructor, even though a call to such a function cannot appear in a constant expression. If no specialization of the template would satisfy the requirements for a constexpr function or constexpr constructor when considered as a non-template function or constructor, the template is ill-formed, no diagnostic required.
The interesting part is:
fail to satisfy the requirements for a ... constexpr constructor, that specialization is still a ... constexpr constructor
So, even if a constructor is marked with constexpr, it may not be used in a constant expression.
Why does this rule exist? Why isn't constexpr removed, when a function doesn't satisfy the requirements?
The current behavior is bad in two ways:
the non-constexpr-ness isn't caught at the closest possible location, but at the actual constexpr expression, where it is used. So we have to find the offending part, where constexpr silently removed.
an object, which is intended to be statically initialized (because it has a constexpr constructor), will be dynamically initialized without any errors/warnings (because the constructor isn't "really" constexpr).
Does this rule have some pros, which balances the cons of it?
This rule allows you to write a templated constructor/function and mark it as constexpr even when it's not always constexpr (only at least sometimes).
For example, std::pair has constexpr constructors, but it is of course usable outside of constant expressions.
This is quite sensible, because otherwise you would have to duplicate all these functions (once with constexpr and once without), even if the code is exactly the same. Let's not even consider ambiguity.
Since it is generally impossible to prove that a template cannot ever satisfy constexpr, no diagnostic is required for it (but it's ill-formed so compilers can complain to you if they can prove this for a given case).
You are correct that this is not very useful if you want to specify "this function shall only be usable in constant expression", but that's not what this wording is aiming for.
Edit: To clarify, constexpr for functions only means "legal to evaluate inside a constant expression" (more precise wording here), not "can only be evaluated at compile-time". By contrast, constexpr variables must be initialized with a constant expression.
Another edit: We have exact wording to discuss, thanks to #JackAidley!
If the instantiated template specialization of a constexpr function template would fail to satisfy the requirements for a constexpr function,
the constexpr specifier is ignored and the specialization is not a constexpr function.
The problem with this is that "there is at least one set of arguments for which the function can be constant-evaluated" is part of the "requirements for a constexpr function". Therefore, compilers cannot implement this clause, since it is not possible to prove (in general) whether such a set exists for a given function (or a function template instantiation). You either have to muddy this requirement further or give up on this aspect. It seems the committee chose the latter.
In earlier versions of the suggestion for the change to the language, it operated as you suggest:
If the instantiated template specialization of a constexpr function
template would fail to satisfy the requirements for a constexpr
function,the constexprspecifier is ignored and the specialization is
not a constexpr function.
But it was later changed. I was unable to track down any definitive answer to your question but I think that it is reasonable to believe that the answer is that constexpr make other semantic changes to the code and these are retained even though the function is no longer usable in other constexpr statements. If you look at defect report 1358 which includes the change to the current wording, you can see an intermediate form of words that includes the a note about retaining const status regardless.
I also think that while the retention of constexpr status is unintuitive, both of your arguments against it are wrong:
Catching the constexpr when the template instantiation is made goes against how C++ templates usually work - you only get an error when you try and use the template in a way it cannot be used for that type; merely not being able to complete the entire signature is not an error. To introduce special case mechanics for constexpr would be unnecessarily confusing and limit usefulness since you'd now need to write different templates for constexprable and un-constexprable types.
Because it maintains the constexpr specifier the fallback isn't to general runtime dynamic initialisation but to dynamic initialisation at the time at which statics are initialised. Which may cause problems because of the Static Initialisation Order Fiasco but does at least happen before the main() function is entered.
C++11 allows functions declared with the constexpr specifier to be used in constant expressions such as template arguments. There are stringent requirements about what is allowed to be constexpr; essentially such a function encapsulates only one subexpression and nothing else. (Edit: this is relaxed in C++14 but the question stands.)
Why require the keyword at all? What is gained?
It does help in revealing the intent of an interface, but it doesn't validate that intent, by guaranteeing that a function is usable in constant expressions. After writing a constexpr function, a programmer must still:
Write a test case or otherwise ensure it's actually used in a constant expression.
Document what parameter values are valid in a constant expression context.
Contrary to revealing intent, decorating functions with constexpr may add a false sense of security since tangential syntactic constraints are checked while ignoring the central semantic constraint.
In short: Would there be any undesirable effect on the language if constexpr in function declarations were merely optional? Or would there be any effect at all on any valid program?
Preventing client code expecting more than you're promising
Say I'm writing a library and have a function in there that currently returns a constant:
awesome_lib.hpp:
inline int f() { return 4; }
If constexpr wasn't required, you - as the author of client code - might go away and do something like this:
client_app.cpp:
#include <awesome_lib.hpp>
#include <array>
std::array<int, f()> my_array; // needs CT template arg
int my_c_array[f()]; // needs CT array dimension
Then should I change f() to say return the value from a config file, your client code would break, but I'd have no idea that I'd risked breaking your code. Indeed, it might be only when you have some production issue and go to recompile that you find this additional issue frustrating your rebuilding.
By changing only the implementation of f(), I'd have effectively changed the usage that could be made of the interface.
Instead, C++11 onwards provide constexpr so I can denote that client code can have a reasonable expectation of the function remaining a constexpr, and use it as such. I'm aware of and endorsing such usage as part of my interface. Just as in C++03, the compiler continues to guarantee client code isn't built to depend on other non-constexpr functions to prevent the "unwanted/unknown dependency" scenario above; that's more than documentation - it's compile time enforcement.
It's noteworthy that this continues the C++ trend of offering better alternatives for traditional uses of preprocessor macros (consider #define F 4, and how the client programmer knows whether the lib programmer considers it fair game to change to say #define F config["f"]), with their well-known "evils" such as being outside the language's namespace/class scoping system.
Why isn't there a diagnostic for "obviously" never-const functions?
I think the confusion here is due to constexpr not proactively ensuring there is any set of arguments for which the result is actually compile-time const: rather, it requires the programmer to take responsibility for that (otherwise §7.1.5/5 in the Standard deems the program ill-formed but doesn't require the compiler to issue a diagnostic). Yes, that's unfortunate, but it doesn't remove the above utility of constexpr.
So, perhaps it's helpful to switch from the question "what's the point of constexpr" to consider "why can I compile a constexpr function that can never actually return a const value?".
Answer: because there'd be a need for exhaustive branch analysis that could involve any number of combinations. It could be excessively costly in compile time and/or memory - even beyond the capability of any imaginable hardware - to diagnose. Further, even when it is practical having to diagnose such cases accurately is a whole new can of worms for compiler writers (who have better uses for their time). There would also be implications for the program such as the definition of functions called from within the constexpr function needing to be visible when the validation was performed (and functions that function calls etc.).
Meanwhile, lack of constexpr continues to forbid use as a const value: the strictness is on the sans-constexpr side. That's useful as illustrated above.
Comparison with non-`const` member functions
constexpr prevents int x[f()] while lack of const prevents const X x; x.f(); - they're both ensuring client code doesn't hardcode unwanted dependency
in both cases, you wouldn't want the compiler to determine const[expr]-ness automatically:
you wouldn't want client code to call a member function on a const object when you can already anticipate that function will evolve to modify the observable value, breaking the client code
you wouldn't want a value used as a template parameter or array dimension if you already anticipated it later being determined at runtime
they differ in that the compiler enforces const use of other members within a const member function, but does not enforce a compile-time constant result with constexpr (due to practical compiler limitations)
When I pressed Richard Smith, a Clang author, he explained:
The constexpr keyword does have utility.
It affects when a function template specialization is instantiated (constexpr function template specializations may need to be instantiated if they're called in unevaluated contexts; the same is not true for non-constexpr functions since a call to one can never be part of a constant expression). If we removed the meaning of the keyword, we'd have to instantiate a bunch more specializations early, just in case the call happens to be a constant expression.
It reduces compilation time, by limiting the set of function calls that implementations are required to try evaluating during translation. (This matters for contexts where implementations are required to try constant expression evaluation, but it's not an error if such evaluation fails -- in particular, the initializers of objects of static storage duration.)
This all didn't seem convincing at first, but if you work through the details, things do unravel without constexpr. A function need not be instantiated until it is ODR-used, which essentially means used at runtime. What is special about constexpr functions is that they can violate this rule and require instantiation anyway.
Function instantiation is a recursive procedure. Instantiating a function results in instantiation of the functions and classes it uses, regardless of the arguments to any particular call.
If something went wrong while instantiating this dependency tree (potentially at significant expense), it would be difficult to swallow the error. Furthermore, class template instantiation can have runtime side-effects.
Given an argument-dependent compile-time function call in a function signature, overload resolution may incur instantiation of function definitions merely auxiliary to the ones in the overload set, including the functions that don't even get called. Such instantiations may have side effects including ill-formedness and runtime behavior.
It's a corner case to be sure, but bad things can happen if you don't require people to opt-in to constexpr functions.
We can live without constexpr, but in certain cases it makes the code easier and intuitive.
For example we have a class which declares an array with some reference length:
template<typename T, size_t SIZE>
struct MyArray
{
T a[SIZE];
};
Conventionally you might declare MyArray as:
int a1[100];
MyArray<decltype(*a1), sizeof(a1)/sizeof(decltype(a1[0]))> obj;
Now see how it goes with constexpr:
template<typename T, size_t SIZE>
constexpr
size_t getSize (const T (&a)[SIZE]) { return SIZE; }
int a1[100];
MyArray<decltype(*a1), getSize(a1)> obj;
In short, any function (e.g. getSize(a1)) can be used as template argument only if the compiler recognizes it as constexpr.
constexpr is also used to check the negative logic. It ensures that a given object is at compile time. Here is the reference link e.g.
int i = 5;
const int j = i; // ok, but `j` is not at compile time
constexprt int k = i; // error
Without the keyword, the compiler cannot diagnose mistakes. The compiler would not be able to tell you that the function is an invalid syntactically as aconstexpr. Although you said this provides a "false sense of security", I believe it is better to pick up these errors as early as possible.
C++11 allows functions declared with the constexpr specifier to be used in constant expressions such as template arguments. There are stringent requirements about what is allowed to be constexpr; essentially such a function encapsulates only one subexpression and nothing else. (Edit: this is relaxed in C++14 but the question stands.)
Why require the keyword at all? What is gained?
It does help in revealing the intent of an interface, but it doesn't validate that intent, by guaranteeing that a function is usable in constant expressions. After writing a constexpr function, a programmer must still:
Write a test case or otherwise ensure it's actually used in a constant expression.
Document what parameter values are valid in a constant expression context.
Contrary to revealing intent, decorating functions with constexpr may add a false sense of security since tangential syntactic constraints are checked while ignoring the central semantic constraint.
In short: Would there be any undesirable effect on the language if constexpr in function declarations were merely optional? Or would there be any effect at all on any valid program?
Preventing client code expecting more than you're promising
Say I'm writing a library and have a function in there that currently returns a constant:
awesome_lib.hpp:
inline int f() { return 4; }
If constexpr wasn't required, you - as the author of client code - might go away and do something like this:
client_app.cpp:
#include <awesome_lib.hpp>
#include <array>
std::array<int, f()> my_array; // needs CT template arg
int my_c_array[f()]; // needs CT array dimension
Then should I change f() to say return the value from a config file, your client code would break, but I'd have no idea that I'd risked breaking your code. Indeed, it might be only when you have some production issue and go to recompile that you find this additional issue frustrating your rebuilding.
By changing only the implementation of f(), I'd have effectively changed the usage that could be made of the interface.
Instead, C++11 onwards provide constexpr so I can denote that client code can have a reasonable expectation of the function remaining a constexpr, and use it as such. I'm aware of and endorsing such usage as part of my interface. Just as in C++03, the compiler continues to guarantee client code isn't built to depend on other non-constexpr functions to prevent the "unwanted/unknown dependency" scenario above; that's more than documentation - it's compile time enforcement.
It's noteworthy that this continues the C++ trend of offering better alternatives for traditional uses of preprocessor macros (consider #define F 4, and how the client programmer knows whether the lib programmer considers it fair game to change to say #define F config["f"]), with their well-known "evils" such as being outside the language's namespace/class scoping system.
Why isn't there a diagnostic for "obviously" never-const functions?
I think the confusion here is due to constexpr not proactively ensuring there is any set of arguments for which the result is actually compile-time const: rather, it requires the programmer to take responsibility for that (otherwise §7.1.5/5 in the Standard deems the program ill-formed but doesn't require the compiler to issue a diagnostic). Yes, that's unfortunate, but it doesn't remove the above utility of constexpr.
So, perhaps it's helpful to switch from the question "what's the point of constexpr" to consider "why can I compile a constexpr function that can never actually return a const value?".
Answer: because there'd be a need for exhaustive branch analysis that could involve any number of combinations. It could be excessively costly in compile time and/or memory - even beyond the capability of any imaginable hardware - to diagnose. Further, even when it is practical having to diagnose such cases accurately is a whole new can of worms for compiler writers (who have better uses for their time). There would also be implications for the program such as the definition of functions called from within the constexpr function needing to be visible when the validation was performed (and functions that function calls etc.).
Meanwhile, lack of constexpr continues to forbid use as a const value: the strictness is on the sans-constexpr side. That's useful as illustrated above.
Comparison with non-`const` member functions
constexpr prevents int x[f()] while lack of const prevents const X x; x.f(); - they're both ensuring client code doesn't hardcode unwanted dependency
in both cases, you wouldn't want the compiler to determine const[expr]-ness automatically:
you wouldn't want client code to call a member function on a const object when you can already anticipate that function will evolve to modify the observable value, breaking the client code
you wouldn't want a value used as a template parameter or array dimension if you already anticipated it later being determined at runtime
they differ in that the compiler enforces const use of other members within a const member function, but does not enforce a compile-time constant result with constexpr (due to practical compiler limitations)
When I pressed Richard Smith, a Clang author, he explained:
The constexpr keyword does have utility.
It affects when a function template specialization is instantiated (constexpr function template specializations may need to be instantiated if they're called in unevaluated contexts; the same is not true for non-constexpr functions since a call to one can never be part of a constant expression). If we removed the meaning of the keyword, we'd have to instantiate a bunch more specializations early, just in case the call happens to be a constant expression.
It reduces compilation time, by limiting the set of function calls that implementations are required to try evaluating during translation. (This matters for contexts where implementations are required to try constant expression evaluation, but it's not an error if such evaluation fails -- in particular, the initializers of objects of static storage duration.)
This all didn't seem convincing at first, but if you work through the details, things do unravel without constexpr. A function need not be instantiated until it is ODR-used, which essentially means used at runtime. What is special about constexpr functions is that they can violate this rule and require instantiation anyway.
Function instantiation is a recursive procedure. Instantiating a function results in instantiation of the functions and classes it uses, regardless of the arguments to any particular call.
If something went wrong while instantiating this dependency tree (potentially at significant expense), it would be difficult to swallow the error. Furthermore, class template instantiation can have runtime side-effects.
Given an argument-dependent compile-time function call in a function signature, overload resolution may incur instantiation of function definitions merely auxiliary to the ones in the overload set, including the functions that don't even get called. Such instantiations may have side effects including ill-formedness and runtime behavior.
It's a corner case to be sure, but bad things can happen if you don't require people to opt-in to constexpr functions.
We can live without constexpr, but in certain cases it makes the code easier and intuitive.
For example we have a class which declares an array with some reference length:
template<typename T, size_t SIZE>
struct MyArray
{
T a[SIZE];
};
Conventionally you might declare MyArray as:
int a1[100];
MyArray<decltype(*a1), sizeof(a1)/sizeof(decltype(a1[0]))> obj;
Now see how it goes with constexpr:
template<typename T, size_t SIZE>
constexpr
size_t getSize (const T (&a)[SIZE]) { return SIZE; }
int a1[100];
MyArray<decltype(*a1), getSize(a1)> obj;
In short, any function (e.g. getSize(a1)) can be used as template argument only if the compiler recognizes it as constexpr.
constexpr is also used to check the negative logic. It ensures that a given object is at compile time. Here is the reference link e.g.
int i = 5;
const int j = i; // ok, but `j` is not at compile time
constexprt int k = i; // error
Without the keyword, the compiler cannot diagnose mistakes. The compiler would not be able to tell you that the function is an invalid syntactically as aconstexpr. Although you said this provides a "false sense of security", I believe it is better to pick up these errors as early as possible.
Any function that consists of a return statement only could be declared
constexpr and thus will allow to be evaluated at compile time if all
arguments are constexpr and only constexpr functions are called in its body. Is there any reason not to declare any such function constexpr ?
Example:
constexpr int sum(int x, int y) { return x + y; }
constexpr i = 10;
static_assert(sum(i, 13) == 23, "sum correct");
Could anyone provide an example where declaring a function constexpr
would do any harm?
Some initial thoughts:
Even if there should be no good reason for ever declaring a function
not constexpr I could imagine that the constexpr keyword has a
transitional role: its absence in code that does not need compile-time
evaluations would allow compilers that do not implement compile-time
evaluations still to compile that code (but to fail reliably on code
that needs them as made explict by using constexpr).
But what I do not understand: if there should be no good reason for
ever declaring a function not constexpr, why is not every function
in the standard library declared constexpr? (You cannot argue
that it is not done yet because there was not sufficient time yet to
do it, because doing it for all is a no-brainer -- contrary to deciding for every single function if to make it constexpr or not.)
--- I am aware that N2976
deliberately not requires cstrs for many standard library types such
as the containers as this would be too limitating for possible
implementations. Lets exclude them from the argument and just wonder:
once a type in the standard library actually has a constexpr cstr, why is not every function operating on it declared constexpr?
In most cases you also cannot argue that you may prefer not to declare a function constexpr simply because you do not envisage any compile-time usage: because if others evtl. will use your code, they may see such a use that you do not. (But granted for type trait types and stuff alike, of course.)
So I guess there must be a good reason and a good example for deliberately not declaring a function constexpr?
(with "every function" I always mean: every function that meets the
requirements for being constexpr, i.e., is defined as a single
return statement, takes only arguments of types with constexpr
cstrs and calls only constexpr functions. Since C++14, much more is allowed in the body of such function: e.g., C++14 constexpr functions may use local variables and loops, so an even wider class of functions could be declared constexpr.)
The question Why does std::forward discard constexpr-ness? is a special case of this one.
Functions can only be declared constexpr if they obey the rules for constexpr --- no dynamic casts, no memory allocation, no calls to non-constexpr functions, etc.
Declaring a function in the standard library as constexpr requires that ALL implementations obey those rules.
Firstly, this requires checking for each function that it can be implemented as constexpr, which is a long job.
Secondly, this is a big constraint on the implementations, and will outlaw many debugging implementations. It is therefore only worth it if the benefits outweigh the costs, or the requirements are sufficiently tight that the implementation pretty much has to obey the constexpr rules anyway. Making this evaluation for each function is again a long job.
I think what you're referring to is called partial evaluation. What you're touching on is that some programs can be split into two parts - a piece that requires runtime information, and a piece that can be done without any runtime information - and that in theory you could just fully evaluate the part of the program that doesn't need any runtime information before you even start running the program. There are some programming languages that do this. For example, the D programming language has an interpreter built into the compiler that lets you execute code at compile-time, provided that it meets certain restrictions.
There are a few main challenges in getting partial evaluation working. First, it dramatically complicates the logic of the compiler because the compiler will need to have the ability to simulate all of the operations that you could put into an executable program at compile-time. This, in the worst case, requires you to have a full interpreter inside of the compiler, making a difficult problem (writing a good C++ compiler) and making it orders of magnitude harder to do.
I believe that the reason for the current specification about constexpr is simply to limit the complexity of compilers. The cases it's limited to are fairly simple to check. There's no need to implement loops in the compiler (which could cause a whole other slew of problems, like what happens if you get an infinite loop inside the compiler). It also avoids the compiler potentially having to evaluate statements that could cause segfaults at runtime, such as following a bad pointer.
Another consideration to keep in mind is that some functions have side-effects, such as reading from cin or opening a network connection. Functions like these fundamentally can't be optimized at compile-time, since doing so would require knowledge only available at runtime.
To summarize, there's no theoretical reason you couldn't partially evaluate C++ programs at compile-time. In fact, people do this all the time. Optimizing compilers, for example, are essentially programs that try to do this as much as possible. Template metaprogramming is one instance where C++ programmers try to execute code inside the compiler, and it's possible to do some great things with templates partially because the rules for templates form a functional language, which the compiler has an easier time implementing. Moreover, if you think of the tradeoff between compiler author hours and programming hours, template metaprogramming shows that if you're okay making programmers bend over backwards to get what they want, you can build a pretty weak language (the template system) and keep the language complexity simple. (I say "weak" as in "not particularly expressive," not "weak" in the computability theory sense).
Hope this helps!
If the function has side effects, you would not want to mark it constexpr. Example
I can't get any unexpected results from that, actually it looks like gcc 4.5.1 just ignores constexpr