Why do we need to mark functions as constexpr?

Question

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:

1. Write a test case or otherwise ensure it's actually used in a constant expression.
2. 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.

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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?

Answer 1

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 compile-time 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 compile-time 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/usage

• 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, and perhaps so that one function definition can happily serve up run-time results for run-time-known arguments/values, but will return a compile-time result when possible and client usage requires).

Answer 2

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.

Answer 3

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.

Answer 4

We can live without constexpr, but in certain cases it makes the code easier and intuitive. The following example shows a class declaring an array with a 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
constexpr int k = i; // error