Whyever **not** declare a function to be `constexpr`?

问题

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.)

The question Why does std::forward discard constexpr-ness? is a special case of this one.

回答1:

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.

回答2:

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!