Why is C++11 constexpr so restrictive?

问题:

As you probably know, C++11 introduces the constexpr keyword.

C++11 introduced the keyword constexpr, which allows the user to guarantee that a function or object constructor is a compile-time constant. [...] This allows the compiler to understand, and verify, that [function name] is a compile-time constant.

My question is why are there such strict restrictions on form of the functions that can be declared. I understand desire to guarantee that function is pure, but consider this:

The use of constexpr on a function imposes some limitations on what that function can do. First, the function must have a non-void return type. Second, the function body cannot declare variables or define new types. Third, the body may only contain declarations, null statements and a single return statement. There must exist argument values such that, after argument substitution, the expression in the return statement produces a constant expression.

That means that this pure function is illegal:

constexpr int maybeInCppC1Y(int a, int b)
{
    if (a>0)
        return a+b;
    else
        return a-b;
  //can be written as   return  (a>0) ? (a+b):(a-b); but that isnt the point
}

Also you cant define local variables... :( So I'm wondering is this a design decision, or do compilers suck when it comes to proving function a is pure?

回答1:

The reason you'd need to write statements instead of expressions is that you want to take advantage of the additional capabilities of statements, particularly the ability to loop. But to be useful, that would require the ability to declare variables (also banned).

If you combine a facility for looping, with mutable variables, with logical branching (as in if statements) then you have the ability to create infinite loops. It is not possible to determine if such a loop will ever terminate (the halting problem). Thus some sources would cause the compiler to hang.

By using recursive pure functions it is possible to cause infinite recursion, which can be shown to be equivalently powerful to the looping capabilities described above. However, C++ already has that problem at compile time - it occurs with template expansion - and so compilers already have to have a switch for "template stack depth" so they know when to give up.

So the restrictions seem designed to ensure that this problem (of determining if a C++ compilation will ever finish) doesn't get any thornier than it already is.

回答2:

The rules for constexpr functions are designed such that it's impossible to write a constexpr function that has any side-effects.

By requiring constexpr to have no side-effects it becomes impossible for a user to determine where/when it was actually evaluated. This is important since constexpr functions are allowed to happen at both compile time and run time at the discretion of the compiler.

If side-effects were allowed then there would need to be some rules about the order in which they would be observed. That would be incredibly difficult to define - even harder than the static initialisation order problem.

A relatively simple set of rules for guaranteeing these functions to be side-effect free is to require that they be just a single expression (with a few extra restrictions on top of that). This sounds limiting initially and rules out the if statement as you noted. Whilst that particular case would have no side-effects it would have introduced extra complexity into the rules and given that you can write the same things using the ternary operator or recursively it's not really a huge deal.

n2235 is the paper that proposed the constexpr addition in C++. It discusses the rational for the design - the relevant quote seems to be this one from a discussion on destructors, but relevant generally:

The reason is that a constant-expression is intended to be evaluated by the compiler at translation time just like any other literal of built-in type; in particular no observable side-effect is permitted.

Interestingly the paper also mentions that a previous proposal suggested the the compiler figured out automatically which functions were constexpr without the new keyword, but this was found to be unworkably complex, which seems to support my suggestion that the rules were designed to be simple.

(I suspect there will be other quotes in the references cited in the paper, but this covers the key point of my argument about the no side-effects)

回答3:

Actually the C++ standardization committee is thinking about removing several of these constraints for c++14. See the following working document http://www.open-std.org/JTC1/SC22/WG21/docs/papers/2013/n3597.html

回答4:

The restrictions could certainly be lifted quite a bit without enabling code which cannot be executed during compile time, or which cannot be proven to always halt. However I guess it wasn't done because

• it would complicate the compiler for minimal gain. C++ compilers are quite complex as is

• specifying exactly how much is allowed without violating the restrictions above would have been time consuming, and given that desired features have been postponed in order to get the standard out of the door, there probably was little incentive to add more work (and further delay of the standard) for little gain

• some of the restrictions would have been either rather arbitrary or rather complicated (especially on loops, given that C++ doesn't have the concept of a native incrementing for loop, but both the end condition and the increment code have to be explicitly specified in the for statement, making it possible to use arbitrary expressions for them)

Of course, only a member of the standards committee could give an authoritative answer whether my assumptions are correct.

回答5:

I think constexpr is just for const objects. I mean; you can now have static const objects like String::empty_string constructs statically(without hacking!). This may reduce time before 'main' called. And static const objects may have functions like .length(), operator==,... so this is why 'expr' is needed. In 'C' you can create static constant structs like below:

static const Foos foo = { .a = 1, .b = 2, };

Linux kernel has tons of this type classes. In c++ you could do this now with constexpr.

note: I dunno but code below should not be accepted so like if version:

constexpr int maybeInCppC1Y(int a, int b) { return (a > 0) ? (a + b) : (a - b); }