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Is compiler allowed to call an immediate (consteval) function during runtime?

This might be a stupid question, but I am confused. I had a feeling that an immediate (consteval) function has to be executed during compile time and we simply cannot see its body in the binary.

This article clearly supports my feeling:

This has the implication that the [immediate] function is only seen at compile time. Symbols are not emitted for the function, you cannot take the address of such a function, and tools such as debuggers will not be able to show them. In this matter, immediate functions are similar to macros.

The similar strong claim might be found in Herb Sutter's publication:

Note that draft C++20 already contains part of the first round of reflection-related work to land in the standard: consteval functions that are guaranteed to run at compile time, which came from the reflection work and are designed specifically to be used to manipulate reflection information.

However, there is a number of evidences that are not so clear about this fact.

From cppreference:

consteval - specifies that a function is an immediate function, that is, every call to the function must produce a compile-time constant.

It does not mean it has to be called during compile time only.

From the P1073R3 proposal:

There is now general agreement that future language support for reflection should use constexpr functions, but since "reflection functions" typically have to be evaluated at compile time, they will in fact likely be immediate functions.

Seems like this means what I think, but still it is not clearly said. From the same proposal:

Sometimes, however, we want to express that a function should always produce a constant when called (directly or indirectly), and a non-constant result should produce an error.

Again, this does not mean the function has to be evaluated during compile time only.

From this answer:

your code must produce a compile time constant expression. But a compile time constant expression is not an observable property in the context where you used it, and there are no side effects to doing it at link or even run time! And under as-if there is nothing preventing that

Finally, there is a live demo, where consteval function is clearly called during runtime. However, I hope this is due to the fact consteval is not yet properly supported in clang and the behavior is actually incorrect, just like in Why does a consteval function allow undefined behavior?

To be more precise, I'd like to hear which of the following statements of the cited article are correct:

  1. An immediate function is only seen at compile time (and cannot be evaluated at run time)
  2. Symbols are not emitted for an immediate function
  3. Tools such as debuggers will not be able to show an immediate function
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Mikhail Avatar asked Oct 19 '19 17:10

Mikhail


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2 Answers

To be more precise, I'd like to hear which of the following statements of the cited article are correct:

  1. An immediate function is only seen at compile time (and cannot be evaluated at run time)
  2. Symbols are not emitted for an immediate function
  3. Tools such as debuggers will not be able to show an immediate function

Almost none of these are answers which the C++ standard can give. The standard doesn't define "symbols" or what tools can show. Almost all of these are dealer's choice as far as the standard is concerned.

Indeed, even the question of "compile time" vs. "run time" is something the standard doesn't deal with. The only question that concerns the standard is whether something is a constant expression. Invoking a constexpr function may produce a constant expression, depending on its parameters. Invoking a consteval function in a way which does not produce a constant expression is il-formed.

The one thing the standard does define is what gets "seen". Though it's not really about "compile time". There are a number of statements in C++20 that forbid most functions from dealing in pointers/references to immediate functions. For example, C++20 states in [expr.prim.id]/3:

An id-expression that denotes an immediate function shall appear only

  • as a subexpression of an immediate invocation, or

  • in an immediate function context.

So if you're not in an immediate function, or you're not using the name of an immediate function to call another immediate function (passing a pointer/reference to the function), then you cannot name an immediate function. And you can't get a pointer/reference to a function without naming it.

This and other statements in the spec (like pointers to immediate function not being valid results of constant expressions) essentially make it impossible for a pointer/reference to an immediate function to leak outside of constant expressions.

So statements about the visibility of immediate functions are correct, to some degree. Symbols can be emitted for immediate functions, but you cannot use immediate functions in a way that would prevent an implementation from discarding said symbols.

And that's basically the thing with consteval. It doesn't use standard language to enforce what must happen. It uses standard language to make it impossible to use the function in a way that will prevent these things from happening. So it's more reasonable to say:

  1. You cannot use an immediate function in a way that would prevent the compiler from executing it at compile time.

  2. You cannot use an immediate function in a way that would prevent the compiler from discarding symbols for it.

  3. You cannot use an immediate function in a way that would force debuggers to be able to see them.

Quality of implementation is expected to take things from there.

It should also be noted that debugging builds are for... debugging. It would be entirely reasonable for advanced compiler tools to be able to debug code that generates constant expressions. So a debugger which could see immediate functions execute is an entirely desirable technology. This becomes moreso as compile-time code grows more complex.

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Nicol Bolas Avatar answered Oct 23 '22 18:10

Nicol Bolas


The proposal mentions:

One consequence of this specification is that an immediate function never needs to be seen by a back end.

So it is definitely the intention of the proposal that calls are replaced by the constant. In other words, that the constant expression is evaluated during translation.

However, it does not say it is required that it is not seen by the backend. In fact, in another sentence of the proposal, it just says it is unlikely:

It also means that, unlike plain constexpr functions, consteval functions are unlikely to show up in symbolic debuggers.


More generally, we can re-state the question as:

Are compilers forced to evaluate constant expressions (everywhere; not just when they definitely need it)?

For instance, a compiler needs to evaluate a constant expression if it is the number of elements of an array, because it needs to statically determine the total size of the array.

However, a compiler may not need to evaluate other uses, and while any decent optimizing compiler will try to do so anyway, it does not mean it needs to.

Another interesting case to think about is an interpreter: while an interpreter still needs to evaluate some constant expressions, it may just do it lazily all the time, without performing any constant folding.

So, as far as I know, they aren't required, but I don't know the exact quotes we need from the standard to prove it (or otherwise). Perhaps it is a good follow-up question on its own, which would answer this one too.

For instance, in [expr.const]p1 there is a note that says they can, not that they are:

[Note: Constant expressions can be evaluated during translation. — end note]

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Acorn Avatar answered Oct 23 '22 18:10

Acorn