Fixing inconsistencies between constexpr and consteval functions

Document Number: P1937R1
Date: 2020-01-12
Author: David Stone (,
Audience: Core Working Group (CWG)


A consteval function is the same as a constexpr function except that:

  1. A consteval function is guaranteed to be evaluated at compile time.
  2. You cannot have a pointer to a consteval function except in the body of another consteval function.
  3. All overrides of a virtual function must be consteval if the base function is consteval, otherwise they must not be consteval.
  4. A consteval function is evaluated even in contexts that would otherwise be unevaluated.

The first point is the intended difference between the two functions. The second point is a consequence of the first. The third point (virtual functions) ends up making sense as a difference between them because of what is fundamentally possible. The final point seems like an unfortunate inconsistency in the language. This paper proposes changing the rules surrounding bullet 4 to unify the behavior of consteval and constexpr functions for C++20.

Revision history

Since R0

Unevaluated contexts

Requiring evaluation of consteval functions in unevaluated contexts does not seem necessary. When I asked about this discrepancy on the core reflector, the answer that I got was essentially that evaluation of a consteval function may have side-effects on the abstract machine, and we would like to preserve these side-effects. However, consteval functions are not special in this ability, so it seems strange to have a rule that makes them special. It seems to me that we should either

  1. not have a special case for consteval, which would mean that it is not evaluated in an unevaluated operand, or
  2. Rethink the concept of "unevaluated operands".

This paper argues in favor of option 1, as option 2 is a breaking change that doesn't seem to bring much benefit.

My understanding is that the status quo means the following is valid:

constexpr auto add1(auto lhs, auto rhs) { return lhs + rhs; } using T = decltype(add1(std::declval<int>(), std::declval<int>()));

but this very similar code is ill-formed

consteval auto add2(auto lhs, auto rhs) { return lhs + rhs; } using T = decltype(add2(std::declval<int>(), std::declval<int>()));


Note: this wording is presumed incomplete. There is a note stating that consteval functions are always evaluated, but I cannot find where the normative wording actually states this.

Modify [expr.const], paragraph 13, as follows:

[ Note: A manifestly constant-evaluated expression is evaluated even in an unevaluated operand. — end note ]