1 Revision history

R2 to R3: Wording changes.

R1 to R2: R1 [P1938R1] of this paper originally mandated the use of braces in if consteval (see why braces, also new in this revision) grammatically. Davis Herring pointed out in an EWG telecon that this causes some issues:

if (cond) 
    if consteval { s0; }
    else s1;

If the grammar for if consteval has compound-statement, then the else here cannot be associated with the if consteval, it must be associated with the outer if. Rather than that, we want this to a compile error due to the lack of braces around s1.

R0 to R1: R0 [P1938R0] of this paper initially contained only a positive form: if consteval. This paper additionally adds a negated form, if not consteval.

2 Introduction

Despite this paper missing both our respective NB comment deadlines and the mailing deadline, we still believe it provides a significant enough improvement to the status quo that it should be considered for C++20.

C++20 will have several new features to aid programmers in writing code during constant evaluation. Two of these are std::is_constant_evaluated() [P0595R2] and consteval [P1073R3], both adopted in San Diego 2018. consteval is for functions that can only be invoked during constant evaluation. is_constant_evaluated() is a magic library function to check if the current evaluation is constant evaluation to provide, for instance, a valid implementation of an algorithm for constant evaluation time and a better implementation for runtime.

However, despite being adopted at the same meeting, these features interact poorly with each other and have other issues that make them ripe for confusion.

3 Problems with Status Quo

There are two problems this paper wishes to address.

3.1 Interaction between constexpr and consteval

The first problem is the interplay between this magic library function and the new consteval. Consider the example:

consteval int f(int i) { return i; }

constexpr int g(int i) {
    if (std::is_constant_evaluated()) {
        return f(i) + 1; // <==
    } else {
        return 42;
    }
}

consteval int h(int i) {
    return f(i) + 1;
}

The function h here is basically a lifted, constant-evaluation-only version of the function g. At constant evaluation time, they do the same thing, except that during runtime, you cannot call h, and g has this extra path. Maybe this code started with just h and someone decided a runtime version would also be useful and turned it into g.

Unfortunately, h is well-formed while g is ill-formed. You cannot make that call to f (that is ominously marked with an arrow) in that location. Even though that call will only happen during constant evaluation, that’s still not enough.

With specific terms, the call to f() inside of g() is an immediate invocation and needs to be a constant expression and it is not. Whereas the call to f() inside of h() is not considered an immediate invocation because it is in an immediate function context (i.e. it’s invoked from another immediate function), so it has a weaker set of restrictions that it needs to follow.

In other words, this kind of construction of conditionally invoking a consteval function from a constexpr function just Does Not Work (modulo the really trivial cases - one could call f(42) for instance, just never f(i)).

We find this lack of composability of features to be problematic and think it can be improved.

3.2 The if constexpr (std::is_constant_evaluated()) problem

The second problem is specific to is_constant_evaluated. Once you learn what this magic function is for, the obvious usage of it is:

constexpr size_t strlen(char const* s) {
    if constexpr (std::is_constant_evaluated()) {
        for (const char *p = s; ; ++p) {
            if (*p == '\0') {
                return static_cast<std::size_t>(p - s);
            }
        }    
    } else {
        __asm__("SSE 4.2 insanity");        
    }
}

This example, inspired by [P1045R0], has a bug: it uses if constexpr to check the conditional is_constant_evaluated() rather than a simple if. You have to really deeply understand a lot about how constant evaluation works in C++ to understand that this is in fact not only not “obviously correct” but is in fact “obviously incorrect,” for some definition of obvious. This is such a likely source of error that Barry submitted bugs to both gcc and clang to encourage the compilers to warn on such improper usage. gcc 10.1 will provide a warning for the simple case:

<source>: In function 'constexpr int f(int)':
<source>:4:45: warning: 'std::is_constant_evaluated' always evaluates to true in 'if constexpr' [-Wtautological-compare]
    4 |     if constexpr (std::is_constant_evaluated()) {
      |                   ~~~~~~~~~~~~~~~~~~~~~~~~~~^~

But then people have to understand why this is a warning, and what this even means. Nevertheless, a compiler warning is substantially better than silently wrong code, but it is problematic to have an API in which many users are drawn to a usage that is tautologically incorrect.

3.2.1 Compiler warnings

When R0 of this paper was presented in Belfast, the implementers assured that all the compilers would properly warn on all tautological uses of std::is_constant_evaluated() - both in the always-true and always-false cases.

As of this writing, for instance, EDG warns on all of the following:

constexpr int f1() {
  if constexpr (!std::is_constant_evaluated() && sizeof(int) == 4) { // warning: always true
    return 0;
  }
  if (std::is_constant_evaluated()) {
    return 42;
  } else {
    if constexpr (std::is_constant_evaluated()) { // warning: always true
      return 0;
    }
  }
  return 7;
}


consteval int f2() {
  if (std::is_constant_evaluated() && f1()) { // warning: always true
    return 42;
  }
  return 7;
}


int f3() {
  if (std::is_constant_evaluated() && f1()) { // warning: always false
    return 42;
  }
  return 7;
}

We expect the other compilers to follow suit.

4 Proposal

We propose a new form of if statement which is spelled:

if consteval { }

The braces (in both the if and the optional else) are mandatory and there is no condition. If evaluation of this statement occurs during constant evaluation, the first substatement is executed. Otherwise, the second substatement (if there is one) is executed.

This behaves exactly as today’s:

if (std::is_constant_evaluated()) { }

except with three differences:

1. No header include is necessary.
2. The syntax is different, which completely sidesteps the confusion over the proper way to check if we’re in constant evaluation. You simply cannot misuse the syntax.
3. We can use if consteval to allow invoking immediate functions.

To explain the last point a bit more, the current language rules allow you to invoke a consteval function from inside of another consteval function ([expr.const]/13) - we can do this by construction:

An expression or conversion is in an immediate function context if it is potentially evaluated and its innermost non-block scope is a function parameter scope of an immediate function. An expression or conversion is an immediate invocation if it is an explicit or implicit invocation of an immediate function and is not in an immediate function context. An immediate invocation shall be a constant expression.

By extending the term immediate function context to also include an if consteval block, we can allow the second example to work:

consteval int f(int i) { return i; }

constexpr int g(int i) {
    if consteval {
        return f(i) + 1; // ok: immediate function context
    } else {
        return 42;
    }
}

consteval int h(int i) {
    return f(i) + 1; // ok: immediate function context
}

Additionally, such a feature would allow for an easy implementation of the original std::is_constant_evaluated():

constexpr bool is_constant_evaluated() {
    if consteval {
        return true;
    } else {
        return false;
    }
}

Although this paper does not suggest removing the library function.

4.1 Negated form

Many people have expressed the view that a negated form is also useful. That form is also proposed here, spelled:

if not consteval { }

or

if ! consteval { }

With the semantics that the first substatement is executed if the context is not manifestly constant evaluated, otherwise the second substatement (if any) is executed.

4.2 Conditioned Form

As proposed, this new form of if does not have a condition - unlike the other two we already have. While there are certainly cases where an added condition would be useful, this paper is deliberately not including such a thing. The vast majority of uses are expected to be just of the if consteval or if not consteval form and we do not want to clutter future design space in this area.

There are currently two uses in libstdc++ that are of the form if (is_constant_evaluated() && cond). One example:

if (std::is_constant_evaluated() && __n < 0)
  throw "attempt to decrement a non-bidirectional iterator";

This usage is perfectly fine and doesn’t necessary need special support from this proposal. Or it could also be written as:

if consteval {
  if (__n < 0) throw "attempt to decrement a non-bidirectional iterator";
}

Or factored into a function like:

if consteval {
  consteval_assert(__n >= 0,
    "attempt to decrement a non-bidirectional iterator");
}

Either way, the condition form doesn’t feel strongly motivated except for consistency with if and if constexpr.

4.3 Deprecating std::is_constant_evaluated()

One of the questions that comes up regularly in discussing this paper is: if we had if consteval, we do we even need std::is_constant_evaluated(), and can we just deprecate it?

This paper proposes no such deprecation. The reason is that this function is actually still occasionally useful (as in the previous section). If the standard library does not provide it, users will write their own. We’re not concerned about the implementation difficulty of it - the users that need this will definitely be able to write it correctly - but we are concerned with a proliferation of exactly this function. The advantage of having the one std::is_constant_evaluated() is both that it becomes actually teachable and also that it becomes warnable: the warnings discussed can happen only because we know what this name means. Maybe it’s still possible to warn on if constexpr (your::is_constant_evaluated()) but that’s a much harder problem.

And note that libstdc++ already has some uses that do require the function form.

4.4 Why braces?

The braces in an if consteval statement are mandatory:

if consteval {
   f(i); // ok
}

if consteval f(i); // ill-formed

There is no technical reason to mandate braces. Our reason for them is to emphasize visually that we’re dropping into an entirely different context. In this sense, it’s similar to try and catch mandating braces. It’s also quite unlike if constexpr in that the language rules in the taken statement actually change, so we think it merits the different rules.

Without braces, we have no separation between the if consteval introducer and the statement being executed. This becomes especially important if the statement being executed itself includes some sort of keyword:

if consteval for (; it != end; ++it) ;
if consteval if (it != end);

A normal if statement has a parenthesized condition, which functions as such a separator, and we think such visual separation is important:

if (cond) f(i);

An alternate syntax suggested was if (consteval), with no requirement on braces, to look more like a regular if statement. But this syntax suggests that other forms might be valid, which we either cannot support due to the implementation heroics necessary to so:

if (consteval && n < 0) {
  // We would need to take an arbitrary boolean condition, C, and
  // be able to prove that (not is_constant_evaluated) implies (not C).
  // That's an arbitrarily complex problem.
  some_consteval_fun(n);
}

Or suggests that other forms might be valid that make little sense to even want to support:

while (consteval) { /* ... */ }
switch (consteval) { /* ... */ }

Overall, we think the fact that if consteval looks different from a regular if statement is arguably a benefit.

This question of syntax was discussed on the October 8th EWG telecon [github.p1938], where the following polls were taken (note that the first two polls are phrased as changes from the status quo as proposed in this paper):

4.5 Examples

Here are a few examples from libstdc++. Today, they’re implemented uses a builtin function, and how they would look with if consteval. It’s not a big difference, just spelling.

if (!__builtin_is_constant_evaluated())
  __glibcxx_assert( __shift_exponent != numeric_limits<_Tp>::digits );

if not consteval {
   __glibcxx_assert( __shift_exponent != numeric_limits<_Tp>::digits );
}

if (__builtin_is_constant_evaluated())
  return __x < __y;
return (__UINTPTR_TYPE__)__x > (__UINTPTR_TYPE__)__y;

if consteval {
  return __x < __y;
} else {
  return (__UINTPTR_TYPE__)__x > (__UINTPTR_TYPE__)__y;
}

As of this writing, libstdc++ has 23 uses that could be replaced by if consteval, 2 that could be replaced by if not consteval, and 2 that require an extra condition on the comparison.

5 History

The initial revision of the std::is_constant_evaluated() proposal [P0595R0] was actually targeted as a language feature rather than a library feature. The original spelling was if (constexpr()). The paper was presented in Kona 2017 and was received very favorably in the form it was presented (17-4). The poll to consider a magic library alternative was only marginally more preferred (17-3). We believe that in the two years since these polls were taken, having a dedicated language feature with an impossible-to-misuse API, that can coexist with the rest of the constant ecosystem, is the right direction.

6 Wording

Extend the definition of immediate function context in 7.7 [expr.const] (and use bullet points):

An expression or conversion is in an immediate function context if it is potentially evaluated and either:

• (13.1) its innermost non-block scope is a function parameter scope of an immediate function. , or
• (13.2) its enclosing statement is enclosed ([stmt.pre]) by the compound-statement of a consteval if statement ([stmt.if]).

Change 8.5 [stmt.select] to add the new grammar:

  selection-statement:
     if constexpropt ( init-statementopt condition ) statement
     if constexpropt ( init-statementopt condition ) statement else statement
+    if !opt consteval compound-statement
+    if !opt consteval compound-statement else statement
     switch ( init-statementopt condition ) statement

Add a new clause to 8.5.2 [stmt.if]:

a An if statement of the form if consteval is called a consteval if statement. The statement, if any, in a consteval if statement shall be a compound-statement. [ Example -

constexpr void f(bool b) {
  if (true)
    if consteval { }
    else ; // error: not a compound-statement
           // "else" not associated with outer if
}

• end example ]

b If a consteval if statement is evaluated in a context that is manifestly constant-evaluated ([expr.const]), the first substatement is executed. [Note: The first substatement is an immediate function context - end note ] Otherwise, if the else part of the selection statement is present, then the second substatement is executed. A case or default label appearing within such an if statement shall be associated with a switch statement within the same if statement. A label declared in a substatement of a consteval if statement shall only be referred to by a statement in the same substatement.

c An if statement of the form if ! consteval compound-statement is not itself a consteval if statement, but is equivalent to the consteval if statement

if consteval { } else compound-statement

An if statement of the form if ! consteval compound-statement1 else statement2 is not itself a consteval if statement, but is equivalent to the consteval if statement

if consteval statement2 else compound-statement1

Change 20.15.11 [meta.const.eval] to use this new functionality:

constexpr bool is_constant_evaluated() noexcept;

1 Returns: true if and only if evaluation of the call occurs within the evaluation of an expression or conversion that is manifestly constant-evaluated ([expr.const]).

1 Effects: Equivalent to:

if consteval {
    return true;
} else {
    return false;
}

6.1 Feature test macro

Add the macro __cpp_if_consteval.

7 Acknowledgments

Thank you to David Stone and Tim Song for working through these examples. Thank you to Davis Herring for helping improve the grammar to remove a potential source of error, and Jens Maurer with help wording it properly.

8 References