5.9 — Constexpr and consteval functions

In lesson 5.6 -- Constexpr variables, we introduced the constexpr keyword, which we used to create compile-time (symbolic) constants. We also introduced constant expressions, which are expressions that can be evaluated at compile-time rather than runtime.

One challenge with constant expressions is that function call to a normal function are not allowed in constant expressions. This means we cannot use such function calls anywhere a constant expression is required.

Consider the following program:

#include <iostream>

int main()
{
    constexpr double radius { 3.0 };
    constexpr double pi { 3.14159265359 };
    constexpr double circumference { 2.0 * radius * pi };
    
    std::cout << "Our circle has circumference " << circumference << "\n";

    return 0;    
}

This produces the result:

Our circle has circumference 18.8496

Having a complex initializer for circumference isn’t great (and requires us to instantiate two supporting variables, radius and pi). So let’s make it a function instead:

#include <iostream>

double calcCircumference(double radius)
{
    constexpr double pi { 3.14159265359 };
    return 2.0 * pi * radius;
}

int main()
{
    constexpr double circumference { calcCircumference(3.0) }; // compile error
    
    std::cout << "Our circle has circumference " << circumference << "\n";

    return 0;    
}

This code is much cleaner. It also doesn’t compile. Constexpr variable circumference requires that its initializer is a constant expression, and the call calcCircumference() isn’t a constant expression.

In this particular case, we could make circumference non-constexpr, and the program would compile. While we’d lose the benefits of constant expressions, at least the program would run.

However, there are other cases in C++ (which we’ll introduce in the future) where we do not have alternate options available, and only a constant expression will do. In those cases, we’d really like to be able to use functions, but calls to normal functions just won’t work. So what are we to do?

Constexpr functions can be used in constant expressions

A constexpr function is a function that is allowed to be called in a constant expression.

To make a function a constexpr function, we simply use the constexpr keyword in front of the function’s return type.

Here’s the same example as above, but using a constexpr function:

#include <iostream>

constexpr double calcCircumference(double radius) // now a constexpr function
{
    constexpr double pi { 3.14159265359 };
    return 2.0 * pi * radius;
}

int main()
{
    constexpr double circumference { calcCircumference(3.0) }; // now compiles
    
    std::cout << "Our circle has circumference " << circumference << "\n";

    return 0;    
}

Because calcCircumference() is now a constexpr function, it can be used in a constant expression, such as the initializer of circumference.

Constexpr functions can be evaluated at compile time

In lesson 5.5 -- Constant expressions, we noted that in contexts that require a constant expression (such as the initialization of a constexpr variable), a constant expression is required to evaluate at compile-time. If a required constant expression contains a constexpr function call, that constexpr function call must evaluate at compile-time.

In our example above, variable circumference is constexpr and thus requires a constant expression initializer. Since calcCircumference() is part of this required constant expression, calcCircumference() must be evaluated at compile-time.

When a function call is evaluated at compile-time, the compiler will calculate the return value of the function call at compile-time, and then replace the function call with the return value.

So in our example, the call to calcCircumference(3.0) is replaced with the result of the function call, which is 18.8496. In other words, the compiler will compile this:

#include <iostream>

constexpr double calcCircumference(double radius)
{
    constexpr double pi { 3.14159265359 };
    return 2.0 * pi * radius;
}

int main()
{
    constexpr double circumference { 18.8496 };
    
    std::cout << "Our circle has circumference " << circumference << "\n";

    return 0;    
}

To evaluate at compile-time, two other things must also be true:

• The call to the constexpr function must have arguments that are known at compile time (e.g. are constant expressions).
• All statements and expressions within the constexpr function must be evaluatable at compile-time.

When a constexpr (or consteval) function is being evaluated at compile-time, any other functions it calls are required to be evaluated at compile-time (otherwise the initial function would not be able to return a result at compile-time).

Constexpr functions can also be evaluated at runtime

Constexpr functions can also be evaluated at runtime, in which case they will return a non-constexpr result. For example:

#include <iostream>

constexpr int greater(int x, int y)
{
    return (x > y ? x : y);
}

int main()
{
    int x{ 5 }; // not constexpr
    int y{ 6 }; // not constexpr

    std::cout << greater(x, y) << " is greater!\n"; // will be evaluated at runtime

    return 0;
}

In this example, because arguments x and y are not constant expressions, the function cannot be resolved at compile-time. However, the function will still be resolved at runtime, returning the expected value as a non-constexpr int.

Constexpr function calls in non-required constant expressions

You might expect that a constexpr function would evaluate at compile-time whenever possible, but unfortunately this is not the case.

In lesson 5.5 -- Constant expressions, we noted that in contexts that do not require a constant expression, the compiler may choose whether to evaluate a constant expression at either compile-time or at runtime. Accordingly, any constexpr function call that is part of a non-required constant expression may be evaluated at either compile-time or runtime.

For example:

#include <iostream>

constexpr int getValue(int x)
{
    return x;
}

int main()
{
    int x { getValue(5) }; // may evaluate at runtime or compile-time
    
    return 0;
}

In the above example, because getValue() is constexpr, the call getValue(5) is a constant expression. However, because variable x is not constexpr, it does not require a constant expression initializer. So even though we’ve provided a constant expression initializer, the compiler is free to choose whether getValue(5) evaluates at runtime or compile-time.

Diagnosis of constexpr functions

The compiler is not required to determine whether a constexpr function is evaluatable at compile-time until it is actually evaluated at compile-time. It is fairly easy to write a constexpr function that compiles successfully for runtime use, but then fails to compile when evaluated at compile-time.

As a silly example of this:

#include <iostream>

int getValue(int x)
{
    return x;
}

// This function can be evaluated at runtime
// When evaluated at compile-time, the function will produce a compilation error
// because the call to getValue(x) cannot be resolved at compile-time
constexpr int foo(int x)
{
    if (x < 0) return 0; // needed prior to adoption of P2448R1 in C++23 (see note below)
    return getValue(x);  // call to non-constexpr function here
}

int main()
{
    int x { foo(5) };           // okay: will evaluate at runtime
    constexpr int y { foo(5) }; // compile error: foo(5) can't evaluate at compile-time

    return 0;
}

In the above example, when foo(5) is used as an initializer for non-constexpr variable x, it will be evaluated at runtime. This works fine, and returns the value 5.

However, when foo(5), is used as an initializer for constexpr variable y, it must be evaluated at compile-time. At that point, the compiler will determine that the call to foo(5) can’t be evaluated at compile-time, as getValue() is not a constexpr function.

Therefore, when writing a constexpr function, always explicitly test that it compiles when evaluated at compile-time (by calling it in a context where a constant expression is required, such as in the initialization of a constexpr variable).

Constexpr/consteval function parameters are not constexpr

The parameters of a constexpr function are not implicitly constexpr, nor may they be declared as constexpr.

Because such parameters are not constexpr, they cannot be used in constant expressions within the function.

consteval int goo(int c)    // c is not constexpr, and cannot be used in constant expressions
{
    return c;
}

constexpr int foo(int b)    // b is not constexpr, and cannot be used in constant expressions
{
    constexpr int b2 { b }; // compile error: constexpr variable requires constant expression initializer

    return goo(b);          // compile error: consteval function call requires constant expression argument
}

int main()
{
    constexpr int a { 5 };

    std::cout << foo(a); // okay: constant expression a can be used as argument to constexpr function foo()
    
    return 0;
}

In the above example, function parameter b is not constexpr (even though argument a is a constant expression). This means b cannot be used anywhere a constant expression is required, such as the the initializer for a constexpr variable (e.g. b2) or in a call to a consteval function (goo(b)).

The parameters of constexpr functions may be declared as const, in which case they are treated as runtime constants.

Constexpr functions are implicitly inline

When a constexpr function is evaluated at compile-time, the compiler must be able to see the full definition of the constexpr function prior to such function calls (so it can perform the evaluation itself). A forward declaration will not suffice in this case, even if the actual function definition appears later in the same compilation unit.

This means that a constexpr function called in multiple files needs to have its definition included into each translation unit -- which would normally be a violation of the one-definition rule. To avoid such problems, constexpr functions are implicitly inline, which makes them exempt from the one-definition rule.

As a result, constexpr functions are often defined in header files, so they can be #included into any .cpp file that requires the full definition.

For constexpr function calls that are only evaluated at runtime, a forward declaration is sufficient to satisfy the compiler. This means you can use a forward declaration to call a constexpr function defined in another translation unit, but only if you invoke it in a context that does not require compile-time evaluation.

#include <iostream>

constexpr int foo(int);

constexpr int goo(int c)
{
	return foo(c);   // note that foo is not defined yet
}

constexpr int foo(int b) // okay because foo is still defined before any calls to goo
{
	return b;
}

int main()
{
	 constexpr int a{ goo(5) }; // this is the outermost invocation

	return 0;
}

Recap

Marking a function as constexpr means it can be used in a constant expression. It does not mean “will evaluate at compile-time”.

A constant expression (which may contain constexpr function calls) is only required to evaluate at compile-time in contexts where a constant expression is required.

In contexts that do not require a constant expression, the compiler may choose whether to evaluate a constant expression (which may contain constexpr function calls) at compile-time or at runtime.

A runtime (non-constant) expression (which may contain constexpr function calls or non-constexpr function calls) will evaluate at runtime.

Another example

Let’s do another examine to explore how a constexpr function is required or likely to evaluate further:

#include <iostream>

constexpr int greater(int x, int y)
{
    return (x > y ? x : y);
}

int main()
{
    constexpr int g { greater(5, 6) };              // case 1: always evaluated at compile-time
    std::cout << g << " is greater!\n";

    std::cout << greater(5, 6) << " is greater!\n"; // case 2: may be evaluated at either runtime or compile-time

    int x{ 5 }; // not constexpr but value is known at compile-time
    std::cout << greater(x, 6) << " is greater!\n"; // case 3: likely evaluated at runtime

    std::cin >> x;
    std::cout << greater(x, 6) << " is greater!\n"; // case 4: always evaluated at runtime

    return 0;
}

In case 1, we’re calling greater() in a context that requires a constant expression. Thus greater() must be evaluated at compile-time.

In case 2, the greater() function is being called in a context that does not require a constant expression, as output statements must execute at runtime. However, since the arguments are constant expressions, the function is eligible to be evaluated at compile-time. Thus the compiler is free to choose whether this call to greater() will be evaluated at compile-time or runtime.

In case 3, we’re calling greater() with one argument that is not a constant expression. So this will typically execute at runtime.

However, this argument has a value that is known at compile-time. Under the as-if rule, the compiler could decide to treat the evaluation of x as a constant expression, and evaluate this call to greater() at compile-time. But more likely, it will evaluate it at runtime.

In case 4, the value of argument x can’t be known at compile-time, so this call to greater() will always evaluate at runtime.

Note that your compiler’s optimization level setting may have an impact on whether it decides to evaluate a function at compile-time or runtime. This also means that your compiler may make different choices for debug vs. release builds (as debug builds typically have optimizations turned off).

For example, both gcc and Clang will not compile-time evaluate a constexpr function called where a constant expression isn’t required unless the compiler told to optimize the code (e.g. using the -O2 compiler option).

Determining if a constexpr function call is evaluating at compile-time or runtime

C++ does not currently provide any reliable mechanisms to do this.

What about std::is_constant_evaluated or if consteval? Advanced

Neither of these capabilities tell you whether a function call is evaluating at compile-time or runtime.

std::is_constant_evaluated() (defined in the <type_traits> header) returns a bool indicating whether the current function is executing in a constant-evaluated context. A constant-evaluated context (also called a constant context) is defined as one in which a constant expression is required (such as the initialization of a constexpr variable). So in cases where the compiler is required to evaluate a constant expression at compile-time std::is_constant_evaluated() will true as expected.

This is intended to allow you to do something like this:

#include <type_traits> // for std::is_constant_evaluated()

constexpr int someFunction()
{
    if (std::is_constant_evaluated()) // if evaluating in constant context
        doSomething();
    else
        doSomethingElse();
}

However, the compiler may also choose to evaluate a constexpr function at compile-time in a context that does not require a constant expression. In such cases, std::is_constant_evaluated() will return false even though the function did evaluate at compile-time. So std::is_constant_evaluated() really means “the compiler is being forced to evaluate this at compile-time”, not “this is evaluating at compile-time”.

Introduced in C++23, if consteval is a replacement for if (std::is_constant_evaluated()) that provides a nicer syntax and fixes some other issues. However, it evaluates the same way.

Forcing a constexpr function to be evaluated at compile-time

There is no way to tell the compiler that a constexpr function should prefer to evaluate at compile-time whenever it can (e.g. in cases where the return value of a constexpr function is used in a non-constant expression).

However, we can force a constexpr function that is eligible to be evaluated at compile-time to actually evaluate at compile-time by ensuring the return value is used where a constant expression is required. This needs to be done on a per-call basis.

The most common way to do this is to use the return value to initialize a constexpr variable (this is why we’ve been using variable ‘g’ in prior examples). Unfortunately, this requires introducing a new variable into our program just to ensure compile-time evaluation, which is ugly and reduces code readability.

However, in C++20, there is a better workaround to this issue, which we’ll present in a moment.

Consteval C++20

C++20 introduces the keyword consteval, which is used to indicate that a function must evaluate at compile-time, otherwise a compile error will result. Such functions are called immediate functions.

#include <iostream>

consteval int greater(int x, int y) // function is now consteval
{
    return (x > y ? x : y);
}

int main()
{
    constexpr int g { greater(5, 6) };              // ok: will evaluate at compile-time
    std::cout << g << '\n';

    std::cout << greater(5, 6) << " is greater!\n"; // ok: will evaluate at compile-time

    int x{ 5 }; // not constexpr
    std::cout << greater(x, 6) << " is greater!\n"; // error: consteval functions must evaluate at compile-time

    return 0;
}

In the above example, the first two calls to greater() will evaluate at compile-time. The call to greater(x, 6) cannot be evaluated at compile-time, so a compile error will result.

Perhaps surprisingly, the parameters of a consteval function are not constexpr (even though consteval functions can only be evaluated at compile-time). This decision was made for the sake of consistency.

Using consteval to make constexpr execute at compile-time C++20

The downside of consteval functions is that such functions can’t evaluate at runtime, making them less flexible than constexpr functions, which can do either. Therefore, it would still be useful to have a convenient way to force constexpr functions to evaluate at compile-time (even when the return value is being used where a constant expression is not required), so that we could have compile-time evaluation when possible, and runtime evaluation when we can’t.

Consteval functions provides a way to make this happen, using a neat helper function:

#include <iostream>

// Uses abbreviated function template (C++20) and `auto` return type to make this function work with any type of value
// See 'related content' box below for more info (you don't need to know how these work to use this function)
consteval auto compileTimeEval(auto value)
{
    return value;
}

constexpr int greater(int x, int y) // function is constexpr
{
    return (x > y ? x : y);
}

int main()
{
    std::cout << greater(5, 6) << '\n';                  // may or may not execute at compile-time
    std::cout << compileTimeEval(greater(5, 6)) << '\n'; // will execute at compile-time

    int x { 5 };
    std::cout << greater(x, 6) << '\n';                  // we can still call the constexpr version at runtime if we wish

    return 0;
}

This works because consteval functions require constant expressions as arguments -- therefore, if we use the return value of a constexpr function as an argument to a consteval function, the constexpr function must be evaluated at compile-time! The consteval function just returns this argument as its own return value, so the caller can still use it.

Note that the consteval function returns by value. While this might be inefficient to do at runtime (if the value was some type that is expensive to copy, e.g. std::string), in a compile-time context, it doesn’t matter because the entire call to the consteval function will simply be replaced with the calculated return value.

Constexpr/consteval functions can use non-const local variables

Within a constexpr or consteval function, we can use local variables that are not constexpr, and the value of these variables can be changed.

As a silly example:

#include <iostream>

consteval int doSomething(int x, int y) // function is consteval
{
    x = x + 2;       // we can modify the value of non-const function parameters

    int z { x + y }; // we can instantiate non-const local variables
    if (x > y)
        z = z - 1;   // and then modify their values

    return z;
}

int main()
{
    constexpr int g { doSomething(5, 6) };
    std::cout << g << '\n';

    return 0;
}

When such functions are evaluated at compile-time, the compiler will essentially “execute” the function and return the calculated value.

Constexpr/consteval functions can use function parameters and local variables as arguments in constexpr function calls

Above, we noted, “When a constexpr (or consteval) function is being evaluated at compile-time, any other functions it calls are required to be evaluated at compile-time.”

Perhaps surprisingly, a constexpr or consteval function can use its function parameters (which aren’t constexpr) or even local variables (which may not be const at all) as arguments in a constexpr function call. When a constexpr or consteval function is being evaluated at compile-time, the value of all function parameters and local variables must be known to the compiler (otherwise it couldn’t evaluate them at compile-time). Therefore, in this specific context, C++ allows these values to be used as arguments in a call to a constexpr function, and that constexpr function call can still be evaluated at compile-time.

#include <iostream>

constexpr int goo(int c) // goo() is now constexpr
{
    return c;
}

constexpr int foo(int b) // b is not a constant expression within foo()
{
    return goo(b);       // if foo() is resolved at compile-time, then `goo(b)` can also be resolved at compile-time
}

int main()
{
    std::cout << foo(5);
    
    return 0;
}

In the above example, foo(5) may or may not be evaluated at compile time. If it is, then the compiler knows that b is 5. And even though b is not constexpr, the compiler can treat the call to goo(b) as if it were goo(5) and evaluate that function call at compile-time. If foo(5) is instead resolved at runtime, then goo(b) will also be resolved at runtime.

Can a constexpr function call a non-constexpr function?

The answer is yes, but only when the constexpr function is being evaluated in a non-constant context. A non-constexpr function may not be called when a constexpr function is evaluating in a constant context (because then the constexpr function wouldn’t be able to produce a compile-time constant value), and doing so will produce a compilation error.

Calling a non-constexpr function is allowed so that a constexpr function can do something like this:

#include <type_traits> // for std::is_constant_evaluated

constexpr int someFunction()
{
    if (std::is_constant_evaluated()) // if evaluating in constant context
        return someConstexprFcn();
    else
        return someNonConstexprFcn();
}

Now consider this variant:

constexpr int someFunction(bool b)
{
    if (b)
        return someConstexprFcn();
    else
        return someNonConstexprFcn();
}

This is legal as long as someFunction(false) is never called in a constant expression.

For best results, we’d advise the following:

1. Avoid calling non-constexpr functions from within a constexpr function if possible.
2. If your constexpr function requires different behavior for constant and non-constant contexts, conditionalize the behavior with if (std::is_constant_evaluated()) (in C++20) or if consteval (C++23 onward).
3. Always test your constexpr functions in a constant context, as they may work when called in a non-constant context but fail in a constant context.

Why not constexpr every function?

There are a few reasons you may not want to constexpr a function:

1. constexpr signals that a function can be used in a constant expression. If your function cannot be evaluated as part of a constant expression, it should not be marked as constexpr.
2. constexpr is part of the interface of a function. Once a function is made constexpr, it can be called by other constexpr functions or used in contexts that require constant expressions. Removing the constexpr later will break such code.
3. constexpr makes functions harder to debug because you can’t inspect them at runtime.

Why constexpr a function when it is not actually evaluated at compile-time?

New programmers sometimes ask, “why should I constexpr a function when it is only evaluated at runtime in my program (e.g. because the arguments in the function call are non-const)”?

There are a few reasons:

1. There’s little downside to using constexpr, and it may help the compiler optimize your program to be smaller and faster.
2. Just because you’re not calling the function in a compile-time evaluatable context right now doesn’t mean you won’t call it in such a context when you modify or extend your program. And if you haven’t constexpr’d the function already, you may not think to when you do start to call it in such a context, and then you’ll miss out on the performance benefits. Or you may be forced to constexpr it later when you need to use the return value in a context that requires a constant expression somewhere.
3. Repetition helps ingrain best practices.

On a non-trivial project, it’s a good idea to implement your functions with the mindset that they may be reused (or extended) in the future. Any time you modify an existing function, you risk breaking it, and that means it needs to be retested, which takes time and energy. It’s often worth spending an extra minute or two “doing it right the first time” so you don’t have to redo (and retest) it again later.

As a general rule, if a function can be evaluated as part of a constant expression, it should be made constexpr. That said, if your program is trivial or a throw-away and you don’t constexpr a function, the world isn’t going to end. Hopefully.