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std::is_constant_evaluated(3) C++ Standard Libary std::is_constant_evaluated(3)

NAME

std::is_constant_evaluated - std::is_constant_evaluated

Synopsis


Defined in header <type_traits>
constexpr bool is_constant_evaluated() noexcept; (since C++20)


Detects whether the function call occurs within a constant-evaluated context.
Returns true if the evaluation of the call occurs within the evaluation of an
expression or conversion that is manifestly constant-evaluated; otherwise returns
false.


While testing whether initializers of following variables are constant initializers,
compilers may first perform a trial constant evaluation of the initializers:


* variables with reference type or const-qualified integral or enumeration type


* if the initializers are constant expressions, the variables are usable in
constant expressions


* static and thread local variables


* if when all subexpressions of the initializers (including constructor calls
and implicit conversions) are constant expressions, static initialization
is performed, which can be asserted by constinit


It is not recommended to depend on the result in this case.


int y;
const int a = std::is_constant_evaluated() ? y : 1;
// Trial constant evaluation fails. The constant evaluation is discarded.
// Variable a is dynamically initialized with 1


const int b = std::is_constant_evaluated() ? 2 : y;
// Constant evaluation with std::is_constant_evaluation() == true succeeds.
// Variable b is statically initialized with 2

Parameters


(none)

Return value


true if the evaluation of the call occurs within the evaluation of an expression or
conversion that is manifestly constant-evaluated; otherwise false.

Possible implementation


// This implementation requires C++23 if consteval.
constexpr bool is_constant_evaluated() noexcept
{
if consteval {
return true;
}
else {
return false;
}
}

Notes


When directly used as the condition of static_assert declaration or constexpr if
statement, std::is_constant_evaluated() always returns true.


Because if consteval is absent in C++20, std::is_constant_evaluated is typically
implemented by a compiler extension.


Feature-test macro: __cpp_lib_is_constant_evaluated

Example

// Run this code


#include <type_traits>
#include <cmath>
#include <iostream>


constexpr double power(double b, int x)
{
if (std::is_constant_evaluated() && !(b == 0.0 && x < 0)) {
// A constant-evaluation context: Use a constexpr-friendly algorithm.
if (x == 0)
return 1.0;
double r = 1.0, p = x > 0 ? b : 1.0 / b;
auto u = unsigned(x > 0 ? x : -x);
while (u != 0) {
if (u & 1) r *= p;
u /= 2;
p *= p;
}
return r;
} else {
// Let the code generator figure it out.
return std::pow(b, double(x));
}
}


int main()
{
// A constant-expression context
constexpr double kilo = power(10.0, 3);
int n = 3;
// Not a constant expression, because n cannot be converted to an rvalue
// in a constant-expression context
// Equivalent to std::pow(10.0, double(n))
double mucho = power(10.0, n);


std::cout << kilo << " " << mucho << "\n"; // (3)
}

Output:


1000 1000

See also


constexpr specifier(C++11) specifies that the value of a variable or function can be
computed at compile time
specifies that a function is an immediate function, that
consteval specifier(C++20) is, every call to the function must be in a constant
evaluation
constinit specifier(C++20) asserts that a variable has static initialization, i.e.
zero initialization and constant initialization

2022.07.31 http://cppreference.com