Effective Modern C++ - Item 15 - Use constexpr whenever possible

3/10/2025 / 5 minutes to read / Tags: cpp

Item 15: Use constexpr whenever possible

Key takeaways:

• constexpr objects are const and initialized with values known at compile time
• constexpr functions can produce compile-time results when called with arguments whose values are known during compilation
• constexpr objects and functions may be used in wider range of contexts than non-constexpr objects and functions
• constexpr is part of an object/function’s interface

constexpr functions are will be evaluated at compile-time if all arguments are known, otherwise it will be treated like a normal function.

constexpr int pow(int base, unsigned exp) noexcept {...}

constexpr auto numConds = 5;
std::array<int, pow(3, numConds)> results; // creates an array with 3^5 elements at compile-time

auto base = readfromDB(...);
auto exp = readFromDB(...);
auto p = pow(base, exp); // Regular run-time usage

C++11 and C++14 restrictions for constexpr functions

In C++11, constexpr functions may contain no more than single return statement. The workaround for that is ternary operators:

constexpr int pow(int base, unsigned exp) noexcept
{
  return (exp == 0 ? 1 : base * pow(base, exp-1));
}

In C++14, you can do more computation before calling return.

Getters

constexpr functions are limited to taking and returning literal types (i.e. non-trivial construction logic). This includes user defined types:

class Point {
public:
  constexpr Point(double xVal = 0, double yVal = 0) noexcept : x(xVal), y(yVal){}

  constexpr double xValue() const noexcept { return x; }
  constexpr double yValue() const noexcept { return y; }

  void setX(double newX) noexcept { x = newX; }
  void setY(double newY) noexcept { y = newY; }
private:
  double x, y;
};

So this way you can declare an object of the class as a constexpr (e.g. constexpr Point p1(1, 2);)

C++11 and C++14 restrictions for constexpr member functions

In C++11, two restrictions prevent Point’s member functions setX and setY from being declared constexpr.

• First, they modify the object they operate on, and in C++11, constexpr member functions are implicitly const.
• Second, they have void return types, and void isn’t a literal type in C++11.

Both these restrictions are lifted in C++14, so in C++14, even Point’s setters can be constexpr:

class Point
{
public:
  constexpr void setX(double newX) noexcept
  { x = newX; }

  constexpr void setY(double newY) noexcept
  { y = newY; }
};