constexpr if
In today’s post, I want to introduce an exciting C++17 feature: constexpr if. constexpr if
enables it to compile source code conditionally and can be used for nice tricks at compile time.
Introducing constexpr if
is straightforward.
template <typename T> auto getValue(T t) { if constexpr (std::is_pointer_v<T>) // (1) return *t; // deduces return type to int for T = int* else // (2) return t; // deduces return type to int for T = int }
The code snippet shows one exciting fact about constexpr if
: Although it is called constexpr if
, it is used as if constexpr
: if constexpr (std::is_pointer_v<T>
).
If T
is a pointer, the if branch in line (1) will be compiled. If not, the else branch in line (2). Two points are important to mention. The function getValue
has two different return types, and both branches of the if
statement have to be valid.
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The expression in constexpr if has to be a compile-time predicate. A compile time predicate is a function that returns a boolean and runs at compile time. I used in the code snippet a function from the type-traits library. Alternatively, in C++20, you can use a concept. Here is the equivalent example using the concept std::integral:
template <typename T> auto get_value(T t) { if constexpr (std::integral<T>) // (1) return *t; // deduces return type to int for T = int* else // (2) return t; // deduces return type to int for T = int }
I see, the two code snippets are not so impressive. Let me continue with template metaprogramming.
Thanks to constexpr if
, template metaprogramming is often easier to write and read.
Template Metaprogramming with constexpr if
Metaprogramming is programming on programs. C++ applies metaprogramming at compile time. It started in C++98 with template metaprogramming, was formalized in C++11 with the type-traits library, and since C++11 has steadily improved.
Here is the “Hello World” of template metaprogramming: calculating the factorial of a number:
// factorial.cpp #include <iostream> template <int N> // (2) struct Factorial{ static int const value = N * Factorial<N-1>::value; }; template <> // (3) struct Factorial<1>{ static int const value = 1; }; int main(){ std::cout << '\n'; std::cout << "Factorial<5>::value: " << Factorial<5>::value << '\n'; // (1) std::cout << "Factorial<10>::value: " << Factorial<10>::value << '\n'; // (4) std::cout << '\n'; }
The call factorial<5>::value
(line 1) causes the instantiation of the primary or general template (line 2). During this instantiation, Factorial<4>::value
will be instantiated. This recursion will end if the fully specialized class template Factorial<1>
kicks in (line 3).
If you want to know more about template metaprogramming, read my previous posts:
- Template Metaprogramming – How it All Started
- Template Metaprogramming – How it Works
- Template Metaprogramming – Hybrid Programming
Let me rewrite the program using constexpr i
f:
// factorialConstexprIf.cpp template <int N> // (1) struct Factorial{ static int const value = N * Factorial<N-1>::value; }; template <> // (2) struct Factorial<1>{ static int const value = 1; }; template <int N> // (3) constexpr int factorial() { if constexpr (N >= 2) return N * factorial<N-1>(); else return N; } int main(){ static_assert(Factorial<5>::value == factorial<5>()); // (4) static_assert(Factorial<10>::value == factorial<10>()); // (4) }
The primary template of Factorial
(line 1) becomes the if condition in the constexpr
function factorial
(line 3), and the full specialization of Factorial
for 1 (line 2) becomes the else case in the constexpr
function factorial (line 3). Of course, the class template Factorial
and the constexpr
function factorial
return the same result and are executed at compile time (line 4). To make it short, I prefer the constexpr function using constexpr if
because it reads almost such as a usual function.
Let’s do it once more. Here is the infamous Fibonacci function-based template metaprogramming (Fibonacci
) and constexpr if (fibonacci).
// fibonacciConstexprIf.cpp template<int N> constexpr int fibonacci() { if constexpr (N>=2) return fibonacci<N-1>() + fibonacci<N-2>(); else return N; } template <int N> // (1) struct Fibonacci{ static int const value = Fibonacci<N-1>::value + Fibonacci<N-2>::value; }; template <> // (2) struct Fibonacci<1>{ static int const value = 1; }; template <> // (3) struct Fibonacci<0>{ static int const value = 0; }; int main() { static_assert(fibonacci<7>() == 13); static_assert(fibonacci<7>() == Fibonacci<7>::value); }
The constexpr
function fibonacci
is straightforward to read. The entire functionality is in one function body. In contrast, the template metaprogram Fibonacci
requires three classes. The primary template (line 1) and the two full specializations for the values 1 and 0 (lines 2 and 3).
More Information about my Mentoring Program, “Fundamentals for C++ Professionals”
I created the platform for my new mentoring on https://www.modernescpp.org/. You can skip through each of the 28 lessons. I also presented the 6th lesson about move semantics and perfect forwarding in the post ‘More Information about my Mentoring Program “Fundamentals for C++ Professionals”‘. Here are the next steps before I start the mentoring program.
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What’s Next?
Templates are a powerful tool and, therefore, provide new design choices. In my next post, I will write about static and dynamic polymorphism.
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