constexpr array size calculation

I was writing an article about adding constexpr to some legacy code generation function when I found myself explaining one feature so detailed I decided to extract it into the separate article.

Basic version

Imagine we have some class, which is responsible for generating an array of data. Simplified code looks like this:

#include <cstdint>
#include <vector>

class Foo {
private:
    std::size_t array_size_seed = 0;

    auto GetArraySize() {
        return array_size_seed * 2;
    }

public:
    Foo(std::size_t val) : array_size_seed(val){}

    auto GetVector() {
        std::vector<int> dst(GetArraySize());
        for(std::size_t i = 0; i < dst.size(); ++i)
            dst[i] = i * 5;

        return dst;        
    }
} foo(12);

Both the size calculation and array filling are simplified since they’re not that important.

First attempt to constexpr

In the current (C++17) standard there is no such thing as the constexpr std::vector<T>, but there is a std::array<T, size>. Let’s add constexpr to everything:

#include <array>
#include <cstdint>
#include <vector>

class Foo {
private:
    std::size_t array_size_seed = 0;

    constexpr auto GetArraySize() const {
        return array_size_seed * 2;
    }

public:
    constexpr Foo(std::size_t val) : array_size_seed(val){}

    constexpr auto GetArray() {
        std::array<int, GetArraySize()> dst{};
        for(std::size_t i = 0; i < dst.size(); ++i)
            dst[i] = i * 5;

        return dst;        
    }
};

There is, however, a drawback: this won’t compile. According to the standard, constexpr functions may be called both at the compile- and runtime. Therefore, arguments of the functions (this pointer in that case) are not constant, so std::array size cannot be resolved at the compile-time.

As I’ve mentioned earlier, there is a proposal for the constexpr! specifier, which forces calculations to be compile-time only. If we’ve had it, we’d be done by now.

To make this work we need to force the calculations to be compile-time. Hopefully, there’s a way to make this: templates.

Force constexpr with templates

There’s a so-called “non-type template parameters” feature, which we’ll be using here. The thing is that with templates you’re not limited only to template <typename T>, you can also pass integral or enumeration values, references etc.

#include <array>
#include <cstdint>
#include <vector>

class Foo {
private:
    std::size_t array_size_seed = 0;

    constexpr auto GetArraySize() const {
        return array_size_seed * 2;
    }

public:
    constexpr Foo(std::size_t val) : array_size_seed(val){}

    template <const Foo& foo>        // <-----
    constexpr static auto GetArray() {
        std::array<int, foo.GetArraySize()> dst{};
        for(std::size_t i = 0; i < dst.size(); ++i)
            dst[i] = i * 5;

        return dst;        
    }
};

Notice, that we’ve made the method static and get the object via a template. One of the good things is that the standard allows static class methods to access the private members of the same class (notice the foo.GetArraySize(), which is declared private).

With that said and done we can create an object during the compile time and generate a variable-size array with its methods.

constexpr Foo foo(12);
constexpr auto arr = Foo::GetArray<foo>();

Just take a look at the disassembly here!


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