Using constexpr to initialize objects in std.array

Question

I have an array called foos with instances of Foo. They are stored in an std.array and I would like to initialize them during compile time. Is that possible by using C++17 and constexpr?

struct Foo
{
    constexpr void setA(int a);
    int _a{0};
};

static std::array<Foo, 100> foos;

static constexpr void initialize()
{
    int i = 0;
    for (auto& e : foos)
    {
        e.setA(i++);
    }
}

It seems the initialization is still done during runtime. Am I missing something?

https://gcc.godbolt.org/z/r4WUbE

I am aware that -O3 will generate better output, but my original example is slightly better and the compiler does not optimize it under this optimization.

Davis Herring · Accepted Answer · 2019-10-18 01:20:16Z

1

constexpr doesn’t mean “schedule this function to run before (normal) execution begins”. In particular, a constant expression cannot modify an object (here, any of the Foo::_a objects) that it did not create. You can, however, make a constexpr function that returns an array and use it as an initializer:

using Foos=std::array<Foo,100>;
constexpr Foos iota_foos() {
  Foos ret;
  int i=0;
  for(auto &f : ret) f.setA(i++);
  return ret;
}

static Foos foos=iota_foos();

answered Oct 18, 2019 at 1:20

Davis Herring

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6 Comments

Jonas Greitemann Over a year ago

Shouldn’t it be static constexpr Foos foos=iota_foos(); to enforce that iota_foos is evaluated in a constexpr context?

Davis Herring Over a year ago

@JonasGreitemann: There was no statement that foos should be const. It’s constant-initialized anyway (but cannot itself be used in constant expressions); C++20 provides constinit to verify that fact without changing the type.

Jonas Greitemann Over a year ago

Agreed. Upon testing, I checked that gcc and clang constant-initialized it even at -O0, but as it stands, there's no way to guarantee this pre-C++20. However, if foos were const, it really should be constexpr. As pointed out by Jason Turner at CppCon, static const is a code smell.

Jarod42 Over a year ago

@JonasGreitemann: We can guarantee that iota_foos is called at compile time and just have no guaranty about copy constructor: static constexpr Foos const_foos=iota_foos(); static Foos foos= const_foos();.

Davis Herring Over a year ago

@JonasGreitemann: Constant initialization isn’t an optimization; it’s just part of the language. So it is guaranteed, but absent constexpr/constinit there’s no way to “guarantee the guarantee”.

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