What is the use of a constexpr on a non-const member function?

Question

The accepted answer in literal class compile error with constexpr constructor and function (differ vc, g++) shows that in C++14 there is a difference in the way constexpr int A::a() and constexpr A::a() const can be used. i.e. constexpr on a member function does not imply that the function does not change the object it acts on.

The given example is:

struct A {
    constexpr A() {}
    constexpr int a() {return 12; }
    constexpr int b() const {return 12; }
};

int main()
{
    constexpr A a;
    // DOES NOT COMPILE as a() is not const
    // constexpr int j = a.a();
    const int k = a.b(); // Fine since b() is const
}

To me the constexpr on a() seems useless. Is there a concrete use for constexpr on a non-const member function?

bolov · Accepted Answer · 2018-05-03 08:30:35Z

17

Question: how do you create a constexpr array of size 1024 with all elements set to 0 except element element 42 which needs to be 11?

#include <array>

constexpr auto make_my_array() -> std::array<int, 1024>
{
    std::array<int, 1024> a{};
    a[42] = 11; // std::array::operator[] is constexpr non-const method since C++17

    return a;
}

auto test()
{
    constexpr std::array<int, 1024> a = make_my_array();
}

Or a better yet suggestion from @michael-anderson a make_iota_array:

template <std::size_t N>
constexpr auto make_iota_array() -> std::array<int, N>
{
    std::array<int, N> a{};

    for (std::size_t i = 0; i < N; ++i)
        a[i] = i;

    return a;
}

edited May 3, 2018 at 8:30

answered May 3, 2018 at 7:34

bolov

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4 Comments

StoryTeller - Unslander Monica Over a year ago

Figures I'd cook up something crazy with templates when there's a perfectly clearer, and simpler example. +1.

bolov Over a year ago

@StoryTeller and I found a practical example

Dean Seo Over a year ago

@StoryTeller I think you added a good value to this question especially regarding constant expression. Please don't delete it this time :)

Michael Anderson Over a year ago

I like this example. I guess you could even do a make_iota_array<N>() -> array<int,N> that initialized a[k]=k in a for loop - the kind of thing I've coded as awful template hackery before.

StoryTeller - Unslander Monica · Accepted Answer · 2018-05-03 07:32:34Z

16

constexpr means "can be used where a constant expression is required". The "implied const" for declared objects doesn't mean we can't have non-const objects in other contexts. For instance, a somewhat contrived example, created from your own:

template<int>
struct foo {
};

struct A {
    int i = 0;
    constexpr A() {}
    constexpr int a() { return i; }
    constexpr int b() const {return 12; }
    constexpr A&  c() { ++i; return *this; }
};

int main()
{
    foo<A{}.c().a()> f1;
}

Obviously the template argument must be a constant expression. Now, A{} is a prvalue of a literal type with a constexpr c'tor, and it's a non-const object. The member function is allowed to modify this "constant" because the entire computation can collapse to a constant expression at compile time. That's the rationale for the rules, on one foot.

edited May 3, 2018 at 7:32

answered May 3, 2018 at 7:26

StoryTeller - Unslander Monica

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1 Comment

Michael Anderson Over a year ago

Wow. I never even realized something like this would be possible. Guess I've gotta drag my mind out of the pre-2000 C++ mindset...

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