What is Allowed in a constexpr Function?

Question

constexpr functions are not supposed to contain:

A definition of a variable of non-literal type

But in this answer a lambda is defined in one: https://stackoverflow.com/a/41616651/2642059

template <typename T>
constexpr auto make_div(const T quot, const T rem)
{
    return [&]() {
        decltype(std::div(quot, rem)) result;
        result.quot = quot;
        result.rem = rem;
        return result;
    }();
}

and in my comment I define a div_t in one: How can I Initialize a div_t Object?

template <typename T>
constexpr decltype(div(T{}, T{})) make_div(const T quot, const T rem)
{
    decltype(div(T{}, T{})) x{};
    x.quot = quot;
    x.rem = rem;
    return x;
}

Exactly what is meant by the prohibition of the "definition of a variable of non-literal type"?

Visual Studio 2015 won't allow my definition of a div_t but I find it nonsensical that it would be allowable to just wrap such illegitimate behavior in a lambda and execute it. I'd like to know which if either of the compilers are behaving correctly with respect to the div_t definition.

What version of the C++ language? The requirements were relaxed quite a bit from 2011. — rubenvb
– rubenvb, Commented Jan 12, 2017 at 16:45
A lambda expression is a expression, not a variable definition. But until C++17, a lambda expression cannot appear in a constant expression. Indeed, when the context requires a constant expression, MSVC will complain. — cpplearner
– cpplearner, Commented Jan 12, 2017 at 16:57
@cpplearner So my (the first) example only compiles as the constexpr function is evaluated as const function? In my opinion the second example should not compile by definition. — Simon Kraemer
– Simon Kraemer, Commented Jan 12, 2017 at 17:12
I don't think VS2015 supports C++14 extended constexpr. That explains your second function template since in C++11 constexpr functions are basically limited to a single return statement; it has nothing to do with "non-literal type" (and div_t is a literal type). The first one is ill-formed NDR pre-C++17. — T.C.
– T.C., Commented Jan 12, 2017 at 21:19

Community · Accepted Answer · 2020-06-20 09:12:55Z

29

It's virtually guaranteed that if there's a discrepancy gcc has the correct behavior, because Visual Studio 2015 doesn't support c++14's extension of constexpr: https://msdn.microsoft.com/en-us/library/hh567368.aspx#C-14-Core-Language-Features

C++11 `constexpr` functions

The function body can only contain:

null statements (plain semicolons)

static_assert declarations

typedef declarations and alias declarations that do not define classes or enumerations

using declarations

using directives

exactly one return statement

So c++11 cannot tolerate the definition of decltype(div(T{}, T{})) x{}. It would however be acceptable to roll the ternary suggested here in a constexpr function to achieve the same results:

template <typename T>
constexpr auto make_div(const T quot, const T rem)
{
    using foo = decltype(div(T{}, T{}));
                         
    return foo{1, 0}.quot != 0 ? foo{quot, rem} : foo{rem, quot};
}

Live Example

C++14 `constexpr` functions

The function body may contain anything but:

an asm declaration

a goto statement

a statement with a label other than case and default

a try-block

a definition of a variable of non-literal type

a definition of a variable of static or thread storage duration

a definition of a variable for which no initialization is performed

Where a "Literal Type" is defined here, specifically for objects though, they may be aggregate types with a trivial destructor. So div_t definitely qualifies. Thus c++14, and by extension gcc, can tolerate the definition of decltype(div(T{}, T{})) x{}.

C++17 `constexpr` functions

C++17 added support for closure types to the definition of "Literal Type", so I find it strange that both gcc and Visual Studio support the use of the lambda in the return statement. I guess that's either forward looking support or the compiler chose to inline the lambda. In either case I don't think that it qualifies as a c++14 constexpr function.

[Source]

edited Jun 20, 2020 at 9:12

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answered Jan 13, 2017 at 13:55

Jonathan Mee

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2 Comments

camino Over a year ago

@jnoathan_mee "The function body may contain anything but: ..." , we cannot call non-constexpr function in the body, right?

Jonathan Mee Over a year ago

@camino Correct, the result of this function must be calculable at compile-time, thus functions or parameters which cannot satisfy this requirement are not acceptable. That point isn't clearly documented though is it :(

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