Must constexpr expressions be captured by a lambda in C++?

Question

Here is a piece of code that won't compile in MSVC 2015 (ignore the uninitialized value access):

#include <array>
int main() {
    constexpr int x = 5;
    auto func = []() {
        std::array<int, x> arr;
        return arr[0];
    };
    func();
}

It complains that:

'x' cannot be implicitly captured because no default capture mode has been specified

But x is a constexpr! x is known at compile time to be 5. Why does MSVC kick up a fuss about this? (Is it yet another MSVC bug?) GCC will happily compile it.

It's just as likely to be a GCC extension that it accepts it. — Martin Bonner supports Monica
– Martin Bonner supports Monica, Commented Mar 5, 2017 at 15:52
GCC has too many extensions nowadays. Clang will not compile this as is proper. — Ivaylo Valchev
– Ivaylo Valchev, Commented Mar 5, 2017 at 15:54
Clang accepts this if you change std::array<int, x> to std::array<int, +x>. — cpplearner
– cpplearner, Commented Mar 5, 2017 at 16:01
This was fixed with VC 16.1: developercommunity.visualstudio.com/content/problem/367326/… — klaus triendl
– klaus triendl, Commented Jan 23, 2021 at 7:39

Barry · Accepted Answer · 2017-03-05 17:26:37Z

The code is well-formed. The rule from [expr.prim.lambda] is:

If a lambda-expression or an instantiation of the function call operator template of a generic lambda odr-uses (3.2) this or a variable with automatic storage duration from its reaching scope, that entity shall be captured by the lambda-expression.

Any variable that is odr-used must be captured. Is x odr-used in the lambda-expression? No, it is not. The rule from [basic.def.odr] is:

A variable x whose name appears as a potentially-evaluated expression ex is odr-used by ex unless applying the lvalue-to-rvalue conversion (4.1) to x yields a constant expression (5.20) that does not invoke any non-trivial functions and, if x is an object, ex is an element of the set of potential results of an expression e, where either the lvalue-to-rvalue conversion (4.1) is applied to e, or e is a discarded-value expression (Clause 5).

x is only used in a context where we apply the lvalue-to-rvalue conversion and end up with a constant expression, so it is not odr-used, so we do not need to capture it. The program is fine. This is the same idea as why this example from the standard is well-formed:

void f(int, const int (&)[2] = {}) { }   // #1
void f(const int&, const int (&)[1]) { } // #2

void test() {
    const int x = 17;
    auto g = [](auto a) {
        f(x); // OK: calls #1, does not capture x
    };
    // ...
}

Sam Varshavchik · Accepted Answer · 2017-03-05 15:53:24Z

-2

Even though x is a constexpr, it is no different from any other object, otherwise, and follows the same rules with regards to scoping. There are no exceptions to scoping rules for constexprs, and a lambda must be coded to explicitly capture it.

answered Mar 5, 2017 at 15:53

Sam Varshavchik

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8 Comments

Bernard Over a year ago

Is there then a way to capture it but allow me to use it as a template parameter as in my code?

Walter Over a year ago

That's another question.

Ivaylo Valchev Over a year ago

clang would allow you to just capture it [x]() {...} and then use it as a template parameter, yet GCC doesn't. I'm confused.

yuri kilochek Over a year ago

@Bernard just make it static, no need to capture.

Barry Over a year ago

No, this isn't the case, -1.

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